JEE Main 2026 Jan 28 Shift 2, Mathematics Q4: Introduction To Probability

JEE Main2026Jan 28, Shift 2Mathematics

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

If $E (X) = \frac{263}{15}$ , then $P (X < 20)$ is equal to :

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

Solution

Compute $E (X) = \sum X_{i} P (X_{i})$ . Collecting all terms over a common denominator of $15 \times 7 = 105$ , the numerator simplifies to $526 k$ , giving $E (X) = \frac{526 k}{105} = \frac{263}{15}$ , so $k = \frac{7}{2}$ .

Substituting $k$ back, the X values become $14, 15, 16, 17, 18, 19, 20, 21$ with their respective probabilities.

$P (X < 20) = P (14) + P (15) + P (16) + P (17) + P (18) + P (19) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{1}{5} + \frac{1}{15} + \frac{2}{15} = \frac{11}{15}$ .

Concept behind this question

Introduction To ProbabilityNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 28 Shift 2, Mathematics Q4: Introduction To Probability

Concept behind this question

More previous year questions on Introduction To Probability

Practice more Mathematics