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 $\frac{1}{2},\frac{1}{4}and\frac{1}{4}$. 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. $P\left(X_1^C|X\right)=\frac{3}{16}$
• B. P (Exactly two engines of the ship are functioning | X) $=\frac{7}{8}$
• C. $P\left(X|X_2\right)=\frac{5}{16}$
• D. $P\left(X|X_1\right)=\frac{7}{16}$

Answer 1

Answer: (B), (D)

The correct options are
B

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

D

$P\left(X|X_1\right)=\frac{7}{16}$

Description of the situation: It is given that ship would work only if at least two of its engines work. If X be the event that the ship is operational. Then, X = either any two of $E_1,E_2,E_3$ work or all three engines $E_1,E_2,E_3$ work.
Given, $P\left(E_1\right)=\frac{1}{2},P\left(E_2\right)=\frac{1}{4},P\left(E_3\right)=\frac{1}{4}\therefore P\left(X\right)=\left\{P\right(E_1\cap E_2\cap \overline{E_3})+P(E_1\cap \overline{E_2}\cap E_3)+P(\overline{E_1}\cap E_2\cap E_3)+P(E_1\cap E_2\cap E_3\left)\right\}(\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4})+(\frac{1}{2}.\frac{1}{4}.\frac{1}{4})=\frac{1}{4}Now,\left(a\right)P\left(X_1^c|X\right)=P\left(\frac{(X_1^c\cap X)}{P\left(X\right)}\right)=\frac{P(\overline{E_1}\cap E_2\cap E_3)}{P\left(X\right)}=\frac{\frac{1}{2}.\frac{1}{4}.\frac{3}{4}}{\frac{1}{4}}=\frac{1}{8}$

(b) P(exactly two engines of the ship are functioning)
$=\frac{P(E_1\cap E_2\cap \overline{E_3})+P(E_1\cap \overline{E_2}\cap E_3)+P(\overline{E_1}\cap E_2\cap E_3)}{P\left(X\right)}=\frac{\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{4}}=\frac{7}{8}$

(c) $P\left(X|X_2\right)=\frac{P(X\cap X_2)}{P\left(X_2\right)}=\frac{P\left(shipisoperatingwith{E}_{2}functioning\right)}{P\left(X_2\right)}$
$=\frac{P(E_1\cap E_2\cap \overline{E_3})+P(\overline{E_1}\cap E_2\cap E_3)+P(E_1\cap E_2\cap E_3)}{P\left(E_2\right)}=\frac{\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{4}}=\frac{5}{8}$

(d) $P\left(X|X_1\right)=\frac{P(X\cap X_1)}{P\left(X_1\right)}=\frac{\frac{1}{2}.\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{3}{4}}{\frac{1}{2}}=\frac{7}{16}$