A fair die is rolled. Consider events E={1,3, | vedclass

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(A) When a fair die is rolled,the sample space $S$ is $S=\{1,2,3,4,5,6\}$.Given events are $E=\{1,3,5\}, F=\{2,3\},$ and $G=\{2,3,4,5\}$.The probabili

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$P(E|G) = \frac{1}{2}, P(G|E) = \frac{2}{3}$

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(A) When a fair die is rolled,the sample space $S$ is $S=\{1,2,3,4,5,6\}$.Given events are $E=\{1,3,5\}, F=\{2,3\},$ and $G=\{2,3,4,5\}$.The probability of event $E$ is $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.The probability of event $G$ is $P(G) = \frac{n(G)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.The intersection of $E$ and $G$ is $E \cap G = \{3,5\}$.The probability of $E \cap G$ is $P(E \cap G) = \frac{n(E \cap G)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.Using the formula for conditional probability $P(A|B) = \frac{P(A \cap B)}{P(B)}$:$P(E | G) = \frac{P(E \cap G)}{P(G)} = \frac{1/3}{2/3} = \frac{1}{2}$.$P(G | E) = \frac{P(E \cap G)}{P(E)} = \frac{1/3}{1/2} = \frac{2}{3}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

$A$ fair die is rolled. Consider events $E=\{1,3,5\}, F=\{2,3\},$ and $G=\{2,3,4,5\}$. Find $P(E | G)$ and $P(G | E)$.

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