If the probability that the random variable X takes the value x is given by , x = 0, 1, 2, 3, ..., where k is a constant, then , is equal to [2025]

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Solution

Q.
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 $P (X \geq 3)$ , is equal to [2025]

1 $\frac{8}{27}$

2 $\frac{4}{9}$

3 $\frac{1}{9}$

4 $\frac{7}{27}$

Ans.
(3)

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

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

$\Rightarrow \frac{1}{k} = 1 + \frac{2}{3} + \frac{3}{3^{2}} + \frac{4}{3^{3}} + . . .$ ... (i)

$\Rightarrow \frac{1}{3 k} = \frac{1}{3} + \frac{2}{3^{2}} + \frac{3}{3^{3}} + . . .$ ... (ii)

Subtracting equation (ii) from (i), we get

$\frac{1}{k} - \frac{1}{3 k} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . .$

$\Rightarrow \frac{2}{3 k} = \frac{1}{1 - \frac{1}{3}} \Rightarrow k = \frac{4}{9}$

Now, $P (X \geq 3) = 1 - [P (X = 0) + P (X = 1) + P (X = 2)]$

$= 1 - k (1 + \frac{2}{3} + \frac{3}{9}) = 1 - \frac{4}{9} (2) = 1 - \frac{8}{9} = \frac{1}{9}$ .

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