91Ó°ÊÓ

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

rolled until a total of $5or7$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.

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 $5or7$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 $5nor7$could have occurred until $(n-1)$turns

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

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

$\underset{n=1}{\overset{∞}{∑}}\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 $0.4$