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.