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 $X$ denotes 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. $P [ X_1^c | X ] = \frac {3} {16}$
• B. $P$ [exactly two engines of the ship are functioning] $= \frac{7}{8}$
• C. $P[ X|X _2]= \frac{5}{16}$
• D. $P[ X|X _1]= \frac{7}{16}$

Answer 1

Answer: (D)

$P[ X|X _1]= \frac{7}{16}$

Answer 2

PLAN It is based on law of to ta l probability and Bay 's Law Description of Situation It is given that ship would
work if atleast two of engines must work. If X be 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
Given , $P(E_1) = \frac{1}{2} , P(E_2) = \frac{1}{4} ,P(E_3) = \frac{1}{4}$
\therefore \ \ \ \ \ \ \ \ \ P(X) = \bigg \\{ \begin{array} \ 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) \\\ \end{array} \bigg \\}.
$= \bigg ( \frac{1}{2} . \frac{1}{4} . \frac{3}{4} + \frac{1}{2}. \frac{3}{4}. \frac{1}{4} + \frac{1}{2}. \frac{1}{4}. \frac{1}{4} \bigg ) + \bigg( \frac{1}{2}. \frac{1}{4}. \frac{1}{4} \bigg) \frac{1}{4}$
Now , $(a) P(X_1^c | X)$
$P\bigg( \frac{ X_1^c \cap X}{P(X)} = \frac{P(\overline {E_1} \cap E_2 \cap E_3 )}{ P(X)} = \frac{ \frac{ 1}{2} . \frac{ 1}{4}. \frac{ 1}{4}}{1} = \frac{ 1}{8}$
$( c ) P \bigg( \frac{ X}{X_2} = \frac{ P(X \cap X_2)}{ P( X_2)}$
$= \frac{ P (ship \ is \ operating \ with E _2 \ function)}{P(X_2)}$
$= \frac{ P(E_1 \cap E_2 \cap \overline{E_3}) + P( \overline{E_1} \cap E_2 \cap E_3) + P(E_1 \cap E_2 \cap E_3) }{ P( E _2)}$
$\frac{ \frac{1}{2}. \frac{1}{4}. \frac{3}{4}+ \frac{1}{2}. \frac{1}{4}. \frac{1}{4}+ \frac{1}{2}. \frac{1}{4}. \frac{1}{4} }{ \frac{1}{4}} = \frac{5}{8}$
(d) $(PX / X_1) = \frac{ P (X \cap X_1)}{PX_1}$
$\frac{ \frac{1}{2}. \frac{1}{4}. \frac{1}{4}+ \frac{1}{2}. \frac{3}{4}. \frac{1}{4}+ \frac{1}{2}. \frac{1}{4}. \frac{3}{4}}{ \frac{1}{2}} = \frac{7}{16}$