Question

A ship is fitted with three engines ${E_1},{E_2}\;and\;{E_3}$​. The engines function independently of each other with respective probabilities \[\dfrac{1}{2},\dfrac{1}{4}\;and\;\dfrac{1}{4}\] . 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?
\[
  A.P[{X_1}^c|X] = \dfrac{3}{{16}} \\
  B.P[Exactly\;two\;engines\;of\;the\;ship\;are\;functioning|X] = \dfrac{7}{8} \\
  C.P[X|{X_2}] = \dfrac{5}{{16}} \\
  D.P[X|{X_1}] = \dfrac{7}{{16}} \\
 \]

Answer 1

Hint: First compute the probability for event X, which is for ship being operational. Then compute each probability as given in each option. We have to use the concept of conditional probability as well as the formula for the probability of independent events occurring together. Remember that exactly two means exclude the case of all three together. While at least two means consider two or more events together one by one.

Complete step by step answer:
There are three engines ${E_1},{E_2}\;and\;{E_3}$ having event ${X_1},{X_2}\;and\;{X_3}$as their operational status. Also, X is the event that the ship is operational.
According to the question, we are having the following data,
$P[{X_1}] = \dfrac{1}{2}$
$P[{X_2}] = \dfrac{1}{4}$ and
$P[{X_3}] = \dfrac{1}{4}$
The ship will be operational if two or more lifts are operational independently.
Therefore,
P[X]= ${X_1},{X_2}$ are operational only + ${X_1},{X_3}$ are operational only + ${X_2},{X_3}$ are operational only+${X_1},{X_3},{X_3}$ are operational
$\therefore $ \[P(X) = P({X_1} \cap {X_2} \cap {X_3}^c) + P({X_1} \cap {X_2}^c \cap {X_3}) + P({X_1}^c \cap {X_2} \cap {X_3}) + P({X_1} \cap {X_2} \cap {X_3})\]
Since all are independent, then we have
\[
  P(X) = P[{X_1}] \times P[{X_2}] \times P[{X_3}^c] + P[{X_1}] \times P[{X_2}^c] \times P[{X_3}] + P[{X_1}^c] \times P[{X_2}] \times P[{X_3}] + P[{X_1}] \times P[{X_2}] \times P[{X_3}] \\
  P(X) = \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} + \dfrac{1}{2} \times \dfrac{3}{4} \times \dfrac{1}{4} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4} \\
  P[X] = \dfrac{1}{4} \\
 \]Now, we will compute each probability one by one.
A) \[ P[{X_1}^c|X] = \dfrac{{P[{X_1}^c \cap X]}}{{P[X]}} = \dfrac{{\dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4}}}{{\dfrac{1}{4}}} \\
    \\
 \]
Solving it, we get
\[
  P[{X_1}^c|X] = \dfrac{1}{8} \\
    \\
 \]
B)\[P[Exactly\;two\;engines\;of\;the\;ship\;are\;functioning|X] \\
   = \dfrac{{P[{X_1}] \times P[{X_2}] \times P[{X_3}^c] + P[{X_1}] \times P[{X_2}^c] \times P[{X_3}] + P[{X_1}^c] \times P[{X_2}] \times P[{X_3}]}}{{P[X]}} \\
   = \dfrac{{\dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} + \dfrac{1}{2} \times \dfrac{3}{4} \times \dfrac{1}{4} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4}}}{{\dfrac{1}{4}}} \\
 \]
Solving the above, we get
\[
  P[Exactly\;two\;engines\;of\;the\;ship\;are\;functioning|X] = \dfrac{7}{8} \\
    \\
 \]
(C) \[P[X|{X_2}] = \dfrac{{P[X \cap {X_2}]}}{{P[{X_2}]}} \\
  P[X|{X_2}] = \dfrac{{\dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{1}{4} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} + \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{4}}}{{\dfrac{1}{4}}} \\
    \\
 \]
Solving the above, we get
\[
  P[X|{X_2}] = \dfrac{5}{8} \\
    \\
 \]
(D) \[P[X|{X_1}]\]
$ = \dfrac{{P[X \cap {X_1}]}}{{P[{X_1}]}}$
\[ = \dfrac{{\dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} + \dfrac{1}{2} \times \dfrac{1}{4} \times \dfrac{3}{4} + \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{4}}}{{\dfrac{1}{4}}}\]
Solving the above, we get
\[P[X|{X_1}]\] = $\dfrac{7}{{16}}$
It is clear that option D is also the correct option.

$\therefore$ Option (B) and option (D) are correct.

Note:
This question has used some popular concepts of the probabilities like conditional probability and complementary probability. Complementary probability is the difference between the main probability of 1. Also, conditional probability is the probability of some event A occurring given that another event B has already occurred. And it is represented as P[A|B]. There is another way to compute it, popular as Bayes’ theorem.