When three coins are tossed simultaneously, what is the probability of getting
- exactly two heads
- exactly two tails
- at least two tails
- at most one head
When three coins are tossed,
sample space $= S = \left\{HHH, \; HHT, \; HTH, \; HTT, \; THH, \; THT, \; TTH, \; TTT \right\}$
$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 8$
Let $A =$ event of getting exactly two heads
Then $\;$ $A = \left\{HHT, \; HTH, \; THH \right\}$
$\therefore \;$ Number of elements in $A = n \left(A\right) = 3$
$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{3}{8}$
Let $B =$ event of getting exactly two tails
Then $\;$ $B = \left\{HTT, \; THT, \; TTH \right\}$
$\therefore \;$ Number of elements in $B = n \left(B\right) = 3$
$\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{3}{8}$
Let $C =$ event of getting at least two tails (i.e. two or more tails)
Then $\;$ $C = \left\{HTT, \; THT, \; TTH, \; TTT \right\}$
$\therefore \;$ Number of elements in $C = n \left(C\right) = 4$
$\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$
Let $D =$ event of getting at most one head (i.e. one or no head)
Then $\;$ $D = \left\{HTT, \; THT, \; TTH, \; TTT \right\}$
$\therefore \;$ Number of elements in $D = n \left(D\right) = 4$
$\therefore \;$ Probability of event $D = P \left(D\right) = \dfrac{n \left(D\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$