aekaakee: June 2020

Probability

In the refectory of a college both tea and coffee are sold.

The number of cups of coffee and tea sold per five minute interval may be considered to be independent Poisson distributions with means $2.7$ and $1.5$ respectively.

Calculate the probabilities that, in a given five minute interval,

exactly one cup of coffee and one cup of tea are sold,
exactly two drinks are sold,
more than five drinks are sold.

Number of cups of coffee sold per five minute interval is a Poisson distribution with mean $= 2.7$

$\therefore \;$ Parameter of Poisson distribution (for coffee) $= \lambda_c = 2.7$

Number of cups of tea sold per five minute interval is a Poisson distribution with mean $= 1.5$

$\therefore \;$ Parameter of Poisson distribution (for tea) $= \lambda_t = 1.5$

$P \left(\text{exactly one cup of coffee and one cup of tea are sold}\right)$
$\hspace{3cm}$ $= P \left(\text{1 cup of coffee sold}\right) \times P \left(\text{1 cup of tea sold}\right)$

$\begin{aligned} P \left(1 \text{ cup of coffee sold}\right) & = P \left(X = 1\right) \\\\ & = \dfrac{e^{- \lambda_c} \times \left(\lambda_c\right)^{1}}{1!} \\\\ & = e^{- 2.7} \times 2.7 \\\\ & = 0.0672 \times 2.7 = 0.1814 \end{aligned}$

$\begin{aligned} P \left(1 \text{ cup of tea sold}\right) & = P \left(X = 1\right) \\\\ & = \dfrac{e^{- \lambda_t} \times \left(\lambda_t\right)^{1}}{1!} \\\\ & = e^{- 1.5} \times 1.5 \\\\ & = 0.2231 \times 1.5 = 0.3347 \end{aligned}$

$\therefore \;$ $P \left(\text{exactly one cup of coffee and one cup of tea are sold}\right)$
$\hspace{3cm}$ $= 0.1814 \times 0.3347 = 0.0607$
Parameter of Poisson distribution for both number of cups of coffee and tea sold per five minute interval $= \lambda = \lambda_c + \lambda_t = 2.7 + 1.5 = 4.2$

$P \left(\text{exactly 2 drinks are sold}\right) = P \left(X = 2\right)$

$\begin{aligned} P \left(X = 2\right) & = \dfrac{e^{- \lambda} \times \lambda^2}{2!} \\\\ & = \dfrac{e^{- 4.2} \times \left(4.2\right)^2}{2} \\\\ & = \dfrac{0.0150 \times 17.64}{2} = 0.1323 \end{aligned}$
$P \left(\text{more than 5 drinks are sold}\right) = P \left(X > 5\right)$

$P \left(X > 5\right) = 1 - P \left(X \leq 5\right)$

$\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 4.2} \times \left[1 + \dfrac{4.2}{1} + \dfrac{4.2^2}{2} + \dfrac{4.2^3}{6} + \dfrac{4.2^4}{24} + \dfrac{4.2^5}{120} \right] \\\\ & = 0.0150 \times 50.2243 = 0.7534 \end{aligned}$

$\therefore \;$ $P \left(X > 5\right) = 1 - 0.7534 = 0.2466$

Probability

The lost property office at a mainline station records items left on trains handed in in two categories: Category A is clothing and Category B is paperwork.

Over a period of time the mean number of items in category A handed in per day is $2$ and the mean number in category B per day is $3$.

These distributions may be assumed to be independent of each other and each may be modeled by the Poisson distribution.

Find the probability that

in one day there are no items handed in;
in one day there is exactly one item handed in;
in one day there are at most 2 items handed in.

Mean number of items in category A handed in per day $= 2$

Mean number of items in category B handed in per day $= 3$

The number of items handed in per day, in category A and category B, are both modeled by the Poisson distribution.

$\therefore \;$ Parameter of Poisson distribution for both categories together $= \lambda = 2 + 3 = 5$

$P\left(\text{No items are handed in}\right)= P \left(X = 0\right)$

$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 5} = 0.0067 \end{aligned}$
$P \left(\text{exactly 1 item handed in}\right) = P \left(X = 1\right)$

$\begin{aligned} P \left(X = 1\right) & = \dfrac{e^{- \lambda} \times \lambda^1}{1!} \\\\ & = e^{- \lambda} \times \lambda \\\\ & = e^{- 5} \times 5 \\\\ & = 0.0067 \times 5 = 0.0335 \end{aligned}$
$P \left(\text{at most 2 items handed in}\right) = P \left(X \leq 2\right)$

$\begin{aligned} P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 5} \times \left[1 + \dfrac{5}{1} + \dfrac{5^2}{2} \right] \\\\ & = 0.0067 \times 18.5 = 0.1240 \end{aligned}$

Probability

The coach arriving into Station A each morning at 0800 has a capacity of $54$ passengers.

On this route standing passengers are not allowed.

Over a period of time the number of passengers alighting at Station A on $20$ days was recorded as follows:

$50, \; 50, \; 50, \; 50, \; 50, \; 51, \; 51, \; 51, \; 51, \; 51$

$51, \; 51, \; 52, \; 52, \; 52, \; 52, \; 52, \; 52, \; 53, \; 54$

$50$ of the passengers were regular commuters. The random variable, $X$, is the number of passengers arriving at Station A on this coach other than the $50$ regulars.

Show that the Poisson distribution is a reasonable fit for these data.

Use this distribution with your mean for the data above to estimate the probability that on any given day someone has been left behind because the coach was full.

The random variable, $X$, is the number of passengers arriving at Station A on this coach other than the $50$ regulars.

$\therefore \;$ Number of passengers (other than regulars) arriving $= X - 50$

i.e. $\;$ $0, \; 0, \; 0, \; 0, \; 0, \; 1, \; 1, \; 1, \; 1, \; 1, \;$ $1, \; 1, \; 2, \; 2, \; 2, \; 2, \; 2, \; 2, \; 3, \; 4$

Frequency table to illustrate the number of passengers arriving at station A (other than the $50$ regulars):

Number of passengers $\left(x_i\right)$	Frequency $\left(f_i\right)$	$x_i \cdot f_i$	$x_i^2$	$x_i^2 \cdot f_i$
$0$	$5$	$0$	$0$	$0$
$1$	$7$	$7$	$1$	$7$
$2$	$6$	$12$	$4$	$24$
$3$	$1$	$3$	$9$	$9$
$4$	$1$	$4$	$16$	$16$

For this set of data,

$n = \sum f_i = 20$; $\;$ $\sum x_i \cdot f_i = 26$; $\;$ $\sum x_i^2 \cdot f_i = 56$

$\therefore \;$ Mean number of passengers arriving at station A (other than the $50$ regulars)

$= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{26}{20} = 1.3$

Variance of number of passengers arriving at station A (other than the $50$ regulars)

$= s^2 = \dfrac{\sum x_i^2 \cdot f_i - n \cdot \overline{x}^2}{n - 1}$

i.e. $\;$ $s^2 = \dfrac{56 - \left(20 \times 1.3^2\right)}{20 - 1} = 1.168$

$\because \;$ The mean and the variance of the modified data are close to one another, Poisson distribution is a reasonable fit to the modified data (passengers other than the $50$ regulars).

$\therefore \;$ Mean of data $=$ Parameter of Poisson distribution $= \lambda = 1.3$

P(someone is left behind on a given day) $= P \left(X > 4\right)$

$P \left(X > 4\right) = 1 - P \left(X \leq 4\right)$

$\begin{aligned} P \left(X \leq 4\right) & = \sum \limits_{x = 0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 1.3} \times \left[1 + \dfrac{1.3}{1} + \dfrac{1.3^2}{2} + \dfrac{1.3^3}{6} + \dfrac{1.3^4}{24} \right] \\\\ & = 0.2725 \times 3.6302 = 0.9892 \end{aligned}$

$\therefore \;$ $P \left(X > 4\right) = 1 - 0.9892 = 0.0108$

Probability

In a college, the number of accidents to students requiring hospitalization in one year of $30$ weeks is recorded as follows:

Number of accidents requiring hospitalization each week	Frequency
$0$	$25$
$1$	$4$
$2$	$1$
$3$ or more	$0$

This data is used to assesses the risk of such accidents.

Assuming that the Poisson distribution is a suitable model calculate the probability that

In any one week there will be $3$ accidents requiring hospitalization,
In a term of $8$ weeks there will be no accidents.

Frequency table to illustrate the number of the number of accidents to students requiring hospitalization in each week:

Number of accidents requiring hospitalization each week $\left(x_i\right)$	Frequency $\left(f_i\right)$	$x_i \cdot f_i$
$0$	$25$	$0$
$1$	$4$	$4$
$2$	$1$	$2$
$\geq 3$	$0$	$0$

For this set of data,

$n = \sum f_i = 30$; $\;$ $\sum x_i \cdot f_i = 6$

$\therefore \;$ Mean number of accidents requiring hospitalization each week

$= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{6}{30} = 0.2$

Mean $=$ Parameter of Poisson distribution $= \lambda = 0.2$

P(in any one week there will be $3$ accidents requiring hospitalization) $= P \left(X = 3\right)$

$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 0.2} \times \left(0.2\right)^{3}}{6} \\\\ & = \dfrac{0.8187 \times 0.008}{6} = 0.0011 \end{aligned}$
P(in $1$ week there will be no accidents) $= P \left(X = 0\right)$

$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!} \\\\ & = e^{- 0.2} = 0.8187 \end{aligned}$

$\therefore \;$ P(in a term of $8$ weeks there will be no accidents)

$= \left[P \left(X = 0\right)\right]^{8} = \left(0.8187\right)^8 = 0.2018$

Probability

The numbers of customers entering a shop in forty consecutive periods of one minute are given below:

$3, \; 0, \; 0, \; 1, \; 0, \; 2, \; 1, \; 0, \; 1, \; 1$
$0, \; 3, \; 4, \; 1, \; 2, \; 0, \; 2, \; 0, \; 3, \; 1 $
$1, \; 0, \; 1, \; 2, \; 0, \; 2, \; 1, \; 0, \; 1, \; 2 $
$3, \; 1, \; 0, \; 0, \; 2, \; 1, \; 0, \; 3, \; 1, \; 2 $

Calculate values of the mean and variance of the number of customers entering the shop in a one minute period.
Fit a Poisson distribution to the data and comment on the degree of agreement between the calculated and observed values.

Frequency table to illustrate the number of customers entering a shop in forty consecutive periods of one minute:

Table 1
Number of customers $\left(x_i\right)$	Number of samples $\left(f_i\right)$	$x_i \cdot f_i$	$x_i^2$	$x_i^2 \cdot f_i$
$0$	$13$	$0$	$0$	$0$
$1$	$13$	$13$	$1$	$13$
$2$	$8$	$16$	$4$	$32$
$3$	$5$	$15$	$9$	$45$
$\geq 4$	$1$	$4$	$\geq 16$	$16$

For this set of data,

$n = \sum f_i = 40$; $\;$ $\sum x_i \cdot f_i = 48$; $\;$ $\sum x_i^2 \cdot f_i = 106$

$\therefore \;$ Mean number of customers entering the shop in a one minute period

$= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{48}{40} = 1.2$

Variance of number of customers entering the shop in a one minute period

$= s^2 = \dfrac{\sum x_i^2 \cdot f_i - n \cdot \overline{x}^2}{n - 1}$

i.e. $\;$ $s^2 = \dfrac{106 - \left(40 \times 1.2^2\right)}{40 - 1} = 1.2410$

From the cumulative probability tables with $\lambda = 1.2$ we have

Table 2
$\left(x_i\right)$	Probability
$0$	$P \left(X = 0\right) = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}$
$0$	$ = e^{-1.2} = 0.3012$
$\leq 1$	$P \left(X \leq 1\right) = \sum \limits_{0}^{1} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
$\leq 1$	$= 0.3012 + \dfrac{e^{-1.2} \times 1.2^1}{1!} = 0.3012 + 0.3614 = 0.6626$
$\leq 2$	$P \left(X \leq 2\right) = \sum \limits_{0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
$\leq 2$	$= 0.6626 + \dfrac{e^{-1.2} \times 1.2^2}{2!} = 0.6626 + 0.2169 = 0.8795$
$\leq 3$	$P \left(X \leq 3\right) = \sum \limits_{0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
$\leq 3$	$= 0.8795 + \dfrac{e^{-1.2} \times 1.2^3}{3!} = 0.8795 + 0.0867 = 0.9662$
$\leq 4$	$P \left(X \leq 4\right) = \sum \limits_{0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!}$
$\leq 4$	$= 0.9662 + \dfrac{e^{-1.2} \times 1.2^4}{4!} = 0.9662 + 0.0260 = 0.9922$

$\therefore \;$ The probabilities are

Table 3
$\left(x_i\right)$	Probability	Expected frequency $=$ Probability $\times \; n$
$0$	$0.3012$	$12.048$
$1$	$0.6626 - 0.3012 = 0.3614$	$14.456$
$2$	$0.8795 - 0.6626 = 0.2169$	$8.676$
$3$	$0.9662 - 0.8795 = 0.0867$	$3.468$
$4$	$0.9922 - 0.9662 = 0.0260$	$1.040$

Comparing the given frequency of number of customers (Table 1) with the frequency of customers calculated using Poisson distribution (Table 3) we have,

Table 4
Number of customers	Actual frequency	Theoretical frequency (1 d.p)
$0$	$13$	$12.0$
$1$	$13$	$14.5$
$2$	$8$	$8.7$
$3$	$5$	$3.5$
$4$	$1$	$1.0$

As seen from Table 4, the fit is very good.

This is because the mean and the variance are very close to one another.

Probability

The number of bacteria in $50$ $100cc$ samples of water are given in the following table.

Number of bacteria per sample	Number of samples
$0$	$23$
$1$	$16$
$2$	$9$
$3$	$2$
$4 \;$ or more	$0$

Find the mean number and the variance of bacteria in a $100cc$ sample.
State whether the conditions for using the Poisson distribution as a model apply.
Using the Poisson distribution with the mean found in part (1), estimate the probability that another $100cc$ sample will contain
1. no bacteria,
2. more than $4$ bacteria.

For the given set of data,

Number of bacteria per sample $\left(x_i\right)$	Number of samples $\left(f_i\right)$	$x_i \cdot f_i$	$x_i^2$	$x_i^2 \cdot f_i$
$0$	$23$	$0$	$0$	$0$
$1$	$16$	$16$	$1$	$16$
$2$	$9$	$18$	$4$	$36$
$3$	$2$	$6$	$9$	$18$
$\geq 4$	$0$	$0$	$\geq 16$	$0$

$n = \sum f_i = 50$; $\;$ $\sum x_i \cdot f_i = 40$; $\;$ $\sum x_i^2 \cdot f_i = 70$

$\therefore \;$ Mean number of bacteria $= \overline{x} = \dfrac{\sum x_i \cdot f_i}{\sum f_i} = \dfrac{40}{50} = 0.8000$

Variance of bacteria $= s^2 = \dfrac{\sum x_i^2 \cdot f_i - n \cdot \overline{x}^2}{n - 1}$

i.e. $\;$ $s^2 = \dfrac{70 - \left(50 \times 0.8^2\right)}{50 - 1} = 0.7755$

The number of bacteria present in a given sample are random and are independent of one another.

The mean is close to the variance.

Hence, Poisson distribution as a model may be applied to the given problem.
$\because \;$ Mean $= \overline{x} = 0.80$,

parameter of Poisson's distribution $= \lambda = 0.8$
1. $P \left(\text{sample contains no bacteria}\right) = P \left(X = 0\right)$
  
  $\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}\\\\ & = e^{- \lambda} \\\\ & = e^{- 0.80} = 0.4493 \end{aligned}$
2. $P \left(\text{sample contains more than 4 bacteria}\right) = P \left(X > 4\right)$
  
  Now, $P \left(X > 4\right) = 1 - P \left(X \leq 4\right)$
  
  $\begin{aligned} P \left(X \leq 4\right) & = \sum \limits_{x = 0}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 0.8} \times \left[1 + \dfrac{0.8}{1} + \dfrac{0.8^2}{2} + \dfrac{0.8^3}{6} + \dfrac{0.8^4}{24} \right] \\\\ & = 0.4493 \times 2.2224 = 0.9985 \end{aligned}$
  
  $\therefore \;$ $P \left(X > 4\right) = 1 - 0.9985 = 0.0015$

Probability

A hire company has two lawnmowers which it hires out by the day. The number of demands per day may be modeled by a Poisson distribution with mean $1.5$. In a period of $100$ working days, how many times do you expect

neither lawnmower to be used;
some requests for a lawnmower to have to be refused?

Given: Mean $= 1.5$

$\therefore \;$ Parameter of Poisson's distribution $= \lambda = 1.5$

Number of lawnmowers $= 2$

$P \left(\text{neither lawnmower is used}\right) = P \left(X = 0\right)$

$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}\\\\ & = e^{- \lambda} \\\\ & = e^{- 1.5} = 0.2231 \end{aligned}$

$\therefore \;$ In $100$ working days, neither lawnmower will be used $= 0.2231 \times 100 = 22.31 \approx 22$ times
$P \left(\text{some requests for a lawnmower to have to be refused}\right) = P \left(X > 2\right)$

Now, $P \left(X > 2\right) = 1 - P \left(X \leq 2\right)$

$\begin{aligned} P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 1.5} \times \left[1 + \dfrac{1.5}{1} + \dfrac{1.5^2}{2} \right] \\\\ & = 0.2231 \times 3.625 = 0.8087 \end{aligned}$

$\therefore \;$ $P \left(X > 2\right) = 1 - 0.8087 = 0.1913$

$\therefore \;$ In $100$ working days, number of times a request for a lawnmower will be refused $= 0.1913 \times 100 = 19.13 \approx 19$ times

Probability

A book of $500$ pages has $500$ misprints. Using the Poisson distribution, estimate the probabilities that a given page contains

exactly $3$ misprints;
more than $3$ misprints.

Mean number of misprints per page $= \dfrac{500}{500} = 1$

$\therefore \;$ Parameter of Poisson's distribution $= \lambda = 1$

$P \left(\text{exactly 3 misprints}\right) = P \left(X = 3\right)$

$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 1} \times 1^3}{6} \\\\ & = \dfrac{0.3679}{6} = 0.06132 \end{aligned}$
$P \left(\text{more than 3 misprints}\right) = P \left(X > 3\right) = 1 - P \left(X \leq 3\right)$

$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 1} \times \left[1 + \dfrac{1}{1} + \dfrac{1^2}{2} + \dfrac{1^3}{6}\right] \\\\ & = 0.3679 \times 2.6667 = 0.9811 \end{aligned}$

$\therefore \;$ $P \left(\text{more than 3 misprints}\right) = 1 - 0.9811 = 0.0189$

Probability

The demand for cars from a car hire firm may be modelled by a Poisson distribution with mean $4$ per day.

Find the probability that in a randomly chosen day the demand is for $3$ cars.
The firm has $5$ cars available for hire. Find the probability that demand exceeds the number of cars available.

Mean of Poisson distribution $= \lambda = 4$

$P \left(\text{Demand is for 3 cars}\right) = P \left(X = 3\right)$

$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 4} \times 4^3}{6} \\\\ & = \dfrac{0.01832 \times 64}{6} = 0.1954 \end{aligned}$
Number of cars available for hire $= 5$

$P \left(\text{Demand exceeds number of available cars}\right) = P \left(X > 5\right)$

$\begin{aligned} P \left(X > 5\right) & = 1 - P \left(\text{Demand is for 5 cars or less}\right) \\\\ & = 1 - P \left(X \leq 5\right) \end{aligned}$

$\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6} + \dfrac{4^4}{24} + \dfrac{4^5}{120}\right] \\\\ & = 0.01832 \times 42.8667 = 0.7853 \end{aligned}$

$\therefore \;$ $P \left(\text{Demand exceeds number of available cars}\right) = 1 - 0.7853 = 0.2147$

Probability

Alpha particles are emitted by a radio active source at an average rate of $5$ in a $20$ minutes interval. Using Poisson distribution find the probability that there will be

$2$ emissions;
at least $2$ emissions

in a particular $20$ minutes interval. $\;\;\;$ Given: $e^{- 5} = 0.0067$

Average rate of emission in $20$ minute interval $= 5$ emissions

i.e. $\;$ $\lambda = 5$

P(there will be $2$ emissions) $= P \left(X = 2\right)$

$\begin{aligned} P \left(X = 2\right) & = \dfrac{e^{- \lambda} \times \lambda^{2}}{2!} \\\\ & = \dfrac{e^{- 5} \times 5^2}{2} \\\\ & = \dfrac{0.0067 \times 25}{2} = 0.08375 \end{aligned}$
P(at least $2$ emissions) $= P \left(\text{2 or more emissions}\right)$

$\begin{aligned} \text{Now, } P \left(\text{2 or more emissions}\right) & = 1 - P \left(\text{less than 2 emissions}\right) \\\\ & = 1 - P \left(X < 2\right) \end{aligned}$

$\begin{aligned} P \left(X < 2\right) & = \sum \limits_{x = 0}^{1} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} \right] \\\\ & = e^{- 5} \times \left[1 + \dfrac{5}{1} \right] \\\\ & = 0.0067 \times 6 = 0.0402 \end{aligned}$

$\therefore \;$ $P \left(\text{at least 2 emissions}\right) = 1 - 0.0402 = 0.9598$

Probability

If the probability of a defective fuse from a manufacturing unit is $2\%$ in a box of $200$ fuses, find the probability that

exactly $4$ fuses are defective;
more than $3$ fuses are defective. $\;\;\;$ Given: $e^{- 4} = 0.0183$

Probability that a fuse is defective $= p = 2 \% = 0.02$

Total number of fuses $= n = 200$

$\therefore \;$ Mean of Poisson distribution $= \lambda = n \times p = 200 \times 0.02 = 4$

$P \left(\text{exactly 4 fuses are defective }\right) = P \left(X = 4\right)$

$\begin{aligned} P \left(X = 4\right) & = \dfrac{e^{- \lambda} \times \lambda^{4}}{4!} \\\\ & = \dfrac{e^{- 4} \times 4^4}{24} \\\\ & = \dfrac{0.0183 \times 256}{24} = 0.1952 \end{aligned}$
$P \left(\text{more than 3 fuses are defective}\right) = P \left(X > 3\right) = 1 - P \left(X \leq 3\right)$ where

$P \left(X \leq 3\right) = P \left(\text{at most 3 fuses are defective}\right)$

$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6}\right] \\\\ & = 0.0183 \times 23.67 = 0.4331 \end{aligned}$

$\therefore \;$ $P \left(\text{more than 3 fuses are defective}\right) = 1 - 0.4331 = 0.5669$

Probability

Let $X$ have a Poisson distribution with mean $4$. Find

$P \left(X \leq 3\right)$
$P \left(2 \leq X < 5\right)$ $\;\;\;$ Given: $e^{- 4} = 0.0183$

Mean of Poisson distribution $= \lambda = 4$

To find $P \left(X \leq 3\right)$:

$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6}\right] \\\\ & = 0.0183 \times 23.67 = 0.4331 \end{aligned}$
To find $P \left(2 \leq X < 5\right)$:

$\begin{aligned} P \left(2 \leq X < 5\right) & = \sum \limits_{x = 2}^{4} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} \right] \\\\ & = e^{- 4} \times \left[\dfrac{4^2}{2} + \dfrac{4^3}{6} + \dfrac{4^4}{24} \right] \\\\ & = 0.0183 \times 29.33 = 0.5368 \end{aligned}$

Probability

The overall percentage of passes in a certain examination is $80$. If $6$ candidates appear in the examination what is the probability that at least $5$ pass the examination.

Probability of passing an exam $= p = 80\% = \dfrac{80}{100} = \dfrac{4}{5}$

$\therefore \;$ Probability of not passing the exam $= q = 1 - p = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

$\begin{aligned} P \left(\text{at least 5 pass }\right) & = P \left(5 \text{ pass}\right) \; OR \; P \left(\text{all 6 pass}\right) \\\\ & = {^{6}}{C}_{5} \times p^5 \times q^{6 - 5} + \left(p\right)^6 \\\\ & = 6 \times \left(\dfrac{4}{5}\right)^5 \times \dfrac{1}{5} + \left(\dfrac{4}{5}\right)^6 \\\\ & = \left(\dfrac{4}{5}\right)^5 \left(\dfrac{6}{5} + \dfrac{4}{5}\right) \\\\ & = \dfrac{2048}{3125} \end{aligned}$

Probability

If on an average $1$ ship out of $10$ does not arrive safely at a port, find the mean and standard deviation of ships returning safely out of a total of 500 ships.

Probability of a ship not arriving safely at a port $= q = \dfrac{1}{10}$

$\therefore \;$ Probability of a ship arriving safely at a port $= p = 1 - q = 1 - \dfrac{1}{10} = \dfrac{9}{10}$

Total number of ships $= n = 500$

Mean number of ships arriving safely at port $= n \times p$

$\therefore \;$ Mean number of ships arriving safely $= 500 \times \dfrac{9}{10} = 450$ ships

Standard deviation of number of ships arriving safely at port $= \sqrt{n \times p \times q}$

$\therefore \;$ Required standard deviation $= \sqrt{500 \times \dfrac{9}{10} \times \dfrac{1}{10}} = 3 \sqrt{5}$

Probability

A die is thrown $120$ times and getting $1$ or $5$ is considered a success. Find the mean and variance of the number of successes.

Probability of getting a $1$ when a dice is thrown $= \dfrac{1}{6}$

Probability of getting a $5$ when a dice is thrown $= \dfrac{1}{6}$

Let $p = \;$ probability of getting a $1$ or a $5$ when a dice is thrown $= \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}$

$\therefore \;$ $q = 1 - p = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

Number of times the dice is rolled $= n = 120$

By definition, Mean $= n \times p$

$\therefore \;$ $\text{Mean } = 120 \times \dfrac{1}{3} = 40$

By definition, Variance $= n \times p \times q$

$\therefore \;$ $\text{Variance } = 120 \times \dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{80}{3}$

Probability

Find the mean and variance for the probability density function $f \left(x\right) = \begin{cases} x e^{- x}, & x > 0 \\ & \\ 0, & \text{ otherwise} \end{cases}$

By definition, mean $= E \left(X\right) = \int \limits_{- \infty}^{\infty} x \; f \left(x\right) \; dx$

$\left[\text{Note: } \int \limits_{0}^{\infty} x^{n} e^{- \alpha x} \; dx = \dfrac{n!}{\alpha^{n + 1}} \;\;\; \text{ n is a positive integer}\right]$

$\begin{aligned} \text{Here, } E \left(X\right) & = \int \limits_{0}^{\infty} x \times \left(x \; e^{- x}\right) \; dx\\\\ & = \int \limits_{0}^{\infty} x^2 e^{- x} \; dx \\\\ & = \dfrac{2!}{1^{2 + 1}} \\\\ & = 2 \end{aligned}$

By definition, mean $= E \left(X^2\right) = \int \limits_{- \infty}^{\infty} x^2 \; f \left(x\right) \; dx$

$\begin{aligned} \text{Here, } E \left(X^2\right) & = \int \limits_{0}^{\infty} x^2 \times \left(x \; e^{- x}\right) \; dx \\\\ & = \int \limits_{0}^{\infty} x^3 \; e^{- x} \; dx \\\\ & = \dfrac{3!}{1^{3 + 1}} \\\\ & = 6 \end{aligned}$

$\begin{aligned} \text{Now, variance } & = E \left(X^2\right) - \left[E \left(X\right)\right]^2 \\\\ & = 6 - \left(2\right)^2 = 2 \end{aligned}$

Probability

Find the mean and variance for the probability density function $f \left(x\right) = \begin{cases} \dfrac{1}{24}, & - 12 \leq x \leq 12 \\ & \\ 0, & \text{ otherwise} \end{cases}$

By definition, mean $= E \left(X\right) = \int \limits_{- \infty}^{\infty} x \; f \left(x\right) \; dx$

$\begin{aligned} \text{Here, } E \left(X\right) & = \int \limits_{- 12}^{12} \dfrac{x}{24} \; dx\\\\ & = \dfrac{1}{24} \times \left[\dfrac{x^2}{2}\right]_{- 12}^{12} \\\\ & = \dfrac{1}{48} \left[\left(12\right)^2 - \left(-12\right)^2\right] \\\\ & = 0 \end{aligned}$

By definition, mean $= E \left(X^2\right) = \int \limits_{- \infty}^{\infty} x^2 \; f \left(x\right) \; dx$

$\begin{aligned} \text{Here, } E \left(X^2\right) & = \int \limits_{- 12}^{12} \dfrac{x^2}{24} \; dx \\\\ & = \dfrac{1}{24} \times \left[\dfrac{x^3}{3}\right]_{- 12}^{12} \\\\ & = \dfrac{1}{72} \times \left[\left(12\right)^3 - \left(- 12\right)^3\right] \\\\ & = \dfrac{3456}{72} = 48 \end{aligned}$

$\begin{aligned} \text{Now, variance } & = E \left(X^2\right) - \left[E \left(X\right)\right]^2 \\\\ & = 48 - \left(0\right)^2 = 48 \end{aligned}$

Probability

In a game, a person wins $10$ points if she gets all heads or all tails and loses $5$ points if she gets $1$ or $2$ heads when $3$ coins are tossed once. Find her expectation of number of points.

Let $X$ be the random variable denoting the number of points a person can win.

Then $X$ takes values $10, \; -5$.

P(getting $1$ head when $3$ coins are tossed) $= \dfrac{3}{8}$

P(getting $2$ heads when $3$ coins are tossed) $= \dfrac{3}{8}$

P(getting $3$ heads when $3$ coins are tossed) $= \dfrac{1}{8}$

P(getting $3$ tails when $3$ coins are tossed) $= \dfrac{1}{8}$

$\begin{aligned} P \left(X = 10\right) & = P \left(\text{probability of getting 10 points}\right) \\\\ & = P \left(\text{getting all heads}\right) \; OR \;\; P \left(\text{getting all tails}\right) \\\\ & = P \left(\text{getting all heads}\right) + P \left(\text{getting all tails}\right) \\\\ & = \dfrac{1}{8} + \dfrac{1}{8} \\\\ & = \dfrac{2}{8} = \dfrac{1}{4} \end{aligned}$

$\begin{aligned} P \left(X = -5\right) & = P \left(\text{loosing 5 points}\right) \\\\ & = P \left(\text{getting 1 head}\right) \; OR \;\; P \left(\text{getting 2 heads}\right) \\\\ & = P \left(\text{getting 1 head}\right) + P \left(\text{getting 2 heads}\right) \\\\ & = \dfrac{3}{8} + \dfrac{3}{8} \\\\ & = \dfrac{6}{8} = \dfrac{3}{4} \end{aligned}$

$X$	$10$	$-5$
$P \left(X = x\right)$	$\dfrac{1}{4}$	$\dfrac{3}{4}$

Mean value of $X = E \left(X\right) = \sum \limits_{i} x_i p_i$

$\therefore \;$ $E \left(X\right) = \left(10 \times \dfrac{1}{4}\right) + \left(-5 \times \dfrac{3}{4}\right) = - \dfrac{5}{4} = - 1.25$

i.e. $\;$ The person is expected to loose $1.25$ points.

Probability

Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.

Let $X$ be a random variable which represents drawing an ace from a well shuffled deck of $52$ cards.

Then $X$ takes values $0, \; 1, \; 2$.

$2$ cards can be selected (with replacement) from a pack of $52$ cards in $52 \times 52$ ways.

Now, $P \left(X = 0\right) = $ P(No ace is selected)

The first card can be selected in $48$ ways; the second card can be selected in $48$ ways.

$\therefore \;$ $P \left(X = 0\right) = \dfrac{48 \times 48}{52 \times 52} = \dfrac{12 \times 12}{13 \times 13} = \dfrac{144}{169}$

$P \left(X = 1\right) = $ P(only one of the two cards is an ace)

$1$ ace can be selected from $4$ aces in $4$ ways; $1$ card (other than ace) can be selected in $48$ ways.

Amongst themselves, the two cards can be selected in $2$ ways.

$\therefore \;$ $P \left(X = 1\right) = 2 \times \dfrac{4 \times 48}{52 \times 52} = 2 \times \dfrac{1}{13} \times \dfrac{12}{13} = \dfrac{24}{169}$

$P \left(X = 2\right) = $ P(both cards are ace cards)

The first card can be selected in $4$ ways; the second card can be selected in $4$ ways.

$\therefore \;$ $P \left(X = 2\right) = \dfrac{4 \times 4}{52 \times 52} = \dfrac{1}{13} \times \dfrac{1}{13} = \dfrac{1}{169}$

$\therefore \;$ We have

$X$	$0$	$1$	$2$
$P \left(X = x\right)$	$\dfrac{144}{169}$	$\dfrac{24}{169}$	$\dfrac{1}{169}$

Mean value of $X = E \left(X\right) = \sum \limits_{i} x_i p_i$

$\therefore \;$ $E \left(X\right) = \left(0 \times \dfrac{144}{169}\right) + \left(1 \times \dfrac{24}{169}\right) + \left(2 \times \dfrac{1}{169}\right) = \dfrac{26}{169} = \dfrac{2}{13}$

Variance of $X = E \left(X^2\right) - \left[E \left(X\right)\right]^2$

Now, $E \left(X^2\right) = \sum \limits_{i} p_i x_{i}^{2}$

$\therefore \;$ $E \left(X^2\right) = \left(0^2 \times \dfrac{144}{169}\right) + \left(1^2 \times \dfrac{24}{169}\right) + \left(2^2 \times \dfrac{1}{169}\right) = \dfrac{28}{169}$

$\left[E \left(X\right)\right]^2 = \left(\dfrac{2}{13}\right)^2 = \dfrac{4}{169}$

$\therefore \;$ Variance of $X = \dfrac{28}{169} - \dfrac{4}{169} = \dfrac{24}{169}$

Probability

Find the expected value of the number on a dice when thrown.

Let $X$ be a random variable which represents the value of a number when a dice is thrown.

Then $X$ takes values from $1$ to $6$.

P(getting any given number on the dice) $= \dfrac{1}{6}$

$\therefore \;$ We have

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P \left(X = x\right)$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$	$\dfrac{1}{6}$

Expected value of $X = E \left(X\right) = \sum \limits_{i} x_i p_i$

$\begin{aligned} \therefore \; E \left(X\right) & = \left(1 \times \dfrac{1}{6}\right) + \left(2 \times \dfrac{1}{6}\right) + \left(3 \times \dfrac{1}{6}\right) + \left(4 \times \dfrac{1}{6}\right) + \left(5 \times \dfrac{1}{6}\right) + \left(6 \times \dfrac{1}{6}\right) \\\\ & = \dfrac{21}{6} = 3.5 \end{aligned}$

Probability

A random variable $X$ has a probability density function $\;$ $f \left(x\right) = \begin{cases} k, & 0 < x < 2 \pi \\ & \\ 0, & \text{ elsewhere} \end{cases}$
Find:

$k$
$P \left(0 < x < \dfrac{\pi}{2}\right)$
$P \left(\dfrac{\pi}{2} < x < \dfrac{3 \pi}{2}\right)$

$\because \;$ $f \left(x\right)$ is a probability density function, $\int \limits_{-\infty}^{+ \infty} f \left(x\right) dx = 1$

$\therefore \;$ $\int \limits_{0}^{2 \pi} f \left(x\right) dx = 1$

i.e. $\;$ $\int \limits_{0}^{2 \pi} k \; dx = 1$

i.e. $\;$ $k \times \left[x\right]_{0}^{2 \pi} = 1$

$\implies$ $k = \dfrac{1}{2 \pi}$
$P \left(0 < x < \dfrac{\pi}{2}\right) = \int \limits_{0}^{\pi / 2} f \left(x\right) dx$

Now, $f \left(x\right) = \begin{cases} \dfrac{1}{2 \pi}, & 0 < x < 2 \pi \\ & \\ 0, & \text{ elsewhere} \end{cases}$

$\begin{aligned} \therefore \; P \left(0 < x < \dfrac{\pi}{2}\right) & = \dfrac{1}{2 \pi} \int \limits_{0}^{\pi / 2} dx \\\\ & = \dfrac{1}{2 \pi} \times \left[x\right]_{0}^{\pi / 2} \\\\ & = \dfrac{1}{2 \pi} \times \dfrac{\pi}{2} = \dfrac{1}{4} \end{aligned}$
$P \left(\dfrac{\pi}{2} < x < \dfrac{3 \pi}{2}\right) = \int \limits_{\pi / 2}^{3 \pi / 2} f \left(x\right) dx$

$\begin{aligned} \therefore \; P \left(\dfrac{\pi}{2} < x < \dfrac{3 \pi}{2}\right) & = \dfrac{1}{2 \pi} \int \limits_{\pi / 2}^{3 \pi / 2} dx \\\\ & = \dfrac{1}{2 \pi} \times \left[x\right]_{\pi / 2}^{3 \pi / 2} \\\\ & = \dfrac{1}{2 \pi} \times \left(\dfrac{3 \pi}{2} - \dfrac{\pi}{2}\right) = \dfrac{1}{2} \end{aligned}$

Probability

For the distribution function given by $\;$ $F \left(x\right) = \begin{cases} 0, & x < 0 \\ & \\ x^2, & 0 \leq x \leq 1 \\ & \\ 1, & x > 1 \end{cases}$
find the density function. Also evaluate

$P \left(0.5 < x < 0.75\right)$

$P \left(x \leq 0.5\right)$

$P \left(x > 0.75\right)$

Probability density function (p.d.f) $= f \left(x\right) = \dfrac{d}{dx} F \left(x\right)$

Now, $\dfrac{d}{dx} \left(x^2\right) = 2 x$ $\;$ and $\;$ $\dfrac{d}{dx} \left(1\right) = 0$

$\therefore \;$ $f \left(x\right) = \begin{cases} 2 x, & 0 \leq x \leq 1 \\ & \\ 0, & \text{ elsewhere} \end{cases}$

$P \left(0.5 < x < 0.75\right) = \int \limits_{0.5}^{0.75} 2 \; x \; dx$

Now,

$\begin{aligned} \int \limits_{0.5}^{0.75} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.5}^{0.75} \\\\ & = \left(0.75\right)^2 - \left(0.5\right)^2 = 0.3125 \end{aligned}$

$\therefore \;$ $P \left(0.5 < x < 0.75\right) = 0.3125$
$P \left(x \leq 0.5\right) = \int \limits_{0}^{0.5} 2 \; x \; dx$

Now,

$\begin{aligned} \int \limits_{0}^{0.5} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0}^{0.5} \\\\ & = \left(0.5\right)^2 = 0.25 \end{aligned}$

$\therefore \;$ $P \left(x \leq 0.5\right) = 0.25$
$P \left(x > 0.75\right) = \int \limits_{0.75}^{1} 2 \; x \; dx$

Now,

$\begin{aligned} \int \limits_{0.75}^{1} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.75}^{1} \\\\ & = \left(1\right)^2 - \left(0.75\right)^2 = 0.4375 \end{aligned}$

$\therefore \;$ $P \left(x > 0.75\right) = 0.4375$

Probability

For the probability distribution function (p.d.f) $\;$ $f\left(x\right)= \begin{cases} c x \left(1 - x\right)^3, & 0 < x < 1 \\ & \\ 0, & \text{elsewhere} \end{cases}$ $\;$ find

the constant $c$
$P \left(x < \dfrac{1}{2}\right)$

By definition of p.d.f, $\;$ $\int \limits_{0}^{1} f \left(x\right) dx = 1$

i.e. $\;$ $\int \limits_{0}^{1} cx \left(1 - x\right)^3 dx = 1$

i.e. $\;$ $\int \limits_{0}^{1} c \left(1 - x\right) \left[1 - \left(1 - x\right)\right]^3 dx = 1$ $\; \left[\because \; \int \limits_{0}^{a} f \left(x\right) dx = \int \limits_{0}^{a} f \left(a - x\right) dx\right]$

i.e. $\;$ $\int \limits_{0}^{1} c \left(1 - x\right) x^3 dx = 1$

i.e. $\;$ $c \left\{\int \limits_{0}^{1} x^3 dx - \int \limits_{0}^{1} x^4 dx \right\} = 1$

i.e. $\;$ $c \left\{\left[\dfrac{x^4}{4}\right]_{0}^{1} - \left[\dfrac{x^5}{5}\right]_{0}^{1} \right\} = 1$

i.e. $\;$ $c \left(\dfrac{1}{4} - \dfrac{1}{5}\right) = 1$

$\implies$ $c = 20$
$P \left(x < \dfrac{1}{2}\right) = \int \limits_{0}^{1/2} f \left(x\right) dx$

Here, $f \left(x\right) = \begin{cases} 20x \left(1 - x\right)^3, & 0 < x < 1 \\ & \\ 0, & \text{elsewhere} \end{cases}$

$\begin{aligned} \therefore \; P \left(x < \dfrac{1}{2}\right) & = 20 \int \limits_{0}^{1/2} x \left(1 - x\right)^3 dx \\\\ & = 20 \int \limits_{0}^{1/2} x \left(1 - 3x + 3x^2 -x^3\right) dx \\\\ & = 20 \int \limits_{0}^{1/2} \left(x - 3x^2 + 3x^3 -x^4\right) dx \\\\ & = 20 \left[\dfrac{x^2}{2} - 3 \left(\dfrac{x^3}{3}\right) + 3 \left(\dfrac{x^4}{4}\right) - \dfrac{x^5}{5}\right]_{0}^{1/2} \\\\ & = 20 \left[\dfrac{1}{8} - \dfrac{1}{8} + \dfrac{3}{64} - \dfrac{1}{160}\right] \\\\ & = 20 \left[\dfrac{15-2}{320}\right] \\\\ & = 20 \times \dfrac{13}{320} = \dfrac{13}{16} \end{aligned}$

Probability

A discrete random variable $X$ has the following probability distribution:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P \left(X\right)$	$a$	$3a$	$5a$	$7a$	$9a$	$11a$	$13a$	$15a$	$17a$

Find the value of $a$
Find $P \left(x < 3\right)$
Find $P \left(3 < x < 7\right)$

$\sum \limits_{i=0}^{8} p_i = 1 $

Here $\;$ $p_0 = a, \; p_1 = 3a, \; p_2 = 5a, \; \cdots \; p_8 = 17a$

Now, $\;$ $a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 81a$

$\therefore \;$ $81 a = 1$ $\implies$ $a = \dfrac{1}{81}$
$P \left(x < 3\right) = P \left(0\right) + P \left(1\right) + P \left(2\right)$

$\begin{aligned} \therefore \; P \left(x < 3\right) & = a + 3a + 5a \\\\ & = 9a \\\\ & = 9 \times \dfrac{1}{81} = \dfrac{1}{9} \end{aligned}$
$P \left(3 < x < 7\right) = P \left(4\right) + P \left(5\right) + P \left(6\right)$

$\begin{aligned} \therefore \; P \left(3 < x < 7\right) & = 9a + 11a + 13a \\\\ & = 33a \\\\ & = 33 \times \dfrac{1}{81} = \dfrac{11}{27} \end{aligned}$

Probability

$2$ bad oranges are accidentally mixed with $10$ good ones. $3$ oranges are drawn at random without replacement from this lot. Obtain the probability distribution for the number of bad oranges.

$3$ oranges can be drawn, without replacement, from $12$ oranges in ${^{12}}{P}_{3} = \dfrac{12!}{9!}$ ways

Let $X$ denote a random variable which is drawing bad oranges when $3$ oranges are selected without replacement.

$X$ can take values $0, \; 1, \; 2$

Now,

$P \left(X = 0\right) = P \left(\text{No bad orange is selected}\right)$

$ = \dfrac{{^{10}}{P}_{3}}{{^{12}}{P}_{3}} = \dfrac{\dfrac{10!}{7!}}{\dfrac{12!}{9!}} = \dfrac{10! \times 9!}{12! \times 7!} = \dfrac{9 \times 8}{12 \times 11} = \dfrac{6}{11}$

$P \left(X = 1\right) = P \left(1\text{ bad orange is selected}\right)$

$ = \dfrac{3 \times {^{2}}{P}_{1} \times {^{10}}{P}_{2}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{8!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! \times 9!}{12! \times 8!} = \dfrac{3 \times 2 \times 9}{12 \times 11} = \dfrac{9}{22}$

$P \left(X = 2\right) = P \left(2\text{ bad oranges are selected}\right)$

$ = \dfrac{3 \times 2 \times {^{10}}{P}_{1}}{{^{12}}{P}_{3}} = \dfrac{3 \times 2 \times\dfrac{10!}{9!}}{\dfrac{12!}{9!}} = \dfrac{3 \times 2 \times 10! }{12!} = \dfrac{3 \times 2}{12 \times 11} = \dfrac{1}{22}$

Hence the probability distribution of the random variable $X$ is given by

$X$	$0$	$1$	$2$
$P \left(X\right)$	$\dfrac{6}{11} = \dfrac{12}{22}$	$\dfrac{9}{22}$	$\dfrac{1}{22}$

Probability

Find the probability distribution of the number of sixes in throwing three dice once.

Let $X$ denote a random variable which is the number of sixes obtained when three dice are thrown at once.

$X$ can take values $0, \; 1, \; 2, \; 3$

Probability of getting a $6$ when a dice is thrown $= \dfrac{1}{6}$

Probability of NOT getting a $6$ when a dice is thrown $= \dfrac{5}{6}$

Now,

$P \left(X = 0\right) = P \left(\text{No } 6 \text{ is obtained in any throw}\right) = \left(\dfrac{5}{6}\right)^3 = \dfrac{125}{216}$

$P \left(X = 1\right) = P \left(\text{One } 6 \text{ is obtained in } 3 \text{ throws}\right) = 3 \times \dfrac{1}{6} \times \left(\dfrac{5}{6}\right)^2 = \dfrac{75}{216}$

$P \left(X = 2\right) = P \left(\text{Two } 6's \text{ are obtained in } 3 \text{ throws} \right) = 3 \times \left(\dfrac{1}{6}\right)^2 \times \dfrac{5}{6} = \dfrac{15}{216}$

$P \left(X = 3\right) = P \left(6's \text{ are obtained in all throws}\right) = \left(\dfrac{1}{6}\right)^3 = \dfrac{1}{216}$

Hence the probability distribution of the random variable $X$ is given by

$X$	$0$	$1$	$2$	$3$
$P \left(X\right)$	$\dfrac{125}{216}$	$\dfrac{75}{216}$	$\dfrac{15}{216}$	$\dfrac{1}{216}$

Probability

Find each probability if three coins are tossed:

P($3$ heads or $3$ tails)
P(exactly $2$ tails)

When a coin is tossed,

probability of getting head $= p = \dfrac{1}{2}$

probability of getting tail $= q = 1- p = \dfrac{1}{2}$

Number of coins $= 3$

P($3$ heads or $3$ tails)

$= P \left(3 \text{ heads}\right) + P \left(3 \text{ tails}\right)$

$= {^{3}}{C}_{3} \times \left(\dfrac{1}{2}\right)^3 \times \left(\dfrac{1}{2}\right)^0 + {^{3}}{C}_{0} \times \left(\dfrac{1}{2}\right)^0 \times \left(\dfrac{1}{2}\right)^3$

$= 2 \times \left(\dfrac{1}{2}\right)^3$

$= \dfrac{1}{4}$
P(exactly $2$ tails)

$= P \left(\text{exactly }1 \text{ head}\right)$

$= {^{3}}{C}_{1} \times \left(\dfrac{1}{2}\right)^1 \times \left(\dfrac{1}{2}\right)^2$

$= 3 \times \dfrac{1}{8} = \dfrac{3}{8}$

Probability

A lady carries lipstick tubes in a bag in her purse. The probability of pulling out the color she wants is $\dfrac{2}{3}$. Suppose she uses her lipstick $4$ times a day. Find the probability

P(never the correct color)
P(no more than $3$ times correct)

Let $p$ be the probability of selecting the right color lipstick.

Let $p$ be the probability of not selecting the right color lipstick.

Given: $\;$ $p = \dfrac{2}{3}$ $\;$ $\therefore \;$ $q = 1 - p = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Total number of times lipstick is selected $= 4$

P(never selecting the correct color) $= q^4 = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$

P(no more than $3$ times correct)

$\begin{aligned} P\left(\text{no more than 3 times correct}\right) & = 1 - P \left(\text{all times correct}\right)\\\\ & = 1 - \left(p\right)^4 \\\\ & = 1 - \left(\dfrac{2}{3}\right)^4 \\\\ & = 1 - \dfrac{16}{81} = \dfrac{65}{81} \end{aligned}$

Probability

A weather reporter for a channel is forecasting a $30\%$ chance of rain for today and the next four days. Find the probability of having rain on exactly one day.

Let $p$ be the probability of raining on a given day.

Let $q$ be the probability of not raining on a given day.

Given: $\;$ $p = 30\% = 0.3$; $\;$ $q = 1- p = 1 - 0.3 = 0.7$

Total number of days $= 4$

$\begin{aligned} P \left(\text{It rains exactly one day}\right) & = {^{4}}{C}_{1} \times \left(p\right)^{1} \times \left(q\right)^{4 - 1} \\\\ & = \dfrac{4!}{1! \times 3!} \times 0.3 \times \left(0.7\right)^{3} \\\\ & = 4 \times 0.3 \times \left(0.7\right)^{3} \\\\ & = 0.4116 \approx 41.2\% \end{aligned}$

Probability

Three urns are given each containing red and white chips as given below:

Urn $I$ : $6$ red $4$ white ; Urn $II$ : $3$ red $5$ white ; Urn $III$ : $4$ red $6$ white

An urn is chosen at random and a chip is drawn from the urn.

Find the probability that it is white.
If the chip is white find the probability that it is from urn $II$.

Let $A_1$ be the event of selecting urn $I$.

Let $A_2$ be the event of selecting urn $II$.

Let $A_3$ be the event of selecting urn $III$.

Let $B$ be the event of selecting a white chip.

To find: $\;$ P(of selecting a white chip) $= P \left(B\right)$

Events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events.

$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)$ $\;\;\; \cdots \; (1)$

Probability of selecting urn $I$ $= P \left(A_1\right) = \dfrac{1}{3}$

Probability of selecting urn $II$ $= P \left(A_2\right) = \dfrac{1}{3}$

Probability of selecting urn $III$ $= P \left(A_3\right) = \dfrac{1}{3}$

P(selecting a white chip when urn $I$ is selected) $= P \left(B | A_1\right)$

$P \left(B | A_1\right) = \dfrac{{^{4}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{4}{10} = \dfrac{2}{5}$

P(selecting a white chip when urn $II$ is selected) $= P \left(B | A_2\right)$

$P \left(B | A_2\right) = \dfrac{{^{5}}{C}_{1}}{{^{8}}{C}_{1}} = \dfrac{5}{8}$

P(selecting a white chip when urn $III$ is selected) $= P \left(B | A_3\right)$

$P \left(B | A_3\right) = \dfrac{{^{6}}{C}_{1}}{{^{10}}{C}_{1}} = \dfrac{6}{10} = \dfrac{3}{5}$

Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(A_3\right)$, $P \left(B | A_1\right)$, $P \left(B | A_2\right)$ and $P \left(B | A_3\right)$ in equation $(1)$ we have,

$P \left(B\right) = \left(\dfrac{1}{3} \times \dfrac{2}{5}\right) + \left(\dfrac{1}{3} \times \dfrac{5}{8}\right) + \left(\dfrac{1}{3} \times \dfrac{3}{5}\right) = \dfrac{13}{24}$
To find: $\;$ P(of selecting urn $II$ when a white chip is selected) $= P \left(A_2 | B\right)$

Since events $A_1$, $A_2$ and $A_3$ are mutually exclusive and exhaustive events, therefore we have by Bayes' theorem,

$\begin{aligned} P \left(A_2 | B\right) & = \dfrac{P \left(A_2\right) \times P \left(B | A_2\right)}{P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right) + P \left(A_3\right) \times P \left(B | A_3\right)} \\\\ & = \dfrac{\dfrac{1}{3} \times \dfrac{5}{8}}{\dfrac{1}{3} \times \dfrac{2}{5} + \dfrac{1}{3} \times \dfrac{5}{8} + \dfrac{1}{3} \times \dfrac{3}{5}} = \dfrac{5/8}{13/8} = \dfrac{5}{13} \end{aligned}$

Probability

A factory has two Machines - $I$ and $II$. Machine-$I$ produces $25\%$ of items and Machine-$II$ produces $75\%$ of the items of the total output. Further $3\%$ of the items produced by Machine-$I$ are defective whereas $4\%$ produced by Machine-$II$ are defective. If an item is drawn at random, what is the probability that it is defective?

Let $A_1$ be the event that the items are produced by machine $I$

Let $A_2$ be the event that the items are produced by machine $II$

Let $B$ be the event of selecting a defective item

To find: $\;$ P(of selecting a defective item) $= P \left(B\right)$

Events $A_1$ and $A_2$ are mutually exclusive and exhaustive events.

$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$

Probability that the items are produced by machine-$I$ $= P \left(A_1\right) = \dfrac{25}{100}$

Probability that the items are produced by machine-$II$ $= P \left(A_2\right) = \dfrac{75}{100}$

P(getting a defective item when the item is selected from machine-$I$)

$= P \left(B | A_1\right) = \dfrac{3}{100}$

P(getting a defective item when the item is selected from machine-$II$)

$= P \left(B | A_2\right) = \dfrac{4}{100}$

Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,

$P \left(B\right) = \left(\dfrac{25}{100} \times \dfrac{3}{100}\right) + \left(\dfrac{75}{100} \times \dfrac{4}{100}\right) = \dfrac{3}{80}$

Probability

Bag $A$ contains $5$ white, $6$ black balls and bag $B$ contains $4$ white, $5$ black balls. One bag is selected at random and one ball is drawn from it. Find the probability that it is white.

Let $A_1$ be the event of selecting bag $A$.

Let $A_2$ be the event of selecting bag $B$.

Let $B$ be the event of selecting a white ball.

To find: P(of selecting a white ball) $= P \left(B\right)$

Now, $A_1$ and $A_2$ are mutually exclusive and exhaustive events.

$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$

Probability of selecting bag $A = P \left(A_1\right) = \dfrac{1}{2}$

Probability of selecting bag $B = P \left(A_2\right) = \dfrac{1}{2}$

P(selecting a white ball when bag $A$ is selected) $= P \left(B | A_1\right)$

$P \left(B | A_1\right) = \dfrac{{^{5}}{C}_{1}}{{^{11}}{C}_{1}} = \dfrac{5}{11}$

P(selecting a white ball when bag $B$ is selected) $= P \left(B | A_2\right)$

$P \left(B | A_2\right) = \dfrac{{^{4}}{C}_{1}}{{^{9}}{C}_{1}} = \dfrac{4}{9}$

Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,

$P \left(B\right) = \left(\dfrac{1}{2} \times \dfrac{5}{11}\right) + \left(\dfrac{1}{2} \times \dfrac{4}{9}\right) = \dfrac{89}{198}$

Probability

In a game played with a standard deck of cards, each face card has a value of $10$ points,, each ace has a value of $1$ point, and each number card has a value equal to its number. Two cards are drawn at random. One card is the queen of diamonds. What is the probability that the sum of the cards is greater than $18$?

The first card can be drawn from a pack of cards in $52$ ways.

The second card can be drawn from the pack of cards in $51$ ways ($\because \;$ cards are drawn without replacement).

Let $A$ be the event that the card drawn is a queen of diamonds.

Then $P\left(A\right) = \dfrac{1}{52}$

Let $B$ be the event of drawing a card from the cards with value $9$ ($4$ cards), value $10$ ($4$ cards), or the face cards ($11$ cards :-- $4$ kings, $3$ queens, $4$ jacks)

(This is done so that the sum of the two cards is greater than $18$.)

$\therefore \;$ Number of elements in $B = n \left(B\right) = 19$

$\therefore \;$ $P \left(B\right) = \dfrac{19}{51}$

$\left(A \cap B\right) \;$ is the event that a card is queen of diamonds AND the sum of the two cards is greater than $18$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{1}{52} \times \dfrac{19}{51}$

$\therefore \;$ P(when one card is queen of diamonds, then the sum of the cards is greater than $18$)

$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{1}{52} \times \dfrac{19}{51}}{\dfrac{1}{52}} = \dfrac{19}{51}$

Probability

In a classroom, $60\%$ of the students have brown hair, $30\%$ have brown eyes, and $10\%$ have both brown hair and eyes. A student is selected at random.

If the student has brown hair, what is the probability that the student also has brown eyes?
If the student does not have brown hair, what is the probability that the student does not have brown eyes?

Given: $60\%$ of the students have brown hair; $30\%$ have brown eyes; $10\%$ have both brown hair and brown eyes

Let $A$ be the event of selecting a student with brown hair

Then, $P \left(A\right) = \dfrac{60}{100} = \dfrac{6}{10}$

Let $B$ be the event of selecting a student with brown eyes

Then, $P \left(B\right) = \dfrac{30}{100} = \dfrac{3}{10}$

$\left(A \cap B\right)$ is the event of selecting a student with brown hair AND brown eyes

Then, $P \left(A \cap B\right) = \dfrac{10}{100} = \dfrac{1}{10}$

P(if the student has brown hair then the student also has brown eyes)

$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{1 / 10}{6 / 10} = \dfrac{1}{6}$
P(student does not have brown hair) $= P\left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{6}{10} = \dfrac{4}{10}$

P(if the student does not have brown hair then the student also does not have brown eyes)

$= P \left(\overline{B} | \overline{A}\right) = \dfrac{P \left(\overline{A} \cap \overline{B}\right)}{P \left(\overline{A}\right)}$

Now,

$\begin{aligned} P \left(\overline{A} \cap \overline{B}\right) & = P \left(\overline{A \cup B}\right) \\\\ & = 1 - P \left(A \cup B\right) \\\\ & = 1 - \left[P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)\right] \\\\ & = 1 - \left[\dfrac{6}{10} + \dfrac{3}{10} - \dfrac{1}{10}\right] \\\\ & = 1 - \dfrac{8}{10} \\\\ & = \dfrac{2}{10} \end{aligned}$

$\therefore \;$ $P \left(\overline{B} | \overline{A}\right) = \dfrac{2 / 10}{4 / 10} = \dfrac{2}{4} = \dfrac{1}{2}$

Probability

A container holds $3$ green marbles and $5$ yellow marbles. One marble is randomly drawn and discarded. Then a second marble is drawn. Find the probability that the second marble is yellow, given that the first marble was green.

Total number of marbles $= 8$

The first marble can be selected in $8$ ways; the second marble can be selected in $7$ ways.

Let $A$ be the event of drawing the first marble (green).

Let $B$ be the event of drawing the second marble (yellow).

$\therefore \;$ P(selecting a green marble) $= P \left(A\right) = \dfrac{3}{8}$

$1$ yellow marble can be selected from $5$ yellow marbles in $5$ ways

$\therefore \;$ P(selecting a yellow marble) $= \dfrac{5}{7}$

$\left(A \cap B\right) = \;$ event of selecting a green \textbf{AND} a yellow marble

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{3}{8} \times \dfrac{5}{7}$

Now, P(selecting a yellow marble given that the first marble is green)

$= P \left(B|A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{\dfrac{3}{8} \times \dfrac{5}{7}}{\dfrac{3}{8}} = \dfrac{5}{7}$

Probability

Two game tiles, numbered $1$ through $9$, are selected at random from a box without replacement. If their sum is even, what is the probability that both the numbers are odd?

$2$ game tiles can be selected from $9$ tiles in ${^{9}}{P}_{2} = \dfrac{9!}{7!} = 9 \times 8 = 72 \;$ ways

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 72$

Let $A$ be the event that both the tiles are odd.

Let $B$ be the event that the sum of numbers on the two tiles is even.

This happens when both the selected tiles are odd \textbf{OR} both are even.

There are $5$ odd tiles (1, 3, 5, 7, 9) and $4$ even tiles (2,4,6,8).

$2$ Odd tiles can be selected from $5$ tiles in ${^{5}}{P}_{2} = \dfrac{5!}{3!} = 5 \times 4 = 20 \;$ ways

$2$ even tiles can be selected from $4$ tiles in ${^{4}}{P}_{2} = \dfrac{4!}{2!} = 4 \times 3 = 12 \;$ ways

$\therefore \;$ Number of elements in $B = n \left(B\right) = 20 + 12 = 32$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{32}{72}$

$\left(A \cap B\right)$ is the event that both the tiles are odd numbered \textbf{AND} their sum is even.

$\therefore \;$ $n \left(A \cap B\right) = 20$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{20}{72}$

Now, P(that both the numbers are odd given that their sum is even)

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{20/72}{32/72} = \dfrac{20}{32} = \dfrac{5}{8}$

Probability

A city council consists of six people of party $P_1$, two of whom are women, and six of party $P_2$, four of whom are men. A member is chosen at random. If the member chosen is a man, what is the probability that he is from party $P_1$?

Total number of members in city council $= 12$

$\therefore \;$ Number of elements in sample space $S = n \left(S\right) = 12$

Total number of men in the council $= 8$; $\;$ total number of women in the council $= 4$

Let $A$ be the event of selecting a member from party $P_1$

Let $B$ be the event that the member chosen is a man

Then, $n \left(B\right) = 8$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{8}{12} = \dfrac{2}{3}$

$\left(A \cap B\right) = \;$ event of selecting party $P_1$ AND a man

Party $P_1$ can be selected in $1$ way;

A man from $P_1$ can be selected in $4$ ways

$\therefore \;$ Number of ways of selecting party $P_1$ $\;$ AND $\;$ a man $= 1 \times 4 = 4$ ways

$\therefore \;$ Number of elements in $\left(A \cap B\right) = 4$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{4}{12} = \dfrac{1}{3}$

$\therefore \;$ P(selecting party $P_1$ when the member is a man)

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/3}{2/3} = \dfrac{1}{2}$

Probability

A pair of number cubes is thrown. Find each probability given that their sum is greater than or equal to $9$.

$P \left(\text{numbers match}\right)$
$P \left(\text{numbers match or sum is even}\right)$

Number of elements in sample space $S = n \left(S\right) = 36$

Let $C$ be the event that the sum of the numbers on the two cubes is greater than or equal to $9$.

i.e. $\;$ $C = \left\{\left(3, \; 6\right), \; \left(4, \; 5\right), \; \left(4, \; 6\right), \; \left(5, \; 4\right), \; \left(5, \; 5\right), \right.$

$\hspace{3cm}$ $\left. \left(5, \; 6\right), \; \left(6, \; 3\right), \; \left(6, \; 4\right), \; \left(6, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $C = n \left(C\right) = 10$

$\therefore \;$ $P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{10}{36}$

Let $A$ be the event that the numbers on the two cubes match.

i.e. $\;$ $A = \left\{\left(1, \; 1\right), \; \left(2, \; 2\right), \; \left(3, \; 3\right), \; \left(4, \; 4\right), \; \left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\left(A \cap C\right) = \;$ event that the numbers on both the cubes match AND the sum of the numbers is greater than or equal to $9$

$\therefore \;$ $\left(A \cap C\right) = \left\{\left(5, \; 5\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $\left(A \cap C\right) = n \left(A \cap C\right) = 2$

$\therefore \;$ $P \left(A \cap C\right) = \dfrac{n \left(A \cap C\right)}{n \left(S\right)} = \dfrac{2}{36}$

$\therefore \;$ P(numbers on the two cubes match given that their sum is greater than or equal to 9)

$= P \left(A | C\right) = \dfrac{P \left(A \cap C\right)}{P \left(C\right)} = \dfrac{2/36}{10/36} = \dfrac{2}{10} = \dfrac{1}{5}$
Let $B$ be the event that the numbers on the two cubes match or the sum is even

i.e. $\;$ $B = \left\{\left(1, \; 1\right), \; \left(2, \; 2\right), \; \left(3, \; 3\right), \; \left(4, \; 4\right), \; \left(5, \; 5\right), \; \left(6, \; 6\right), \right. $

$\hspace{2cm}$ $\left. \left(1, \; 3\right), \; \left(1, \; 5\right), \; \left(2, \; 4\right), \; \left(2, \; 6\right), \; \left(3, \; 1\right), \; \left(3, \; 5\right) \right. $

$\hspace{3cm}$ $\left. \left(4, \; 2\right), \; \left(4, \; 6\right), \; \left(5, \; 1\right), \; \left(5, \; 3\right), \; \left(6, \; 2\right), \; \left(6, \; 4\right) \right\}$

$\left(B \cap C\right) = \;$ event that the numbers on the two cubes match or the sum is even AND the sum of the numbers is greater than or equal to $9$

$\therefore \;$ $\left(B \cap C\right) = \left\{\left(4, \; 6\right), \; \left(5, \; 5\right), \; \left(6, \; 4\right), \; \left(6, \; 6\right) \right\}$

$\therefore \;$ Number of elements in $\left(B \cap C\right) = n \left(B \cap C\right) = 4$

$\therefore \;$ $P \left(B \cap C\right) = \dfrac{n \left(B \cap C\right)}{n \left(S\right)} = \dfrac{4}{36}$

$\therefore \;$ P(numbers on the two cubes match or sum is even given that their sum is greater than or equal to 9)

$= P \left(B | C\right) = \dfrac{P \left(B \cap C\right)}{P \left(C\right)} = \dfrac{4/36}{10/36} = \dfrac{4}{10} = \dfrac{2}{5}$

Probability

Three coins are tossed. Find the probability that they all land heads up when at least one coin shows a head.

Number of elements in sample space $S = n \left(S\right) = 2^{3} = 8$

Let $A$ be the event that all 3 coins land heads up.

Then $A = \left\{H, \; H, \; H\right\}$

Let $B$ be the event that at least one coin shows a head.

Then $\overline{B}$ is the event that all coins land tails up.

Now, $\overline{B} = \left\{T, \; T, \; T\right\}$ $\implies$ $n \left(\overline{B}\right) = 1$

$\therefore \;$ $P \left(\overline{B}\right) = \dfrac{n \left(\overline{B}\right)}{n \left(S\right)} = \dfrac{1}{8}$

$\therefore \;$ $P \left(B\right) = 1 - P \left(\overline{B}\right) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$

$\left(A \cap B\right) = $ event that all coins show heads AND at least one coin shows a head

Now, $\left(A \cap B\right) = \left\{H, \; H, \; H\right\}$ $\implies$ $n \left(A \cap B\right) = 1$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{1}{8}$

$\therefore \;$ Probability that all coins land heads up when at least one coin shows a head

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{1/8}{7/8} = \dfrac{1}{7}$

Probability

One card is drawn from a standard deck of cards. What is the probability that it is a queen if it is known to be a face card?

Number of elements in sample space $S = n \left(S\right) = 52$

Let $A$ be the event of drawing a queen.

Let $B$ be the event of drawing a face card.

Number of elements in $B = n \left(B\right) = 12$

$\therefore \;$ $P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{12}{52}$

$\left(A \cap B\right) = $ event that the face card is a queen

$\because \;$ There are $4$ queen cards, $n \left(A \cap B\right) = 4$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{n \left(A \cap B\right)}{n \left(S\right)} = \dfrac{4}{52}$

$\therefore \;$ Probability of drawing a queen if the card selected is a face card

$= P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)} = \dfrac{4/52}{12/52} = \dfrac{4}{12} = \dfrac{1}{3}$

Probability

A can hit a target $4$ times in $5$ shots, B $3$ times in $4$ shots and C $2$ times in $3$ shots. They fire a volley. What is the chance that the target is damaged by exactly $2$ hits?

Let $A$ be the event that A hits the target.

Let $B$ be the event that B hits the target.

Let $C$ be the event that C hits the target.

Then, $P \left(A\right) = \dfrac{4}{5}$, $\;$ $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

$P \left(B\right) = \dfrac{3}{4}$, $\;$ $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$P \left(C\right) = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - P \left(C\right) = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Now, probability that the target is damaged by exactly $2$ hits

$= P \left[\left(A \cap B \cap \overline{C}\right) \cup \left(A \cap \overline{B} \cap C\right) \cup \left(\overline{A} \cap B \cap C\right)\right]$

$= P \left(A\right) \times P \left(B\right) \times P \left(\overline{C}\right) + P \left(A\right) \times P \left(\overline{B}\right) \times P \left(C\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(C\right)$

[$\because \;$ $A$, $B$, $C$ are independent events, $\overline{A}$, $\overline{B}$, $\overline{C}$ are also independent events]

$= \left(\dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{1}{3}\right) + \left(\dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{2}{3}\right) + \left(\dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{2}{3}\right) $

$= \dfrac{26}{60}$

$= \dfrac{13}{30}$

Probability

A problem in Mathematics is given to three students whose chances of solving it are $\dfrac{1}{2}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$.

What is the probability that the problem is solved?
What is the probability that exactly one of them will solve it?

Let $A$ be the event that the problem is solved by the first student.

Let $B$ be the event that the problem is solved by the second student.

Let $C$ be the event that the problem is solved by the third student.

Given: $\;$ $P \left(A\right) = \dfrac{1}{2}$, $\;$ $P \left(B\right) = \dfrac{1}{3}$, $\;$ $P \left(C\right) = \dfrac{1}{4}$

Then, $P \left(\overline{A}\right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}$, $\;$ $P \left(\overline{B}\right) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$

Probability that the problem is solved:

$\begin{aligned} P \left(\text{problem is solved}\right) & = P \left(\text{problem is solved by at least one of them}\right) \\\\ & = P \left(A \cup B \cup C\right) \\\\ & = 1 - P \left(\overline{A \cup B \cup C}\right) \\\\ & = 1 - P \left(\overline{A} \cap \overline{B} \cap \overline{C}\right) \;\;\; \left[\text{By DeMorgan's law}\right] \\\\ & = 1 - P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) \\\\ & \left[\because \; \text{A, B, C are independent events, } \right. \\ & \hspace{2cm} \left. \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] \\\\ & = 1 - \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \\\\ & = \dfrac{3}{4} \end{aligned}$
Exactly one of them solves the problem:

$P \left(\text{exctly one solves the problem}\right)$

$= P \left[\left(A \cap \overline{B} \cap \overline{C}\right) \cup \left(\overline{A} \cap B \cap \overline{C} \right) \cup\left(\overline{A} \cap \overline{B} \cap C\right)\right] $

$ = P \left(A \cap \overline{B} \cap \overline{C}\right) + P \left(\overline{A} \cap B \cap \overline{C}\right) + P \left(\overline{A} \cap \overline{B} \cap C\right) $

$ \left[\because \; \left(A \cap \overline{B} \cap \overline{C}\right), \; \left(\overline{A} \cap B \cap \overline{C}\right), \; \left(\overline{A} \cap \overline{B} \cap C\right) \right.$
$\hspace{3cm} \left. \text{ are mutually exclusive events} \right]$

$= P \left(A\right) \times P \left(\overline{B}\right) \times P \left(\overline{C}\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(\overline{C}\right) $
$\hspace{5cm} + P \left(\overline{A}\right) \times P \left(\overline{B}\right) \times P \left(C\right) $

$ \left[\because \; \text{A, B, C are independent events, } \overline{A}, \overline{B}, \overline{C} \text{ are also independent events}\right] $

$ = \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{3}{4}\right) + \left(\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{1}{4}\right) $

$ = \dfrac{11}{24} $

Probability

One bag contains $5$ white and $3$ black marbles. Another bag contains $4$ white and $6$ black marbles. If one marble is drawn from each bag, find the probability that

both are white;
both are black;
one is white and one is black.

Let bag $B_1$ contain $5$ white and $3$ black marbles.

Let bag $B_2$ contain $4$ white and $6$ black marbles.

Let $A$ be the event of drawing a white marble from bag $B_1$

Let $B$ be the event of drawing a white marble from bag $B_2$

Then, probability of drawing white marbles from both the bags $= P \left(A \cap B\right)$

$\because \;$ $A$ and $B$ are independent events, $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$

Now, $P \left(A\right) = \dfrac{5}{8}$, $\;$ $P \left(B\right) = \dfrac{4}{10}$

$\therefore \;$ $P \left(A \cap B\right) = \dfrac{5}{8} \times \dfrac{4}{10} = \dfrac{1}{4}$
Let $C$ be the event of drawing a black marble from bag $B_1$

Let $D$ be the event of drawing a black marble from bag $B_2$

Then, probability of drawing black marbles from both the bags $= P \left(C \cap D\right)$

$\because \;$ $C$ and $D$ are independent events, $P \left(C \cap D\right) = P \left(C\right) \times P \left(D\right)$

Now, $P \left(C\right) = \dfrac{3}{8}$, $\;$ $P \left(D\right) = \dfrac{6}{10}$

$\therefore \;$ $P \left(C \cap D\right) = \dfrac{3}{8} \times \dfrac{6}{10} = \dfrac{9}{40}$
Let $E$ be the event of drawing a white marble from bag $B_1$

Let $F$ be the event of drawing a black marble from bag $B_2$

Let $G$ be the event of drawing a black marble from bag $B_1$

Let $H$ be the event of drawing a white marble from bag $B_2$

Then, probability of drawing a white marble from one bag and a black marble from the other bag $= P \left(E \cap F\right) + P \left(G \cap H\right)$

$\because \;$ $E$, $F$ are independent events and $G$, $H$ are independent events, we have

$= P \left(E \cap F\right) + P \left(G \cap H\right) = P \left(E\right) \times P \left(F\right) + P \left(G\right) \times P \left(H\right)$

Now, $P \left(E\right) = \dfrac{5}{8}$, $\;$ $P \left(F\right) = \dfrac{6}{10}$, $\;$ $P \left(G\right) = \dfrac{3}{8}$, $\;$ $P \left(F\right) = \dfrac{4}{10}$

$\begin{aligned} \therefore \; P \left(E \cap F\right) + P \left(G \cap H\right) & = \dfrac{5}{8} \times \dfrac{6}{10} + \dfrac{3}{8} \times \dfrac{4}{10} \\\\ & = \dfrac{42}{80} \\\\ & = \dfrac{21}{40} \end{aligned}$

Probability

Two cards are drawn one by one at random from a deck of $52$ playing cards. What is the probability of getting two jacks if

the first card is replaced before the second is drawn;
the first card is not replaced before the second card is drawn.

Let $A$ be the event of drawing a jack in the first draw.

Let $B$ be the event of drawing a jack in the second draw.

When the first card is replaced before the second card is drawn:

Number of elements in sample space $S = n \left(S\right) = 52$

$\because \;$ there are $4$ jacks

$n \left(A\right) = 4$, $n \left(B\right) = 4$

Event $A$ will not affect the probability of occurrence of event $B$

i.e. $\;$ $A$ and $B$ are independent events.

$\begin{aligned} \therefore \; \text{Required probability} = P \left(A \cap B\right) & = P \left(A\right) \times P \left(B\right) \\\\ & = \dfrac{n \left(A\right)}{n \left(S\right)} \times \dfrac{n \left(B\right)}{n \left(S\right)} \\\\ & = \dfrac{4}{52} \times \dfrac{4}{52} \\\\ & = \dfrac{1}{169} \end{aligned}$
When the first card is not replaced before the second card is drawn:

In the first draw, there are $4$ jacks and $52$ cards in total.

$\because \;$ The card drawn is not replaced, for the second draw there are only $3$ jacks and $51$ cards in total.

$\therefore \;$ We have $P \left(A\right) = \dfrac{4}{52}$, $P \left(B | A\right) = \dfrac{3}{51}$

By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

$\begin{aligned} \therefore \; P \left(A \cap B\right) & = P \left(A\right) \times P \left(B | A\right) \\\\ & = \dfrac{4}{52} \times \dfrac{3}{51} \\\\ & = \dfrac{1}{221} \end{aligned}$

Probability

Given $P \left(A\right) = 0.45$ and $P \left(A \cup B\right) = 0.75$

Find $P \left(B\right)$ if

$A$ and $B$ are mutually exclusive;
$A$ and $B$ are independent events;
$P \left(A | B\right) = 0.50$;
$P \left(B | A\right) = 0.50$

If $A$ and $B$ are mutually exclusive events, then $P \left(A \cap B\right) = 0$

By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

$\because \;$ $A$ and $B$ are mutually exclusive events,

$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right)$

$\therefore \;$ $P \left(B\right) = P \left(A \cup B\right) - P \left(A\right) = 0.75 - 0.45 = 0.30 = \dfrac{3}{10}$
If $A$ and $B$ are independent events, then $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$

By definition, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

$\because \;$ $A$ and $B$ are independent events,

$P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A\right) \times P \left(B\right)$

$\implies$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A\right)}$

i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.45} = \dfrac{0.30}{0.55} = \dfrac{30}{55} = \dfrac{6}{11}$
By definition, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$

i.e. $\;$ $P \left(A | B\right) \times P \left(B\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (1)$

Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$,

$P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(A | B\right) \times P \left(B\right)$

i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A\right) - P \left(A \cup B\right)}{P \left(A | B\right) - 1}$

i.e. $\;$ $P \left(B\right) = \dfrac{P \left(A \cup B\right) - P \left(A\right)}{1 - P \left(A | B\right)}$

i.e. $\;$ $P \left(B\right) = \dfrac{0.75 - 0.45}{1 - 0.50} = \dfrac{0.30}{0.50} = \dfrac{3}{5} = 0.60$
By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

i.e. $\;$ $P \left(B | A\right) \times P \left(A\right) = P \left(A \cap B\right)$ $\;\;\; \cdots \; (3)$

Now, $P \left(A \cup B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)$

i.e. $\;$ $P \left(A \cap B\right) = P \left(A\right) + P \left(B\right) - P \left(A \cup B\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ We have from equations $(3)$ and $(4)$,

$P \left(A\right) + P \left(B\right) - P \left(A \cup B\right) = P \left(B | A\right) \times P \left(A\right)$

i.e. $\;$ $P \left(B\right) = P \left(B | A\right) \times P \left(A\right) - P \left(A\right) + P \left(A \cup B\right)$

i.e. $\;$ $P \left(B\right) = 0.50 \times 0.45 - 0.45 + 0.75 = 0.525$

Probability

Given $P \left(A\right) = 0.50$, $P \left(B\right) = 0.40$ and $P \left(A \cap B\right) = 0.20$

Verify that

$P \left(A | B\right) = P \left(A\right)$
$P \left(A | \overline{B}\right) = P \left(A\right)$
$P \left(B | A\right) = P \left(B\right)$
$P \left(B | \overline{A}\right) = P \left(B\right)$

By definition, $P \left(A | B\right) = \dfrac{P \left(A \cap B\right)}{P \left(B\right)}$

i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.40} = \dfrac{1}{2} = 0.50 = P \left(A\right)$
By definition, $P \left(A | \overline{B}\right) = \dfrac{P \left(A \cap \overline{B}\right)}{P \left(\overline{B}\right)}$

Now, $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - 0.40 = 0.60$

and $P \left(A \cap \overline{B}\right) = P \left(A\right) - P \left(A \cap B\right) = 0.50 - 0.20 = 0.30$

$\therefore \;$ $P \left(A | \overline{B}\right) = \dfrac{0.30}{0.60} = \dfrac{1}{2} = 0.50 = P \left(A\right)$
By definition, $P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)}$

i.e. $\;$ $P \left(A | B\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$
By definition, $P \left(B | \overline{A}\right) = \dfrac{P \left(B \cap \overline{A}\right)}{P \left(\overline{A}\right)}$

Now, $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - 0.50 = 0.50$

and $P \left(B \cap \overline{A}\right) = P \left(B\right) - P \left(A \cap B\right) = 0.40 - 0.20 = 0.20$

$\therefore \;$ $P \left(B | \overline{A}\right) = \dfrac{0.20}{0.50} = \dfrac{2}{5} = 0.40 = P \left(B\right)$

Table 1

Table 2

Table 3

Table 4