AMU 2026 — Mathematics PYQ

1. Poisson Distribution Formula

For a Poisson distribution with parameter $\lambda$, the probability mass function is given by:

$$P(X=k)=\frac{e^{-\lambda }{\lambda }^k}{k!}$$

where:

• $\lambda$ is the mean of the distribution.

• In a Poisson distribution, the variance is equal to the mean, so $\text{Var}(X)=\lambda$.

2. Set up the Equation

Given the condition $2P(X=1)=P(X=2)$, we substitute the values into the probability mass function:

• For $k=1$: $P(X=1)=\frac{e^{-\lambda }{\lambda }^1}{1!}=\lambda e^{-\lambda }$

• For $k=2$: $P(X=2)=\frac{e^{-\lambda }{\lambda }^2}{2!}=\frac{{\lambda }^2{e}^{-\lambda }}{2}$

Substitute these into the given equation:

$$2(\lambda e^{-\lambda })=\frac{{\lambda }^2{e}^{-\lambda }}{2}$$

3. Solve for $\lambda$

Assuming $\lambda \ne 0$ (as the distribution must exist), we can divide both sides by $e^{-\lambda }$ and by $\lambda$:

$$2\lambda =\frac{{\lambda }^2}{2}$$

Multiply both sides by $2$:

$$4\lambda ={\lambda }^2$$

$${\lambda }^2-4\lambda =0$$

$$\lambda (\lambda -4)=0$$

Since $\lambda$ cannot be $0$, we must have:

$$\lambda =4$$

Conclusion

As established, in a Poisson distribution, the variance is equal to the parameter $\lambda$.

$$\text{Variance}=\lambda =4$$

This directly matches option (b).