Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}, P(Y \mid X)=\frac{1}{3}$ and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?

(A) $P(X \cup Y)=\frac{2}{3}$; (B) $X$ and $Y$ are independent

(C) $X$ and $Y$ are not independent

(D) $P\left(X^{\mathrm{c}} \cap Y\right)=\frac{1}{3}$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Probability XII

Let

X

and

Y

be two events such that

P (X ∣ Y) = \frac{1}{2}, P (Y ∣ X) = \frac{1}{3}

and

P (X \cap Y) = \frac{1}{6}

. Which of the following is (are) correct?

Answer: (A), (B)

From the given:

P (X ∣ Y) = \frac{P ( X \cap Y )}{P ( Y )} = \frac{1}{2}

With

P (X \cap Y) = \frac{1}{6}

P (Y) = \frac{1}{3}

.

Similarly:

P (Y ∣ X) = \frac{P ( X \cap Y )}{P ( X )} = \frac{1}{3}

P (X) = \frac{1}{2}

.

Check independence:

P (X) P (Y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} = P (X \cap Y)

✓ (B)

Therefore (C) is false.

P (X \cup Y) = P (X) + P (Y) - P (X \cap Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}

✓ (A)

For (D):

P (X^{c} \cap Y) = P (Y) - P (X \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6} \neq = \frac{1}{3}

, so (D) is false.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · One or more correct options