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(B) Let $A$ be the event that a sum of $5$ occurs,$B$ be the event that a sum of $7$ occurs,and $C$ be the event that neither a sum of $5$ nor a sum o

Vedclass · Accepted Answer

$\frac{2}{5}$

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(B) Let $A$ be the event that a sum of $5$ occurs,$B$ be the event that a sum of $7$ occurs,and $C$ be the event that neither a sum of $5$ nor a sum of $7$ occurs.For a pair of dice,the total number of outcomes is $36$.The outcomes for sum $5$ are $(1,4), (2,3), (3,2), (4,1)$,so $P(A) = \frac{4}{36} = \frac{1}{9}$.The outcomes for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$,so $P(B) = \frac{6}{36} = \frac{1}{6}$.The probability of neither $5$ nor $7$ is $P(C) = 1 - (P(A) + P(B)) = 1 - (\frac{1}{9} + \frac{1}{6}) = 1 - (\frac{2+3}{18}) = 1 - \frac{5}{18} = \frac{13}{18}$.The probability that $A$ occurs before $B$ is given by the sum of the infinite geometric series:$P = P(A) + P(C)P(A) + P(C)^2 P(A) + \dots = \frac{P(A)}{1 - P(C)}$.Substituting the values: $P = \frac{\frac{1}{9}}{1 - \frac{13}{18}} = \frac{\frac{1}{9}}{\frac{5}{18}} = \frac{1}{9} 	imes \frac{18}{5} = \frac{2}{5}$.Thus,the correct option is $B$.

$A$ pair of fair dice is rolled together till a sum of either $5$ or $7$ is obtained. Then the probability that $5$ comes before $7$ is

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