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?

Options:
A $P\left[X_{1}^{c} \mid X\right]=\frac{3}{16}$
B $P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$
C $P\left[X \mid X_{2}\right]=\frac{5}{16}$
D $P\left[X \mid X_{1}\right]=\frac{7}{16}$

The correct answers are:
$P$ [Exactly two engines of the ship are functioning $[X]=\frac{7}{8}$, $P\left[X \mid X_{1}\right]=\frac{7}{16}$

Solution:
Given that $P\left(X_{1}\right)=\frac{1}{2}, \mathrm{P}\left(X_{2}\right)=\frac{1}{4}, \mathrm{P}\left(X_{3}\right)=\frac{1}{4}$ $P(X)=P$ (at least 2 engines are functioning)

$\begin{aligned}=P\left(X_{1} \cap X_{2}\right.&\left.\cap X_{3}^{C}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right) \\ &+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}\right) \end{aligned}$

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

(a) $P\left(X_{1}^{C} / X\right)=\frac{P\left(X_{1}^{C} \cap X\right)}{P(X)}=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)}{P(X)}$

$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$

$\therefore$ (a) is not true.

(b) $P$ [Exactly two engines are functioning $/ X]$

$=\frac{P[(\text { Exactly two engines are functioning }) \cap X]}{P(X)}$

$=\frac{P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P(X)}$

$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{7}{8}$

$\therefore$ (b) is true.

(c) $P\left(X / X_{2}\right)=\frac{P\left(X \cap X_{2}\right)}{P\left(X_{2}\right)}$

$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1}^{C} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{2}\right)}$

$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{4}}=\frac{5}{8}$

$\therefore$ (c) is not true.

(d) $P\left(X / X_{1}\right)=\frac{P\left(X \cap X_{1}\right)}{P\left(X_{1}\right)}$

$=\frac{P\left(X_{1} \cap X_{2} \cap X_{3}\right)+P\left(X_{1} \cap X_{2}^{C} \cap X_{3}\right)+P\left(X_{1} \cap X_{2} \cap X_{3}^{C}\right)}{P\left(X_{1}\right)}$

$=\frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{3}{4} \times \frac{1}{4}+\frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$

$\therefore$ (d) is true.

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2012 (Paper 1)