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

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(A) The sample space of rolling a fair die is $S = \{1, 2, 3, 4, 5, 6\}$,so $n(S) = 6$.Given events are $E = \{1, 3, 5\}$,$F = \{2, 3\}$,and $G = \{2,

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$3/4$ and $1/4$

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(A) The sample space of rolling a fair die is $S = \{1, 2, 3, 4, 5, 6\}$,so $n(S) = 6$.Given events are $E = \{1, 3, 5\}$,$F = \{2, 3\}$,and $G = \{2, 3, 4, 5\}$.The probability of event $G$ is $P(G) = \frac{n(G)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.First,we find $(E \cup F) = \{1, 2, 3, 5\}$.Then,$(E \cup F) \cap G = \{1, 2, 3, 5\} \cap \{2, 3, 4, 5\} = \{2, 3, 5\}$.Thus,$P((E \cup F) \cap G) = \frac{n(\{2, 3, 5\})}{n(S)} = \frac{3}{6} = \frac{1}{2}$.Using the conditional probability formula: $P((E \cup F) | G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{1/2}{2/3} = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$.Next,we find $(E \cap F) = \{3\}$.Then,$(E \cap F) \cap G = \{3\} \cap \{2, 3, 4, 5\} = \{3\}$.Thus,$P((E \cap F) \cap G) = \frac{n(\{3\})}{n(S)} = \frac{1}{6}$.Using the conditional probability formula: $P((E \cap F) | G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{1/6}{2/3} = \frac{1}{6} 	imes \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$.Therefore,the required probabilities are $3/4$ and $1/4$.

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

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