JEE Main 2026 (Online) 28th January Evening ShiftQuestion 1
The probability distribution of a random variable $X$ is given below :
$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline X&4k&\frac{30}{7}k&\frac{32}{7}k&\frac{34}{7}k&\frac{36}{7}k&\frac{38}{7}k&\frac{40}{7}k&6k\\\hline P(X)&\frac2{15}&\frac1{15}&\frac2{15}&\frac15&\frac1{15}&\frac2{15}&\frac15&\frac1{15}\\\hline\end{array}$
If $E(X)=\frac{263}{15}$, then $P(X<20)$ is equal to :
Probability
Solution
$E(X)=\frac k{15}\left(8+\frac{30+64+102+36+76}7+\frac{120}7+6\right)=\frac{526k}{105}=\frac{263}{15}$, so $k=\frac72$. Values become $14,15,16,17,18,19,20,21$; $P(X<20)=\frac{2+1+2+3+1+2}{15}=\frac{11}{15}$.