If two unbiased dice are rolled simultaneousl | vedclass

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Question

(B) Let $E_1$ be the event that the sum is $7$ and $E_2$ be the event that the sum is $11$.The total number of outcomes when two dice are rolled is $6

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(B) Let $E_1$ be the event that the sum is $7$ and $E_2$ be the event that the sum is $11$.The total number of outcomes when two dice are rolled is $6 	imes 6 = 36$.The outcomes for sum $7$ are $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$, so $P(E_1) = \frac{6}{36} = \frac{1}{6}$.The outcomes for sum $11$ are $(5,6), (6,5)$, so $P(E_2) = \frac{2}{36} = \frac{1}{18}$.We are interested in the event that $E_1$ occurs before $E_2$. This means in any trial, we ignore outcomes where the sum is neither $7$ nor $11$.The probability of getting either $7$ or $11$ is $P(E_1 \cup E_2) = P(E_1) + P(E_2) = \frac{6}{36} + \frac{2}{36} = \frac{8}{36} = \frac{2}{9}$.Given that the sum is either $7$ or $11$, the conditional probability that the sum is $7$ is $P(E_1 | E_1 \cup E_2) = \frac{P(E_1)}{P(E_1) + P(E_2)} = \frac{6/36}{8/36} = \frac{6}{8} = \frac{3}{4}$.Thus, the probability that $7$ appears before $11$ is $\frac{3}{4}$.

If two unbiased dice are rolled simultaneously until a sum of the numbers appearing on these dice is either $7$ or $11$, then the probability that $7$ comes before $11$ is:

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