JEE Main 2025 — Probability Question with Solution

To find $P(X\ge 3)$, we first determine the constant $k$ using the total probability for $X$.

The probability $P(X=x)$ is given by:

$P(X=x)=k(x+1)\cdot 3^{-x},x=0,1,2,3,\dots$

The total probability must equal 1:

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

Calculating that series:

$s=\frac{k}{3^0}+\frac{2k}{3}+\frac{3k}{3^2}+\dots$

Therefore, dividing the series by 3:

$\frac{s}{3}=\frac{k}{3}+\frac{2k}{3^2}+\dots$

Subtracting these:

$s-\frac{s}{3}=k+\frac{k}{3}+\frac{k}{3^2}+\dots$

The resulting series is a geometric series:

$\frac{2s}{3}=k\left(1+\frac{1}{3}+\frac{1}{3^2}+\dots \right)$

The sum of the infinite geometric series is:

$\frac{2s}{3}=k\cdot \frac{1}{1-\frac{1}{3}}=\frac{3k}{2}$

Equating:

$s=\frac{9k}{4}=1$

Thus, solving for $k$:

$k=\frac{4}{9}$

Next, compute $P(X\ge 3)$:

$P(X\ge 3)=1-(P(X=0)+P(X=1)+P(X=2))$

Calculating these:

$P(X=0)=k=\frac{4}{9}$

$P(X=1)=\frac{2k}{3}=\frac{8}{27}$

$P(X=2)=\frac{3k}{9}=\frac{4}{27}$

Adding these probabilities:

$P(X=0)+P(X=1)+P(X=2)=\frac{4}{9}+\frac{8}{27}+\frac{4}{27}=\frac{8}{9}$

Finally, calculate $P(X\ge 3)$:

$P(X\ge 3)=1-\frac{8}{9}=\frac{1}{9}$