The probability distribution of a random variable X is given below : X 4K 30/7 k 32/7 k 34/7 k 36/7 k 38/7 k 40/7 k 6K --- --- --- --- --- --- --- --- --- p(X) 2/15 1/15 2/15 1/5 1/15 2/15 1/5 1/15 If E ( X )=263/15 , then P ( X <20) is equal to :

Mathematics · Probability

JEE Main 2026 — 28 January, Evening Shift — Question 4

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

X	4K	$\frac{30}{7} \mathrm{k}$	$\frac{32}{7} \mathrm{k}$	$\frac{34}{7} \mathrm{k}$	$\frac{36}{7} \mathrm{k}$	$\frac{38}{7} \mathrm{k}$	$\frac{40}{7} \mathrm{k}$	6K
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 :

Option A:
$\frac{3}{5}$
Option B:
$\frac{8}{15}$
Option C:
$\frac{11}{15}$
Correct
Option D:
$\frac{14}{15}$

Answer: C

Step-by-step solution

$\mathrm{E}(\mathrm{X})=\sum \mathrm{X}_{\mathrm{i}} \mathrm{P}\left(\mathrm{X}_{\mathrm{i}}\right)=\frac{526 \mathrm{k}}{15 \times 7}=\frac{263}{15}$

$\Rightarrow \mathrm{k}=\frac{7}{2}$

X	14	15	16	17	18	19	20	21
p(X)	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$\mathrm{P}(\mathrm{X}<20)=\sum_{\mathrm{X}=14}^{19} \mathrm{P}(\mathrm{X})=\frac{11}{15}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Probability
Topic: Random Variables, Binomial & Poission Distribution