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Question

(D) Let $P(A) = \frac{2}{3}$ and $P(B) = \frac{3}{4}$.Assuming the events are independent,the probability of passing both is $P(A \cap B) = P(A) 	ime

Vedclass · Accepted Answer

$\frac{6}{11}$

Vedclass · Answer

(D) Let $P(A) = \frac{2}{3}$ and $P(B) = \frac{3}{4}$.Assuming the events are independent,the probability of passing both is $P(A \cap B) = P(A) 	imes P(B) = \frac{2}{3} 	imes \frac{3}{4} = \frac{1}{2}$.The probability of passing at least one is $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{3} + \frac{3}{4} - \frac{1}{2} = \frac{8+9-6}{12} = \frac{11}{12}$.We need to find the conditional probability $P(A \cap B | A \cup B) = \frac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} = \frac{P(A \cap B)}{P(A \cup B)}$.Substituting the values,we get $\frac{1/2}{11/12} = \frac{1}{2} 	imes \frac{12}{11} = \frac{6}{11}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$2k$	$k$

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

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