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

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Probability

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

Answer: (A)

Since

\sum_{x = 0}^{\infty} P (X = x) = 1

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

Let

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

Using standard series formulas:

x = 0 \sum \infty 3^{- x} = \frac{1}{1 - 1/3} = \frac{3}{2}

x = 0 \sum \infty x \cdot 3^{- x} = \frac{1/3}{( 1 - 1/3 ) ^{2}} = \frac{3}{4}

S = \frac{3}{4} + \frac{3}{2} = \frac{9}{4}

, giving

k = \frac{4}{9}

.

Now:

P (X \geq 2) = 1 - P (X = 0) - P (X = 1)

= 1 - k (1) - k (2) \cdot \frac{1}{3} = 1 - k - \frac{2 k}{3} = 1 - \frac{5 k}{3}

= 1 - \frac{5}{3} \cdot \frac{4}{9} = 1 - \frac{20}{27} = \frac{7}{27}

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option