Let X and Y be two events such that PX=1/3, PXY=1/2 and PY|X=2/5. Then

Correct answer: (A) PY=4/15; (B) P X ¯ | Y=1/2. Using formula PX | Y=(P( X∩Y ))/(P( Y )) and PY | X=(P( Y∩X ))/(P( X )) PX=1/3;(PX∩Y)/PY=1/2;(PY∩X)/PX=2/5 From this information, we get PX∩Y=2/15;PY=4/15 Using formula P(X∪Y)=P(X)+P(Y)−P(X∩Y) ∴ PX∪Y=1/3+4/15-2/15=7/15 P X ¯ | Y=(PX-∩Y)/PY=(PY-PX∩Y)/PY ⇒ P X ¯ | Y=1-2/154/15=1/2

Q20 / 54 · JEE Advanced 2017 (Paper 1) · Mathematics

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

Correct answer
A
$P (Y) = \frac{4}{15}$
Correct answer
B
$P (\bar{X} | Y) = \frac{1}{2}$
C
$P (X \cap Y) = \frac{1}{5}$
D
$P (X \cup Y) = \frac{2}{5}$

Solution

Using formula

P (X | Y) = \frac{P (X \cap Y)}{P (Y)}

and

P (Y | X) = \frac{P (Y \cap X)}{P (X)}

P (X) = \frac{1}{3}; \frac{P (X \cap Y)}{P (Y)} = \frac{1}{2}; \frac{P (Y \cap X)}{P (X)} = \frac{2}{5}

From this information, we get

P (X \cap Y) = \frac{2}{15}; P (Y) = \frac{4}{15}

Using formula

P (X \cup Y) = P (X) + P (Y) - P (X \cap Y)

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

P (\bar{X} | Y) = \frac{P (\bar{X} \cap Y)}{P (Y)} = \frac{P (Y) - P (X \cap Y)}{P (Y)}

\Rightarrow P (\bar{X} | Y) = 1 - \frac{\frac{2}{15}}{\frac{4}{15}} = \frac{1}{2}

Hence, the correct answer is options: 'A', 'B'.

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Solution

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