A ship is fitted with three engines $E_1$, $E_2$ and $E_3$. The engines function independently of each other with respective probabilities $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{4}$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_1$, $X_2$ and $X_3$ denote respectively the events that the engines $E_1$, $E_2$ and $E_3$ are functioning. Which of the following is(are) true?

A ship is fitted with three engines E₁, E₂ and E₃. The engines function independently of each other with respective probabilities 1/2, 1/4 and 1/4. For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X₁, X₂ and X₃ denote respectively the events that the engines E₁, E₂ and E₃ are functioning. Which of the following is(are) true?

$P(X_1) = \frac{1}{2}, \quad P(X_2) = \frac{1}{4}, \quad P(X_3) = \frac{1}{4}$

$X_1,\ X_2,\ X_3 \text{ independent}$

$X = \text{ship operational} \iff \text{at least two of } X_1,X_2,X_3$

$\text{To find: which of } (1),\ (2),\ (3),\ (4) \text{ are TRUE}$

$\text{Cases for } X: \text{ exactly 2, or all 3}$

$P(X) = P(\text{exactly 2}) + P(\text{all 3})$

$P(\text{exactly 2}) = P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3)$

$= \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{3}{4} + \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4}$

$= \frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{7}{32}$

$P(\text{all 3}) = \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4} = \frac{1}{32}$

$P(X) = \frac{7}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$

$P(X_1^c \cap X) = \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{1}{4} = \frac{1}{32}$

$P(X_1^c \mid X) = \frac{1/32}{1/4} = \frac{1}{8} \neq \frac{3}{16} \Rightarrow (1)\ \text{FALSE}$

$P(\text{exactly 2} \mid X) = \frac{7/32}{1/4} = \frac{7}{8} \Rightarrow (2)\ \text{TRUE}$

$P(X \cap X_2) = P(X_2)\left[1 - P(X_1^c)P(X_3^c)\right] = \frac{1}{4}\cdot\frac{5}{8} = \frac{5}{32}$

$P(X \mid X_2) = \frac{5/32}{1/4} = \frac{5}{8} \neq \frac{5}{16} \Rightarrow (3)\ \text{FALSE}$

$P(X \cap X_1) = P(X_1)\left[1 - P(X_2^c)P(X_3^c)\right] = \frac{1}{2}\cdot\frac{7}{16} = \frac{7}{32}$

$P(X \mid X_1) = \frac{7/32}{1/2} = \frac{7}{16} \Rightarrow (4)\ \text{TRUE}$

$\boxed{\text{True: } (2)\ \text{and}\ (4)}$