JEE Main 2020 (Online) 9th January Evening ShiftQuestion 1
A random variable $X$ has the following probability distribution : $X$: $1,2,3,4,5$ and $P(X)$: $K^2,2K,K,2K,5K^2$. Then $P(X>2)$ is equal to :
Probability
Solution
$K^2+2K+K+2K+5K^2=1\Rightarrow6K^2+5K-1=0\Rightarrow K=\frac16$ ($K>0$). $P(X>2)=K+2K+5K^2=\frac12+\frac5{36}=\frac{23}{36}$. Option C.