Question:

Let $P(E_i)$ denote the probability that engine $E_i$ works, and $P(\bar{E_i})$ denote the probability that engine $E_i$ fails. The given probabilities are: $P(E_1) = p_1 = \frac{1}{2}$ $P(E_2) = p_2 = \frac{1}{4}$ $P(E_3) = p_3 = \frac{1}{4}$ The probabilities of the engines failing are: $P(\bar{E_1}) = q_1 = 1 - p_1 = 1 - \frac{1}{2} = \frac{1}{2}$ $P(\bar{E_2}) = q_2 = 1 - p_2 = 1 - \frac{1}{4} = \frac{3}{4}$ $P(\bar{E_3}) = q_3 = 1 - p_3 = 1 - \frac{1}{4} = \frac{3}{4}$ For the ship to be operational, at least two of its engines must work. This means either exactly two engines work, or all three engines work. These are mutually exclusive events. **Case 1: Exactly two engines work.** There are three distinct scenarios for exactly two engines to work: 1. Engine $E_1$ and $E_2$ work, and $E_3$ fails: $P(E_1 \cap E_2 \cap \bar{E_3}) = p_1 p_2 q_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right) = \frac{3}{32}$ 2. Engine $E_1$ and $E_3$ work, and $E_2$ fails: $P(E_1 \cap \bar{E_2} \cap E_3) = p_1 q_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{1}{4}\right) = \frac{3}{32}$ 3. Engine $E_2$ and $E_3$ work, and $E_1$ fails: $P(\bar{E_1} \cap E_2 \cap E_3) = q_1 p_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) = \frac{1}{32}$ The total probability of exactly two engines working is the sum of these probabilities: $$P(\text{exactly 2 engines work}) = \frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{7}{32}$$ **Case 2: All three engines work.** The probability that all three engines work is: $$P(E_1 \cap E_2 \cap E_3) = p_1 p_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) = \frac{1}{32}$$ **Probability of the ship being operational:** The probability that the ship is operational is the sum of the probabilities of exactly two engines working and all three engines working: $$P(\text{operational}) = P(\text{exactly 2 engines work}) + P(\text{all 3 engines work})$$ $$P(\text{operational}) = \frac{7}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$$