1. **Problem statement:** A machine produces on average 2.5 defects per hour. We want to find probabilities for different numbers of defects in one hour. 2. **Formula used:** This is a Poisson distribution problem. The probability of exactly $k$ events when the average rate is $\lambda$ is given by: $$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$ where $\lambda = 2.5$ defects/hour. 3. **Calculate each probability:** **(a) Exactly 4 defects:** $$P(X=4) = \frac{2.5^4 e^{-2.5}}{4!}$$ Calculate factorial and powers: $$4! = 24$$ $$2.5^4 = 2.5 \times 2.5 \times 2.5 \times 2.5 = 39.0625$$ So, $$P(X=4) = \frac{39.0625 e^{-2.5}}{24}$$ **(b) No more than 2 defects:** This means $X \leq 2$, so sum probabilities for $k=0,1,2$: $$P(X \leq 2) = P(0) + P(1) + P(2)$$ Calculate each: $$P(0) = \frac{2.5^0 e^{-2.5}}{0!} = e^{-2.5}$$ $$P(1) = \frac{2.5^1 e^{-2.5}}{1!} = 2.5 e^{-2.5}$$ $$P(2) = \frac{2.5^2 e^{-2.5}}{2!} = \frac{6.25 e^{-2.5}}{2} = 3.125 e^{-2.5}$$ Sum: $$P(X \leq 2) = e^{-2.5} + 2.5 e^{-2.5} + 3.125 e^{-2.5} = (1 + 2.5 + 3.125) e^{-2.5} = 6.625 e^{-2.5}$$ **(c) At least 3 defects:** This means $X \geq 3$, which is the complement of $X \leq 2$: $$P(X \geq 3) = 1 - P(X \leq 2) = 1 - 6.625 e^{-2.5}$$ 4. **Final answers:** $$P(X=4) = \frac{39.0625 e^{-2.5}}{24} \approx 0.133$$ $$P(X \leq 2) = 6.625 e^{-2.5} \approx 0.544$$ $$P(X \geq 3) = 1 - 6.625 e^{-2.5} \approx 0.456$$ These probabilities describe the likelihood of defects occurring in one hour given the average rate of 2.5 defects/hour.