A fair die is thrown at random. Let A be the | vedclass

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(B) The sample space of a fair die is $S = \{1, 2, 3, 4, 5, 6\}$, so $n(S) = 6$.Event $A$: Number is a non-even prime number. The only non-even prime

Vedclass · Accepted Answer

$T, F$

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(B) The sample space of a fair die is $S = \{1, 2, 3, 4, 5, 6\}$, so $n(S) = 6$.Event $A$: Number is a non-even prime number. The only non-even prime number on a die is $3$ and $5$. Thus, $A = \{3, 5\}$ and $n(A) = 2$.$P(A) = \frac{n(A)}{n(S)} = \frac{2}{6} = \frac{1}{3}$. Thus, statement $p$ is True $(T)$.Event $B$: Number is an odd number. Thus, $B = \{1, 3, 5\}$ and $n(B) = 3$.$P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.Intersection $A \cap B$: The numbers in both $A$ and $B$ are $\{3, 5\}$, so $n(A \cap B) = 2$.$P(A \cap B) = \frac{2}{6} = \frac{1}{3}$.For $A$ and $B$ to be independent, $P(A \cap B) = P(A) 	imes P(B)$ must hold.$P(A) 	imes P(B) = \frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6}$.Since $\frac{1}{3} 
eq \frac{1}{6}$, $P(A \cap B) 
eq P(A) 	imes P(B)$. Thus, statement $q$ is False $(F)$.The truth values are $T, F$.

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