The probability distribution of a random variable $X$ is given below : $ \begin{array}{|c|c|c|c|c|c|c|c|c|}…

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) & \frac{2}{15} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} \\ \hline \end{array} $ If $E(X)=\frac{263}{15}$, then $P(X<20)$ is equal to :
  1. $\frac{11}{15}$
  2. $\frac{3}{5}$
  3. $\frac{14}{15}$
  4. $\frac{8}{15}$

Solution

$E(X) = \sum X_i P(X_i) = \dfrac{526k}{15 \times 7} = \dfrac{263}{15}$
$\Rightarrow k = \dfrac{7}{2}$
Substituting $k = \dfrac{7}{2}$, the values of $X$ become:
$X$: 14, 15, 16, 17, 18, 19, 20, 21
$P(X)$: $\dfrac{2}{15}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$
$P(X < 20) = \sum_{X=14}^{19} P(X)$
$= \dfrac{2}{15} + \dfrac{1}{15} + \dfrac{2}{15} + \dfrac{1}{5} + \dfrac{1}{15} + \dfrac{2}{15} = \dfrac{11}{15}$

Asked in: JEE Main 2026 (28 Jan Shift 2)

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