JEE Advanced 2017 · Paper 1 · Mathematics Q7

MathematicsMultiple correct+4 / -2Question 43 of 54

Let $X$ and $Y$ be two events such that $P(X) = \frac{1}{3}$ , $P(X|Y) = \frac{1}{2}$ and $P(Y|X) = \frac{2}{5}$ . Then

P(Y) = \frac{4}{15}

P(X'|Y) = \frac{1}{2}

P(X \cap Y) = \frac{1}{5}

P(X \cup Y) = \frac{2}{5}

Answer and worked solution

Official answer: A, B

Intersection. $P(X\cap Y)=P(Y|X)P(X)=\tfrac25\cdot\tfrac13=\tfrac2{15}$ . So (C) is false.

$P(Y)$ . $P(Y)=\dfrac{P(X\cap Y)}{P(X|Y)}=\dfrac{2/15}{1/2}=\dfrac4{15}$ . (A) is true.

$P(X'|Y)$ . $1-P(X|Y)=\tfrac12$ . (B) is true.

Union. $P(X\cup Y)=\tfrac13+\tfrac4{15}-\tfrac2{15}=\tfrac7{15}$ . (D) is false.

(A), (B)

Topics:Conditional probability

JEE Advanced 2017 · Paper 1, Mathematics question 7