JEE 2023 — Mathematics PYQ

As, we know that sum of all the probabilities $=1$

So, ${\sum }_{x=1}^{\infty }P(X=x)=1$

$\Rightarrow k[1+2\cdot 3^{-1}+3\cdot 3^{-2}+\dots \infty ]=1$

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

$\Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\dots \infty$

On subtracting, we get

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

$\Rightarrow \frac{2S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}$

$\Rightarrow \frac{2S}{3}=\frac{3}{2}$

$\Rightarrow S=\frac{9}{4}$

So, $k\times \frac{9}{4}=1\Rightarrow k=\frac{4}{9}$

Now, P(X \ge 2) = 1 - P(X < 2)

$=1-P(X=0)-P(X=1)$

$=1-\frac{4}{9}(1)-\frac{4}{9}\times \frac{2}{3}$

$=1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27}$

Sum of probabilities $=1$