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Question

(B) Let $A$ be the event that the sum is $7$ and $B$ be the event that the sum is $11$. The total number of outcomes when two dice are thrown is $36$.

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(B) Let $A$ be the event that the sum is $7$ and $B$ be the event that the sum is $11$. The total number of outcomes when two dice are thrown is $36$. The number of outcomes for sum $7$ is $n(A) = \{(1,6), (6,1), (2,5), (5,2), (3,4), (4,3)\} = 6$. The number of outcomes for sum $11$ is $n(B) = \{(5,6), (6,5)\} = 2$. The probabilities are $P(A) = \frac{6}{36} = \frac{1}{6}$ and $P(B) = \frac{2}{36} = \frac{1}{18}$. We are interested in the event that $A$ occurs before $B$. This can happen in the first trial,or in the third trial (if the first two trials result in neither $A$ nor $B$),and so on. Let $p = P(A) = \frac{1}{6}$ and $q = P(B) = \frac{1}{18}$. The probability of neither $A$ nor $B$ occurring is $r = 1 - (p + q) = 1 - (\frac{1}{6} + \frac{1}{18}) = 1 - \frac{4}{18} = 1 - \frac{2}{9} = \frac{7}{9}$. The probability that $A$ occurs before $B$ is given by the sum of the infinite geometric series: $P = p + rp + r^2p + \dots = \frac{p}{1-r} = \frac{1/6}{1 - 7/9} = \frac{1/6}{2/9} = \frac{1}{6} 	imes \frac{9}{2} = \frac{3}{4}$.

If two unbiased six-faced dice are thrown simultaneously until a sum of either $7$ or $11$ occurs,then the probability that $7$ comes before $11$ is

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