Question:

Step 1: Define the probabilities of engines working and failing. Let $P(E_1)$, $P(E_2)$, $P(E_3)$ be the probabilities that engines $E_1$, $E_2$, $E_3$ work, respectively. Let $P(\bar{E_1})$, $P(\bar{E_2})$, $P(\bar{E_3})$ be the probabilities that engines $E_1$, $E_2$, $E_3$ fail, respectively. 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: $$P(\bar{E_1}) = q_1 = 1 - p_1 = 1 - \frac{1}{2} = \frac{1}{2}$$ $$P(\bar{E_2}) = q_2 = 1 - p_2 = 1 - \frac{1}{4} = \frac{3}{4}$$ $$P(\bar{E_3}) = q_3 = 1 - p_3 = 1 - \frac{1}{4} = \frac{3}{4}$$ Step 2: Calculate the probability that the ship is operational. The ship is operational if at least two of its engines work. This means either exactly two engines work or all three engines work. We calculate the probabilities for these mutually exclusive scenarios: 1. **All three engines work ($E_1 \cap E_2 \cap E_3$):** $$P(E_1 \cap E_2 \cap E_3) = p_1 p_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) = \frac{1}{32}$$ 2. **Exactly two engines work:** * $E_1$ and $E_2$ work, $E_3$ fails ($E_1 \cap E_2 \cap \bar{E_3}$): $$P(E_1 \cap E_2 \cap \bar{E_3}) = p_1 p_2 q_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right) = \frac{3}{32}$$ * $E_1$ and $E_3$ work, $E_2$ fails ($E_1 \cap \bar{E_2} \cap E_3$): $$P(E_1 \cap \bar{E_2} \cap E_3) = p_1 q_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{1}{4}\right) = \frac{3}{32}$$ * $E_2$ and $E_3$ work, $E_1$ fails ($\bar{E_1} \cap E_2 \cap E_3$): $$P(\bar{E_1} \cap E_2 \cap E_3) = q_1 p_2 p_3 = \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right) = \frac{1}{32}$$ The total probability that the ship is operational is the sum of these probabilities: $$P(\text{ship operational}) = \frac{1}{32} + \frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$$ Step 3: Evaluate Option 1. Option 1 states: $P(\text{ship operational}) = 7/32$. From Step 2, we calculated $P(\text{ship operational}) = 1/4 = 8/32$. Since $8/32 \neq 7/32$, Option 1 is FALSE. Step 4: Evaluate Option 2. Option 2 states: $P(E_1 \text{ works} | \text{ship is operational}) = 4/7$. We need to calculate the conditional probability $P(E_1 | \text{Operational}) = \frac{P(E_1 \cap \text{Operational})}{P(\text{Operational})}$. The event "$E_1$ works AND ship is operational" means $E_1$ works and at least one other engine works. This includes the following scenarios where $E_1$ works: * $E_1$, $E_2$, and $E_3$ work ($p_1 p_2 p_3 = 1/32$) * $E_1$ and $E_2$ work, $E_3$ fails ($p_1 p_2 q_3 = 3/32$) * $E_1$ and $E_3$ work, $E_2$ fails ($p_1 q_2 p_3 = 3/32$) The probability of ($E_1$ works AND ship is operational) is: $$P(E_1 \cap \text{Operational}) = \frac{1}{32} + \frac{3}{32} + \frac{3}{32} = \frac{7}{32}$$ Now, we calculate the conditional probability: $$P(E_1 | \text{Operational}) = \frac{P(E_1 \cap \text{Operational})}{P(\text{Operational})} = \frac{7/32}{1/4} = \frac{7/32}{8/32} = \frac{7}{8}$$ Since $7/8 \neq 4/7$, Option 2 is FALSE. Step 5: Evaluate Option 3. Option 3 states: $P(\text{only } E_1 \text{ works}) = 3/8$. "Only $E_1$ works" means $E_1$ works, and $E_2$ fails, and $E_3$ fails ($E_1 \cap \bar{E_2} \cap \bar{E_3}$). $$P(E_1 \cap \bar{E_2} \cap \bar{E_3}) = p_1 q_2 q_3 = \left(\frac{1}{2}\right)\left(\frac{3}{4}\right)\left(\frac{3}{4}\right) = \frac{9}{32}$$ Since $9/32 \neq 3/8$ (which is $12/32$), Option 3 is FALSE. Step 6: Evaluate Option 4. Option 4 states: $P(\text{at least one engine fails}) = 1 - 1/32$. The event "at least one engine fails" is the complement of the event "all three engines work". $$P(\text{at least one engine fails}) = 1 - P(\text{all three engines work})$$ From Step 2, $P(\text{all three engines work}) = P(E_1 \cap E_2 \cap E_3) = 1/32$. Therefore, $$P(\text{at least one engine fails}) = 1 - \frac{1}{32}$$ This matches the statement in Option 4. So, Option 4 is TRUE. Step 7: Conclusion. Based on the calculations, Option 4 is TRUE, while Options 1, 2, and 3 are FALSE. The true option is Option 4.