Conditional Probability of Operational Engines in a Ship

To determine which of the statements are true, we begin by calculating the probabilities of the relevant events.

Step 1: Probability of individual engine operations

The events $X_1$, $X_2$, and $X_3$ denote the functioning of engines $E_1$, $E_2$, and $E_3$ respectively. Since the engines operate independently:

• $P(X_1) = \frac{1}{2} \implies P(X_1^c) = 1 - \frac{1}{2} = \frac{1}{2}$
• $P(X_2) = \frac{1}{4} \implies P(X_2^c) = 1 - \frac{1}{4} = \frac{3}{4}$
• $P(X_3) = \frac{1}{4} \implies P(X_3^c) = 1 - \frac{1}{4} = \frac{3}{4}$

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Step 2: Probability of the ship being operational, $P(X)$

The ship remains operational (event $X$) if at least two engines function. This event is the union of two mutually exclusive events:

1. Exactly two engines function.
2. All three engines function.

Case 1: All three engines function $P(X_1 \cap X_2 \cap X_3) = P(X_1) P(X_2) P(X_3) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}$

Case 2: Exactly two engines function

• Only $E_1$ and $E_2$ function: $P(X_1 \cap X_2 \cap X_3^c) = \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{32}$
• Only $E_1$ and $E_3$ function: $P(X_1 \cap X_2^c \cap X_3) = \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}$
• Only $E_2$ and $E_3$ function: $P(X_1^c \cap X_2 \cap X_3) = \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}$

Thus, the probability that exactly two engines function is: $P(\text{Exactly two}) = \frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{7}{32}$

The total probability that the ship is operational is: $P(X) = P(\text{Exactly two}) + P(\text{All three}) = \frac{7}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$

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Step 3: Evaluation of the Options

• Option A: The event $X_1^c \cap X$ means engine $E_1$ does not function, but the ship is operational. This occurs only when both $E_2$ and $E_3$ function: $P(X_1^c \cap X) = P(X_1^c \cap X_2 \cap X_3) = \frac{1}{32}$ Therefore: $P[X_1^c \mid X] = \frac{P(X_1^c \cap X)}{P(X)} = \frac{1/32}{1/4} = \frac{1}{8}$ Hence, Option A is incorrect.

• Option B: $P[\text{Exactly two engines of the ship are functioning} \mid X] = \frac{P(\text{Exactly two} \cap X)}{P(X)} = \frac{P(\text{Exactly two})}{P(X)}$ $P[\text{Exactly two engines of the ship are functioning} \mid X] = \frac{7/32}{8/32} = \frac{7}{8}$ Hence, Option B is correct.

• Option C: Given that $E_2$ is functioning, the ship will be operational if at least one of the remaining two engines ($E_1$ or $E_3$) functions. Since the engines operate independently: $P[X \mid X_2] = P(X_1 \cup X_3) = 1 - P(X_1^c \cap X_3^c) = 1 - P(X_1^c)P(X_3^c)$ $P[X \mid X_2] = 1 - \left(\frac{1}{2} \times \frac{3}{4}\right) = 1 - \frac{3}{8} = \frac{5}{8}$ Hence, Option C is incorrect.

• Option D: Given that $E_1$ is functioning, the ship will be operational if at least one of the remaining two engines ($E_2$ or $E_3$) functions. Since the engines operate independently: $P[X \mid X_1] = P(X_2 \cup X_3) = 1 - P(X_2^c \cap X_3^c) = 1 - P(X_2^c)P(X_3^c)$ $P[X \mid X_1] = 1 - \left(\frac{3}{4} \times \frac{3}{4}\right) = 1 - \frac{9}{16} = \frac{7}{16}$ Hence, Option D is correct.

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Conclusion

The correct statements are B and D.