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

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Probability

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\dots

, where k is a constant, then

P (X \geq 3)

is equal to

Answer: (D)

For a valid distribution

\sum_{x = 0}^{\infty} k (x + 1) 3^{- x} = 1

. Using

\sum_{x = 0}^{\infty} (x + 1) t^{x} = \frac{1}{( 1 - t ) ^{2}}

with

t = \frac{1}{3}

, the sum is

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

, so

k = \frac{4}{9}

. Then

P (X = 0) = \frac{4}{9}

P (X = 1) = \frac{4}{9} \cdot \frac{2}{3} = \frac{8}{27}

and

P (X = 2) = \frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27}

, with total

\frac{24}{27} = \frac{8}{9}

. Therefore

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

, which is

\frac{1}{9}

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option