Step 1: Define the probabilities of individual events

Let $X_1$, $X_2$, and $X_3$ be the events that the engines $E_1$, $E_2$, and $E_3$ are functioning, respectively. The given probabilities are:

• $P(X_1) = {1 \over 2}$
• $P(X_2) = {1 \over 4}$
• $P(X_3) = {1 \over 4}$

Since the engines function independently, the probabilities of them not functioning are:

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

Step 2: Define the event X and calculate its probability

Let $X$ be the event that the ship is operational. The ship is operational if at least two of its engines function. This can happen in two mutually exclusive ways:

1. Exactly two engines function. The possible combinations are:

   • $E_1$ and $E_2$ function, $E_3$ does not: $X_1 \cap X_2 \cap X_3^c$
   • $E_1$ and $E_3$ function, $E_2$ does not: $X_1 \cap X_2^c \cap X_3$
   • $E_2$ and $E_3$ function, $E_1$ does not: $X_1^c \cap X_2 \cap X_3$

   The probability of exactly two engines functioning, let's call it $P(E_2)$, is: $P(E_2) = P(X_1)P(X_2)P(X_3^c) + P(X_1)P(X_2^c)P(X_3) + P(X_1^c)P(X_2)P(X_3)$ $P(E_2) = \left({1 \over 2}\right)\left({1 \over 4}\right)\left({3 \over 4}\right) + \left({1 \over 2}\right)\left({3 \over 4}\right)\left({1 \over 4}\right) + \left({1 \over 2}\right)\left({1 \over 4}\right)\left({1 \over 4}\right)$ $P(E_2) = {3 \over {32}} + {3 \over {32}} + {1 \over {32}} = {7 \over {32}}$

2. All three engines function.

   • $E_1$, $E_2$, and $E_3$ all function: $X_1 \cap X_2 \cap X_3$

   The probability of all three engines functioning, $P(E_3)$, is: $P(E_3) = P(X_1)P(X_2)P(X_3) = \left({1 \over 2}\right)\left({1 \over 4}\right)\left({1 \over 4}\right) = {1 \over {32}}$

The total probability of event $X$ is the sum of these probabilities: $P(X) = P(E_2) + P(E_3) = {7 \over {32}} + {1 \over {32}} = {8 \over {32}} = {1 \over 4}$

Step 3: Evaluate each option

Option A: $P\left[ {X_1^c|X} \right] = {3 \over {16}}$ Using the conditional probability formula, $P(A|B) = {{P(A \cap B)} \over {P(B)}}$. $P(X_1^c|X) = {{P(X_1^c \cap X)} \over {P(X)}}$ The event $X_1^c \cap X$ means that engine $E_1$ is not functioning AND the ship is operational. For the ship to be operational with $E_1$ not functioning, both $E_2$ and $E_3$ must function. So, $X_1^c \cap X$ is the event $X_1^c \cap X_2 \cap X_3$. $P(X_1^c \cap X) = P(X_1^c)P(X_2)P(X_3) = \left({1 \over 2}\right)\left({1 \over 4}\right)\left({1 \over 4}\right) = {1 \over {32}}$ $P(X_1^c|X) = {{{1 \over {32}}} \over {{1 \over 4}}} = {1 \over {32}} \times 4 = {1 \over 8}$ Since ${1 \over 8} \neq {3 \over {16}}$, option A is incorrect.

Option B: $P$[exactly two engines of the ship are functioning $|X] = {7 \over 8}$ Let $E_2$ be the event that exactly two engines are functioning. We need to find $P(E_2|X)$. $P(E_2|X) = {{P(E_2 \cap X)} \over {P(X)}}$ If exactly two engines are functioning (event $E_2$), the ship is operational (event $X$). Therefore, $E_2$ is a subset of $X$, and $E_2 \cap X = E_2$. $P(E_2|X) = {{P(E_2)} \over {P(X)}} = {{{7 \over {32}}} \over {{8 \over {32}}}} = {7 \over 8}$ Option B is correct.

Option C: $P\left[ {X|{X_2}} \right] = {5 \over {16}}$ We need to calculate $P(X|X_2) = {{P(X \cap X_2)} \over {P(X_2)}}$. The event $X \cap X_2$ means the ship is operational AND engine $E_2$ is functioning. This requires $E_2$ to be functioning and at least one of $E_1$ or $E_3$ to be functioning. So, $X \cap X_2 = X_2 \cap (X_1 \cup X_3)$. $P(X \cap X_2) = P(X_2) \times P(X_1 \cup X_3) \quad \text{(due to independence)}$ $P(X_1 \cup X_3) = P(X_1) + P(X_3) - P(X_1 \cap X_3) = {1 \over 2} + {1 \over 4} - \left({1 \over 2}\right)\left({1 \over 4}\right) = {3 \over 4} - {1 \over 8} = {5 \over 8}$ $P(X \cap X_2) = \left({1 \over 4}\right) \times \left({5 \over 8}\right) = {5 \over {32}}$ $P(X|X_2) = {{{5 \over {32}}} \over {P(X_2)}} = {{{5 \over {32}}} \over {{1 \over 4}}} = {5 \over 8}$ Since ${5 \over 8} \neq {5 \over {16}}$, option C is incorrect.

Option D: $P\left[ {X|{X_1}} \right] = {7 \over {16}}$ We need to calculate $P(X|X_1) = {{P(X \cap X_1)} \over {P(X_1)}}$. The event $X \cap X_1$ means the ship is operational AND engine $E_1$ is functioning. This requires $E_1$ to be functioning and at least one of $E_2$ or $E_3$ to be functioning. So, $X \cap X_1 = X_1 \cap (X_2 \cup X_3)$. $P(X \cap X_1) = P(X_1) \times P(X_2 \cup X_3) \quad \text{(due to independence)}$ $P(X_2 \cup X_3) = P(X_2) + P(X_3) - P(X_2 \cap X_3) = {1 \over 4} + {1 \over 4} - \left({1 \over 4}\right)\left({1 \over 4}\right) = {1 \over 2} - {1 \over {16}} = {7 \over {16}}$ $P(X \cap X_1) = \left({1 \over 2}\right) \times \left({7 \over {16}}\right) = {7 \over {32}}$ $P(X|X_1) = {{{7 \over {32}}} \over {P(X_1)}} = {{{7 \over {32}}} \over {{1 \over 2}}} = {7 \over {16}}$ Option D is correct.

Step 4: Conclusion

Based on the calculations, options B and D are the correct statements.