Q.2.25 - Problems - A First Course in Probability

Question

A pair of dice is rolled until a sum of either $5$or $7$ appears. Find the probability that a $5$ occurs first.

 Hint: Let $E_n$ denote the event that a $5$ occurs on the nth roll and no $5 or 7$ occurs on the first $n-1$ rolls. Compute $P\left(E_n\right)$ and argue that $\sum _{n=1}^{\infty }P\left(E_n\right)$is the desired probability 

Step-by-Step Solution

Answer

Probability that $5$ occurs before $7 is 0.4$

Step-1 Given Information

Given in the question that a pair of dice is given. the dice are 

rolled until a total of $5 or 7$occurs

The sample space is total number of possible combination of dice,i.e,$36$. we have to Find the probability that a $5$ occurs first.

Step-2 Explanation

Formula used:$P\left(A\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$P\left(A\right)$ is the probability of an event $A$.

$n\left(E\right)$ is the favorable outcomes.

$n\left(S\right)$is the total number of events in the sample space.

$5$can occur in $4$ ways. Hence, probability that the total is $5$ is equal to $\frac{1}{9}$ in each turn.

$7$can occur in $6$ways. Hence, the probability that the total is $7$ is equal to $\frac{1}{6}$ in each turn.

Probability that total of either $5 or 7$does not occur in a given turn $=1-\frac{1}{6}-\frac{1}{9}=\frac{13}{18}$

For total of $5$ to occur on $n^{th}$urn, neither $5 nor 7$ could have occurred until  $(n-1)$turns

Therefore, if $5$ occurs on $n^{th}$ turn, the probability is ${\left(\frac{13}{18}\right)}^{n-1}\times \frac{1}{9}$

Summing up over all $n$gives probability that $5$ occurs first as

$\sum _{n=1}^{\infty }\frac{1}{9}{\left(\frac{13}{18}\right)}^{n-1}$$=\frac{{\displaystyle \frac{1}{9}}}{1-{\displaystyle \frac{13}{18}}}=\frac{2}{5}$ 

Probability that $5$occurs before $7$ , therefore, is $0.4$