Q41 / 90 · JEE Main 2022 (24 Jun Shift 2) · Mathematics

A random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$

$P (X)$ $k$ $2 k$ $4 k$ $6 k$ $8 k$

The value of $P (\frac{1 < x < 4}{x \leq 2})$ is equal to

Solution

To find $P (\frac{1 < x < 4}{x \leq 2})$ or $P (\frac{A}{B})$

We know that

$P (\frac{A}{B}) = \frac{P (A \cap B)}{P (B)}$

Given

$x$	$0$	$1$	$2$	$3$	$4$
$P (x)$	$k$	$2 k$	$4 k$	$6 k$	$8 k$

$P (A) = \{2, 3\}$

$P (B) = \{0, 1, 2\}$

$P (A \cap B) = P (x = 2)$

$P (B) = P (x = 0) + P (x = 1) + P (x = 2)$

So $P (\frac{A}{B}) = \frac{P (x = 2)}{P (x = 0) + P (x = 1) + P (x = 2)}$

$= \frac{4 k}{k + 2 k + 4 k} = \frac{4 k}{7 k} = \frac{4}{7}$

Hence option $A$ is correct.

Hence, the correct answer is option: 'A'.