A ship is fitted with three engines $E_{1}, E_{2}$ and $E_{3} .$ The engines function independently of each other with respective probabilities $1 / 2, 1 / 4 and 1 / 4 .$ For the ship to be operational atleast two of its engines must function. Let X denote the event that the ship is operational and let $X_{1}, X_{2}, X_{3}$ dneotes, respectively the events that the engines $E_{1}, E_{2} and E_{3}$ are functioning. Which of the following is/are correct?

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$P (X_{1}^{c} / X) = 3 / 16$

$P [exactly two engines of the ship are functioning X]$ $= \frac{7}{8}$

$P (X / X_{2}) = \frac{5}{16}$

$P (X / X_{1}) = \frac{7}{16}$

answer is B, D.

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Detailed Solution

It is based on law of total probability and Baye's Law

It is given that ship would work, if atleast two of engines must work. If X be event that the ship works. Then $X \Rightarrow$ either any two of $E_{1}, E_{2}, E_{3}$ works or all three engines $E_{1}, E_{2}, E_{3}$ works.

Given, $P (E_{1}) = \frac{1}{2}, P (E_{2}) = \frac{1}{4}, P (E_{3}) = \frac{1}{4}$

$∴ P (X) = \{\begin{array}{l} P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) \\ + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3}) \end{array}\}$

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

$P (X_{1}^{c} / X) = P (\frac{X_{1}^{c} \cap X}{P (X)}) = \frac{P ({\bar{E}}_{1} \cap E_{2} \cap E_{3})}{P (X)} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{1}{8}$

P (exactly two enqines otthe ship are fwnGtioolng} ~

$\begin{array}{l} = \frac{P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3})}{P (X)} \\ = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{7}{8} \end{array}$

$\begin{aligned} P (\frac{X}{X_{2}}) = \frac{P (X \cap X_{2})}{P (X_{2})} \\ = \frac{P (Ship is operating with E_{2} function)}{P (X_{2})} \\ = \frac{P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3})}{P (E_{2})} \\ = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{5}{8} \end{aligned}$

$P (X / X_{1}) = \frac{P (X \cap X_{1})}{P (X_{1})} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{1 / 2} = \frac{7}{16}$