Suppose that in a population the probability of survival of an individual would survive is 99/100. If 5 people in the population were to buy an insurance, Using Poisson distribution what is the probability that exactly 2 of such person in the population would die
Thought: $X=$ number of people dying $$\begin{align} X & \sim \mathcal{Pois} (1/100 \cdot 5) \\[2ex] \mathsf P (X = 2) & = \frac{[0.05^2 \cdot e^{-0.05}] }2 \end{align}$$ Am I right? Thanks
You want the distribution of the count of successes in $n=5$ trials, where each trial has an independent and identical rate of success ($p=1/100$).
This is a Binomial Distribution.
$$\begin{align} \mathsf P(X=2) & = \dbinom{5}{2} \dfrac {99^3}{100^5} \\ & \approx {0.00097} \end{align}$$
The Poison Approximation to Binomial is not really accurate at such a small sample size. But if you must use it, then indeed:
$$\begin{align} \mathsf P(X=2) &\;\dot\approx\; \frac{0.05^2\cdot\mathsf e^{-0.05}}{2!} \\ & \approx 0.00119 \end{align}$$