Let X and Y be two events such that P(X Y)=(1)/(2) , P(Y / X)=(1)/(3) and P(X Y)=(1)/(6) . Which of the following is (are) correct?

Correct answer: A. P(X Y)=(2)/(3) P(X / Y)=(P(X Y))/(P(Y)) (1)/(2)=(1 / 6)/(P(Y)) P(Y)=(1)/(3) Similarly, P(Y / X)=(P(X Y))/(P(X)) (1)/(3)=(1 / 6)/(P(X)) P(X)=(1)/(2) (a) P(X Y)=P(X)+P(Y)-P(X Y)=(1)/(2)+(1)/(3)-(1)/(6)=(2)/(3) (a) is true. (b) P(X Y)=P(X) P(Y) X and Y are independent events. (b) is true. But (c) is not true. (d) P(X^ C Y)=P(X^ C ) × P(Y)=(1)/(2) × (1)/(3)=(1)/(6) (d) is not true.

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Mathematics
Chapter: Probability

JEE Advanced 2012 · Paper 2 · Mathematics · Probability

Let X and Y be two events such that P(X Y)=(1)/(2) , P(Y /...

Let

X

and

Y

be two events such that

P(X \mid 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?

Options

A $P(X \cup Y)=\frac{2}{3}$
B $X$ and $Y$ are independent
C $X$ and $Y$ are not independent
DP $\left(X^{c} \cap Y\right)=\frac{1}{3}$

Correct answer

A. $P(X \cup Y)=\frac{2}{3}$

Step-by-step solution

\because P(X / Y)=\frac{P(X \cap Y)}{P(Y)} \Rightarrow \frac{1}{2}=\frac{1 / 6}{P(Y)} \Rightarrow P(Y)=\frac{1}{3}

Similarly,

\mathrm{P}(Y / X)=\frac{P(X \cap Y)}{P(X)}

\Rightarrow \frac{1}{3}=\frac{1 / 6}{P(X)} \Rightarrow P(X)=\frac{1}{2}

(a)

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

\therefore

(a) is true. (b)

\because P(X \cap Y)=P(X) P(Y)

\Rightarrow X

and

Y

are independent events.

\therefore

(b) is true. But (c) is not true. (d)

\mathrm{P}\left(X^{C} \cap Y\right)=P\left(X^{C}\right) \times P(Y)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}

\therefore

(d) is not true.

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