Two persons P and Q are considering applying | vedclass

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(A) Given: $P(P) = 1/4$,$P(P|Q) = 1/2$,and $P(Q|P) = 1/3$. We need to find $P(\bar{P}|\bar{Q})$. First,we find $P(P \cap Q)$ using the definition of c

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(A) Given: $P(P) = 1/4$,$P(P|Q) = 1/2$,and $P(Q|P) = 1/3$. We need to find $P(\bar{P}|\bar{Q})$. First,we find $P(P \cap Q)$ using the definition of conditional probability: $P(Q|P) = \frac{P(P \cap Q)}{P(P)} \implies P(P \cap Q) = P(Q|P) 	imes P(P) = \frac{1}{3} 	imes \frac{1}{4} = \frac{1}{12}$. Next,we find $P(Q)$: $P(P|Q) = \frac{P(P \cap Q)}{P(Q)} \implies P(Q) = \frac{P(P \cap Q)}{P(P|Q)} = \frac{1/12}{1/2} = \frac{1}{6}$. Now,calculate the probabilities of the complements: $P(\bar{P}) = 1 - P(P) = 1 - 1/4 = 3/4$. $P(\bar{Q}) = 1 - P(Q) = 1 - 1/6 = 5/6$. Using De Morgan's Law,$P(\bar{P} \cup \bar{Q}) = 1 - P(P \cap Q) = 1 - 1/12 = 11/12$. Then,$P(\bar{P} \cap \bar{Q}) = P(\bar{P}) + P(\bar{Q}) - P(\bar{P} \cup \bar{Q}) = \frac{3}{4} + \frac{5}{6} - \frac{11}{12} = \frac{9 + 10 - 11}{12} = \frac{8}{12} = \frac{2}{3}$. Finally,the conditional probability is: $P(\bar{P}|\bar{Q}) = \frac{P(\bar{P} \cap \bar{Q})}{P(\bar{Q})} = \frac{2/3}{5/6} = \frac{2}{3} 	imes \frac{6}{5} = \frac{4}{5}$.

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

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