1. **State the problem:** We have a fleet of planes with probabilities: no wing cracks $=0.7$, detectable wing cracks $=0.25$, and critical wing cracks $=0.05$. For the next four planes inspected, we want to find: - The probability that exactly one plane has a critical crack, exactly one has a detectable crack, and exactly two have no cracks. - The probability that at least one plane has a critical crack. 2. **Formula and rules:** This is a multinomial probability problem. The probability of a specific combination of outcomes in $n$ independent trials is given by: $$P = \frac{n!}{n_1! n_2! n_3!} p_1^{n_1} p_2^{n_2} p_3^{n_3}$$ where $n=4$, $n_1=1$ (critical), $n_2=1$ (detectable), $n_3=2$ (no cracks), and $p_1=0.05$, $p_2=0.25$, $p_3=0.7$. 3. **Calculate the first probability:** $$P = \frac{4!}{1! 1! 2!} (0.05)^1 (0.25)^1 (0.7)^2$$ Calculate factorials: $$4! = 24, \quad 1! = 1, \quad 2! = 2$$ So: $$P = \frac{24}{1 \times 1 \times 2} \times 0.05 \times 0.25 \times 0.7^2 = 12 \times 0.05 \times 0.25 \times 0.49$$ Calculate stepwise: $$12 \times 0.05 = 0.6$$ $$0.6 \times 0.25 = 0.15$$ $$0.15 \times 0.49 = 0.0735$$ So the probability is $0.0735$. 4. **Calculate the second probability (at least one critical crack):** The complement is that no planes have critical cracks, so all 4 planes have either detectable or no cracks. Probability no critical cracks: $$P(\text{no critical}) = (0.95)^4$$ Calculate: $$0.95^4 = 0.8145$$ Therefore, $$P(\text{at least one critical}) = 1 - 0.8145 = 0.1855$$ **Final answers:** - Probability exactly one critical, one detectable, two no cracks: $0.0735$ - Probability at least one critical crack: $0.1855$