JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 90

A random variable $X$ has the following probability distribution:
$\begin{array}{|c|c|c|c|c|c|}\hline X&0&1&2&3&4\\\hline P(X)&k&2k&4k&6k&8k\\\hline\end{array}$
The value of $P(1<X<4\mid X\le2)$ is equal to :

\dfrac47

\dfrac23

\dfrac37

\dfrac45

Answer and worked solution

Official answer: A

Normalise. $k(1+2+4+6+8)=21k=1$ .

Conditional probability. $P(X\le2)=7k$ and $P(1<X<4,\ X\le2)=P(X=2)=4k$ .

Value $\dfrac47$ .

(A)

Topics:Conditional probability Random variables and distributions

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 12