The probability distribution of a random variable X is given below : | X | 4 k | $\frac{30}{7} \mathrm{k}$ | $\frac{32}{7} \mathrm{k}$ | $\frac{34}{7} \mathrm{k}$ | $\frac{36}{7} \mathrm{k}$ | $\frac{38}{7} \mathrm{k}$ | $\frac{40}{7} \mathrm{k}$ | 6 k | | P(X) | $\frac{2}{15}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{1}{5}$ | $\frac{1}{15}$ | If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Probability

The probability distribution of a random variable X is given below :

X	4 k	$\frac{30}{7} k$	$\frac{32}{7} k$	$\frac{34}{7} k$	$\frac{36}{7} k$	$\frac{38}{7} k$	$\frac{40}{7} k$	6 k
P(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

E (X) = \frac{263}{15}

, then

P (X < 20)

is equal to :

Answer: (C)

The listed probabilities already total

1

. Writing each

X

-value over

7

, the expectation is

E (X) = \frac{1}{105} (28 k \cdot 2 + 30 k \cdot 1 + 32 k \cdot 2 + 34 k \cdot 3 + 36 k \cdot 1 + 38 k \cdot 2 + 40 k \cdot 3 + 42 k \cdot 1) = \frac{526 k}{105} .

Setting

\frac{526 k}{105} = \frac{263}{15}

gives

k = \frac{7}{2}

. The values of

X

then become

14, 15, 16, 17, 18, 19, 20, 21

, so

P (X < 20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{2}{15} = \frac{11}{15} .

Hence

P (X < 20) = \frac{11}{15}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option