Discussion Forum

A ship is fitted with 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 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 $X_{3}$ denotes,respectively the  events that engines $E_{1},E_{2}$ and $E_{3}$ are functioning .Which of the following is/are true?

Answer:

Explanation:

Concept involved It is based on the law of total probability and Bay's Law.

Description of Situation It is given that
ship would work if at least two of engineers must work. If X be an 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 

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

$\therefore$ 

$P(X)=$

$\left\{P(E_{1}\cap E_{2}\cap \overline{E_{3}})+P(E_{1}\cap \overline{E_{2}}\cap E_{3}) + P(\overline{E_{1}}\cap E_{2}\cap E_{3})+P(E_{1}\cap E_{2}\cap E_{3})\right\}$

 =$\left(\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}\right)+\left(\frac{1}{2}.\frac{1}{4}.\frac{1}{4}\right)$

    =$\frac{1}{4}$

 Now,  (a)  $P(X_{1}^{c}/X)$

 $=P\left(\frac{X_{1}^{c}\cap X}{P(X)}\right)=\frac{P(\overline {E_{1}} \cap E_{2}\cap E_{3})}{P(X)}$

 $=\frac{\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{4}}=\frac{1}{8}$

(b)  

P (Exactly two engines of the ship are functioning)

 $=\frac{P(E_{1}\cap E_{2}\cap\overline{E_{3}})+P(E_{1}\cap \overline{E_{2}}\cap E_{3})+P(\overline{E_{1}}\cap E_{2}\cap E_{3})}{P(X)}$

=$\frac{\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{4}}=\frac{7}{8}$

(c) $P\left(\frac{X}{X_{2}}\right)=\frac{P(X \cap X_{2})}{P(X_{2})}$

=P( ship is operating with $E_{2}$ function )/ $P{X_{2}}$

 =$\frac{P(E_{1}\cap E_{2}\cap\overline{E_{3}})+P(\overline{E_{1}}\cap E_{2}\cap{E_{3}})+P(E_{1}\cap E_{2}\cap{E_{3}})}{P(E_{2})}$

$=\frac{\frac{1}{2}.\frac{1}{4}.\frac{3}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{4}}=\frac{5}{8}$

(d)   $P(X/X_{1})=\frac{P(X\cap X_{1})}{P(X_{1})}$

$=\frac{\frac{1}{2}.\frac{1}{4}.\frac{1}{4}+\frac{1}{2}.\frac{3}{4}.\frac{1}{4}+\frac{1}{2}.\frac{1}{4}.\frac{3}{4}}{\frac{1}{4}}=\frac{7}{16}$