If the probability that the random variable X takes values 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 2 ) is equal to [2023]

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Solution

Q.
If the probability that the random variable X takes values $x$ is given by $P (X = x) = k (x + 1) 3^{- x}, x = 0, 1, 2, 3, \dots$ , where $k$ is a constant, then $P (X \geq 2)$ is equal to [2023]

1 $\frac{7}{27}$

2 $\frac{7}{18}$

3 $\frac{11}{18}$

4 $\frac{20}{27}$

Ans.
(1)

$\sum_{x = 0}^{\infty} P (X = x) = 1 \Rightarrow k (1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + \dots \infty) = 1$

Let $S = 1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + \dots \infty \Rightarrow \frac{S}{3} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + \dots \infty$

$\Rightarrow \frac{2 S}{3} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + \dots \infty \Rightarrow \frac{2 S}{3} = \frac{1}{1 - \frac{1}{3}}$

$\Rightarrow \frac{2 S}{3} = \frac{3}{2} \Rightarrow S = \frac{9}{4} \Rightarrow k = \frac{4}{9}$

Now, $P (X \geq 2) = 1 - P (X = 0) - P (X = 1) = 1 - \frac{4}{9} (1 + \frac{2}{3}) = \frac{7}{27}$

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