1. **Problem statement:** Ravi sells an average of 3 life insurance policies per week. We use Poisson distribution to find probabilities for different scenarios. 2. **Poisson distribution formula:** $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda$ is the average number of events (policies sold) in the interval (week), and $k$ is the number of events. 3. **Given:** $\lambda = 3$ policies/week. 4. **(a) Probability of selling some policies (i.e., at least one):** This is $P(X \geq 1) = 1 - P(X=0)$. Calculate $P(X=0)$: $$P(X=0) = \frac{3^0 e^{-3}}{0!} = e^{-3}$$ So, $$P(X \geq 1) = 1 - e^{-3} \approx 1 - 0.0498 = 0.9502$$ 5. **(b) Probability of selling two or more but less than five policies:** This is $P(2 \leq X < 5) = P(X=2) + P(X=3) + P(X=4)$. Calculate each: $$P(X=2) = \frac{3^2 e^{-3}}{2!} = \frac{9 e^{-3}}{2} = 4.5 e^{-3}$$ $$P(X=3) = \frac{3^3 e^{-3}}{3!} = \frac{27 e^{-3}}{6} = 4.5 e^{-3}$$ $$P(X=4) = \frac{3^4 e^{-3}}{4!} = \frac{81 e^{-3}}{24} = 3.375 e^{-3}$$ Sum: $$P = (4.5 + 4.5 + 3.375) e^{-3} = 12.375 e^{-3} \approx 12.375 \times 0.0498 = 0.616$$ 6. **(c) Probability of selling one policy in one day:** Since 5 working days per week, average per day $\lambda_{day} = \frac{3}{5} = 0.6$. Calculate $P(X=1)$ for one day: $$P(X=1) = \frac{0.6^1 e^{-0.6}}{1!} = 0.6 e^{-0.6} \approx 0.6 \times 0.5488 = 0.3293$$ **Final answers:** - (a) $P(X \geq 1) \approx 0.9502$ - (b) $P(2 \leq X < 5) \approx 0.616$ - (c) $P(X=1 \text{ in one day}) \approx 0.3293$