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 , where k is a constant, then P(X ≥ 3) is equal to

Correct answer: D. (1)/(9) ∑_ x=0 ^ ∞ k(x+1) 3^ -x =1 1 k =1+(2)/(3)+(3)/(3^2)+(4)/(3^3)+ ...(i) 1 3 k =(1)/(3)+(2)/(3^2)+(3)/(3^3)+ ...(ii) (i)- (ii) 1 k - 1 3 k =1+(1)/(3)+(1)/(3^2)+ aligned & k=(4)/(9) & P(x ≥ 3)=1-P(x=0)-P(x=1)-P(x=2) & =1-k(1+(2)/(3)+(3)/(9))=(1)/(9) aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Probability

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Probability

If the probability that the random variable X takes the value x is...

If the probability that the random variable X takes the value

x

is given by

P(X=x)=k(x+1) 3^{-x}

\mathrm{x}=0,1,2,3 \ldots \ldots

, where k is a constant, then

\mathrm{P}(\mathrm{X} \geq 3)

is equal to

Options

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

Correct answer

D. $\frac{1}{9}$

Step-by-step solution

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

\Rightarrow \frac{1}{\mathrm{k}}=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+

...(i)

\frac{1}{3 \mathrm{k}}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots

...(ii)
(i)- (ii)

\Rightarrow \frac{1}{\mathrm{k}}-\frac{1}{3 \mathrm{k}}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots

\begin{aligned} & \Rightarrow \mathrm{k}=\frac{4}{9} \\ & \mathrm{P}(\mathrm{x} \geq 3)=1-\mathrm{P}(\mathrm{x}=0)-\mathrm{P}(\mathrm{x}=1)-\mathrm{P}(\mathrm{x}=2) \\ & =1-\mathrm{k}\left(1+\frac{2}{3}+\frac{3}{9}\right)=\frac{1}{9}\end{aligned}

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