Question

A pair of dice is thrown $4$ times. If getting a doublet is considered as a success,
(1) find the probability of getting a doublet.
(2) hence, find the probability of two successes.

Answer 1

(a)

Possible outcomes is $S=\left\{\right(1,1),(1,2)....(1,6),(2,1),....(2,6),(3,1),....(3,6),.......(6,6\left)\right\}$

Total no. of outcomes is $n\left(S\right)=36$

Outcomes with doublets is $A=\left\{\right(1,1),(2,2),(3,3),(4,4),(5,5),(6,6\left)\right\}$

Then probability of getting a doublet is $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{6}{36}=\frac{1}{6}=a$

(b)

Let $B$ an event of not getting successes

Then $P\left(B\right)=1-P\left(A\right)=1-\frac{1}{6}=\frac{5}{6}=b$

Let $X$ be the number of success.

If we get two successes, then the no. of failures is $2$ as a pair of dice is thrown $4$ times.

Then $P(X=2,2$ successes, $2$ failures$)={}^4{C}_2{a}^2{b}^2$

$=6\times \frac{1}{36}\times \frac{25}{36}=\frac{25}{216}$