The probability mass function of a Poisson variate $X \sim \text{Poisson}(\lambda)$ is:
$$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \dots$$
Given $P(X=1) = P(X=2)$:
$$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^2}{2!}$$
$$\lambda e^{-\lambda} = \frac{\lambda^2 e^{-\lambda}}{2}$$
Dividing both sides by $\lambda e^{-\lambda}$ (since $\lambda \gt 0$):
$$1 = \frac{\lambda}{2} \implies \lambda = 2$$
Now, calculate $P(X=4)$:
$$P(X=4) = \frac{e^{-2} (2)^4}{4!} = \frac{e^{-2} \times 16}{24} = \frac{2e^{-2}}{3}$$
Thus, Option 1 is the correct answer.
प्वासों बंटन का प्रायिकता द्रव्यमान फलन:
$$P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$$
दिया है $P(X=1) = P(X=2)$:
$$\frac{e^{-\lambda} \lambda}{1} = \frac{e^{-\lambda} \lambda^2}{2} \implies 1 = \frac{\lambda}{2} \implies \lambda = 2$$
अब $P(X=4)$ ज्ञात करने पर:
$$P(X=4) = \frac{e^{-2} (2)^4}{4!} = \frac{16 e^{-2}}{24} = \frac{2e^{-2}}{3}$$
अतः सही उत्तर $\frac{2e^{-2}}{3}$ (विकल्प 1) है।
