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

(A) $P(Y)=\frac{4}{15}$; (B) $P\left(X^{\prime} \mid Y\right)=\frac{1}{2}$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Probability XII

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

Answer: (A), (B)

From

P (Y ∣ X) = \frac{2}{5}

and

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

, we get

P (X \cap Y) = P (Y ∣ X) \cdot P (X) = \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}

. From

P (X ∣ Y) = \frac{1}{2}

, we have

P (X \cap Y) = P (X ∣ Y) \cdot P (Y)

, so

\frac{2}{15} = \frac{1}{2} \cdot P (Y)

, giving

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

(option A correct). Also

P (X^{'} ∣ Y) = 1 - P (X ∣ Y) = 1 - \frac{1}{2} = \frac{1}{2}

(option B correct). Option C states

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

, but we found

\frac{2}{15}

, so C is incorrect.

P (X \cup Y) = P (X) + P (Y) - P (X \cap Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{5}{15} + \frac{2}{15} = \frac{7}{15}

, so D is incorrect.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options