JEE Main 2025 Apr 3 Shift 2, Mathematics Q13: Introduction To Probability

JEE Main2025Apr 3, Shift 2Mathematics

If the probability that the random variable $X$ takes the value $x$ is given by $P (X = x) = k (x + 1) 3^{- x}$ , $x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 3)$ is equal to

A
$\frac{7}{27}$
B
$\frac{4}{9}$
C
$\frac{8}{27}$
D
$\frac{1}{9}$

Solution

The total probability is $1$ :

$x = 0 \sum \infty k (x + 1) 3^{- x} = 1$ .

Use the closed form $x = 0 \sum \infty (x + 1) r^{x} = \frac{1}{( 1 - r ) ^{2}}$ for $∣ r ∣ < 1$ , with $r = 1/3$ :

$\frac{1}{( 1 - 1/3 ) ^{2}} = \frac{1}{( 2/3 ) ^{2}} = \frac{9}{4}$ .

So $k \cdot \frac{9}{4} = 1 \Rightarrow k = \frac{4}{9}$ .

Compute $P (X \geq 3) = 1 - P (X = 0) - P (X = 1) - P (X = 2)$ :

$P (X = 0) = k \cdot 1 = \frac{4}{9}$ , $P (X = 1) = k \cdot 2 \cdot \frac{1}{3} = \frac{8}{27}$ , $P (X = 2) = k \cdot 3 \cdot \frac{1}{9} = \frac{4}{27}$ .

Sum $= \frac{12}{27} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27} = \frac{8}{9}$ .

$P (X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9}$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q13: Introduction To Probability

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