Two dice are thrown. What is the probability | vedclass

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(B) Let $A$ be the event that the sum of the numbers is $11$. The possible outcomes for a sum of $11$ are $(5, 6)$ and $(6, 5)$.Let $B$ be the event t

Vedclass · Accepted Answer

$\frac{1}{6}$

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(B) Let $A$ be the event that the sum of the numbers is $11$. The possible outcomes for a sum of $11$ are $(5, 6)$ and $(6, 5)$.Let $B$ be the event that $5$ appears on the first die. The possible outcomes for $B$ are $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$.We need to find the conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$.The intersection $A \cap B$ is the event where the sum is $11$ $AND$ $5$ is on the first die,which is only the outcome $(5, 6)$.Thus,$n(A \cap B) = 1$ and $n(B) = 6$.The required probability is $\frac{n(A \cap B)}{n(B)} = \frac{1}{6}$.

Two dice are thrown. What is the probability that the sum of the numbers appearing on the two dice is $11$,given that $5$ appears on the first die?

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