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

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

$P\left(X\right)=P(X_1\cap X_2\cap \overline{X_3})+P(X_1\cap \overline{X_2}\cap X_3)+P(\overline{X_1}\cap X_2\cap X_3)+P(X_1\cap X_2\cap X_3)=\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}+\frac{1}{2}.\frac{1}{4}\cdot \frac{1}{4}=\frac{1}{4}$

Now, $P\left(X_1^c|X\right)$
$=\frac{P(X_1^c\cap X)}{P\left(X\right)}=\frac{P(\overline{X_1}\cap X_2\cap X_3)}{P\left(X\right)}=\frac{\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$

$P\left(\text{exactly two engines of the ship are functioning}\right)=\frac{P(X_1\cap X_2\cap \overline{X_3})+P(X_1\cap \overline{X_2}\cap X_3)+P(\overline{X_1}\cap X_2\cap X_3)}{P\left(X\right)}=\frac{\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}}{\frac{1}{4}}=\frac{7}{8}$

$P\left(X|X_2\right)=\frac{P(X\cap X_2)}{P\left(X_2\right)}=\frac{P\left(\text{ship is operating with}{E}_2\text{functioning}\right)}{P\left(X_2\right)}$
$=\frac{P(X_1\cap X_2\cap \overline{X_3})+P(\overline{X_1}\cap X_2\cap X_3)+P(X_1\cap X_2\cap X_3)}{P\left(X_2\right)}=\frac{\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{3}{4}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{1}{4}}{\frac{1}{4}}=\frac{5}{8}$

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