1. **Problem Statement:** An insurance company receives on average two claims per week from a factory. The number of claims follows a Poisson distribution with parameter $\lambda = 2$ claims/week. We need to find: - (a) Probability of exactly 3 claims in a week. - (b) Probability of more than 4 claims in a week. - (c) Probability of between 2 and 5 claims inclusive in a week. - (d) Probability of no claims on a given day (5-day week). Then, for a binomial random variable $X \sim B(6,0.42)$, find: - (e) $P(X=6)$ - (f) $P(X=4)$ - (g) $P(X \leq 2)$ - (h) $P(1 \leq X \leq 5)$ --- 2. **Poisson Distribution Formula:** $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $k$ is the number of claims, $\lambda$ is the average rate. 3. **Binomial Distribution Formula:** $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=6$, $p=0.42$. --- 4. **Calculations for Poisson:** (a) $P(X=3) = \frac{2^3 e^{-2}}{3!} = \frac{8 e^{-2}}{6} = \frac{4}{3} e^{-2} \approx 0.180$. (b) $P(X>4) = 1 - P(X \leq 4) = 1 - \sum_{k=0}^4 \frac{2^k e^{-2}}{k!}$. Calculate each term: - $P(0) = e^{-2} \approx 0.1353$ - $P(1) = 2 e^{-2} \approx 0.2707$ - $P(2) = \frac{2^2}{2} e^{-2} = 2 e^{-2} \approx 0.2707$ - $P(3) = 0.180$ (from above) - $P(4) = \frac{2^4}{24} e^{-2} = \frac{16}{24} e^{-2} = \frac{2}{3} e^{-2} \approx 0.0902$ Sum: $0.1353 + 0.2707 + 0.2707 + 0.180 + 0.0902 = 0.947$ So, $P(X>4) = 1 - 0.947 = 0.053$. (c) $P(2 \leq X \leq 5) = \sum_{k=2}^5 P(X=k)$. Calculate $P(5)$: $P(5) = \frac{2^5 e^{-2}}{120} = \frac{32}{120} e^{-2} = \frac{4}{15} e^{-2} \approx 0.0361$ Sum $P(2)+P(3)+P(4)+P(5) = 0.2707 + 0.180 + 0.0902 + 0.0361 = 0.577$. (d) For no claims in a day, daily rate $\lambda_{day} = \frac{2}{5} = 0.4$. $P(X=0) = e^{-0.4} \approx 0.6703$. --- 5. **Calculations for Binomial:** (e) $P(X=6) = \binom{6}{6} 0.42^6 (1-0.42)^0 = 1 \times 0.42^6 \approx 0.0055$. (f) $P(X=4) = \binom{6}{4} 0.42^4 (0.58)^2 = 15 \times 0.0311 \times 0.3364 \approx 0.157$. (g) $P(X \leq 2) = P(0)+P(1)+P(2)$: - $P(0) = \binom{6}{0} 0.42^0 0.58^6 = 1 \times 1 \times 0.0585 = 0.0585$ - $P(1) = \binom{6}{1} 0.42^1 0.58^5 = 6 \times 0.42 \times 0.1009 = 0.254$ - $P(2) = \binom{6}{2} 0.42^2 0.58^4 = 15 \times 0.1764 \times 0.174 = 0.460$ Sum: $0.0585 + 0.254 + 0.460 = 0.772$. (h) $P(1 \leq X \leq 5) = 1 - P(X=0) - P(X=6) = 1 - 0.0585 - 0.0055 = 0.936$. --- **Final answers:** - (a) $\approx 0.180$ - (b) $\approx 0.053$ - (c) $\approx 0.577$ - (d) $\approx 0.670$ - (e) $\approx 0.0055$ - (f) $\approx 0.157$ - (g) $\approx 0.772$ - (h) $\approx 0.936$