A ship is fitted with three engines $E_{1}, E_{2} a n d E_{3} .$ The engines function independently of each other with respective probabilities $\frac{1}{2}, \frac{1}{4} a n d \frac{1}{4} .$ 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 $X_{1}, X_{2} a n d X_{3}$ denotes respectively the events that the engines $E_{1}, E_{2} a n d E_{3}$ are functioning. Which of the following is (are) true?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

$P [X_{1}^{C} | X] = \frac{3}{16}$

$P [X | X_{2}] = \frac{5}{16}$

$P [X | X_{1}] = \frac{7}{16}$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$A) P [X_{1}^{C} | X] = \frac{P ({\bar{X}}_{1} \cap X)}{P (X)} = \frac{(\frac{1}{2}) (\frac{1}{4}) (\frac{1}{4})}{\frac{1}{2} . \frac{1}{4} . \frac{1}{4} + \frac{1}{2} . \frac{1}{4} . \frac{1}{4} + \frac{1}{2} . \frac{3}{4} . \frac{1}{4} + \frac{1}{2} . \frac{1}{4} . \frac{3}{4}} = \frac{1}{8}$

B) Required probability = $P [X_{1}^{C} | X] = \frac{P ({\bar{X}}_{1} \cap X)}{P (X)} = \frac{\frac{3}{2.4.4} + \frac{3}{2.4.4} + \frac{1}{2.4.4}}{\frac{1}{2} . \frac{1}{4} . \frac{3}{4} + \frac{1}{2} . \frac{3}{4} . \frac{1}{4} + \frac{1}{2} . \frac{1}{4} . \frac{1}{4} + \frac{1}{2} . \frac{1}{4} . \frac{1}{4}} = \frac{7}{8}$

$C) P [X | X_{2}] = \frac{P (X \cap X_{2})}{P (X_{2})} = \frac{(\frac{1}{4}) [\frac{1}{2} . \frac{1}{4} + \frac{1}{2} . \frac{3}{4} + (\frac{1}{2}) (\frac{1}{4})]}{(\frac{1}{4})} = \frac{\frac{1}{4} . \frac{5}{8}}{\frac{1}{4}} = \frac{5}{8} D) \frac{P (X \cap X_{1})}{P (X_{1})} = \frac{(\frac{1}{2}) [\frac{1}{4} . \frac{1}{4} + \frac{3}{4} . \frac{1}{4} + \frac{1}{4} . \frac{3}{4}]}{\frac{1}{2}} = \frac{7}{16}$