Q31 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
The probability distribution of a random variable $X$ is given below :
\(\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \dfrac{30}{7}k & \dfrac{32}{7}k & \dfrac{34}{7}k & \dfrac{36}{7}k & \dfrac{38}{7}k & \dfrac{40}{7}k & 6k \\ \hline P(X) & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} \\ \hline \end{array}\)
If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to :
\(\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \dfrac{30}{7}k & \dfrac{32}{7}k & \dfrac{34}{7}k & \dfrac{36}{7}k & \dfrac{38}{7}k & \dfrac{40}{7}k & 6k \\ \hline P(X) & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} \\ \hline \end{array}\)
If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to :
- Correct answerA$\frac{11}{15}$
- B$\frac{3}{5}$
- C$\frac{14}{15}$
- D$\frac{8}{15}$
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
$P(X)$: $\dfrac{2}{15}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$
$P(X < 20) = \sum_{X=14}^{19} P(X)$
$= \dfrac{2}{15} + \dfrac{1}{15} + \dfrac{2}{15} + \dfrac{1}{5} + \dfrac{1}{15} + \dfrac{2}{15} = \dfrac{11}{15}$
$\Rightarrow k = \dfrac{7}{2}$
Substituting $k = \dfrac{7}{2}$, the values of $X$ become:
$X$: 14, 15, 16, 17, 18, 19, 20, 21
$P(X)$: $\dfrac{2}{15}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$
$P(X < 20) = \sum_{X=14}^{19} P(X)$
$= \dfrac{2}{15} + \dfrac{1}{15} + \dfrac{2}{15} + \dfrac{1}{5} + \dfrac{1}{15} + \dfrac{2}{15} = \dfrac{11}{15}$
Hence, the correct answer is option: 'A'.