Probability Ques 115
- A ship is fitted 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 atleast 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?
(2012)
(a) $P\left[X _1^{c} \mid X\right]=3 / 16$
(b) $P$ [exactly two engines of the ship are functioning] $=\frac{7}{8}$
(c) $P\left[X \mid X _2\right]=\frac{5}{16}$
(d) $P\left[X \mid X _1\right]=\frac{7}{16}$
Assertion and Reason
For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II is a correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
Show Answer
Answer:
Correct Answer: 115.(b, d)
Solution:
Formula:
PLAN It is based on law of total probability and Bayes’ 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, $P\left(E _1\right)=\frac{1}{2}, P\left(E _2\right)=\frac{1}{4}, P\left(E _3\right)=\frac{1}{4}$
$$ \begin{aligned} \therefore \quad P(X)= & P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(E _1 \cap \bar{E} _2 \cap E _3\right) \\ & +P\left(\bar{E} _1 \cap E _2 \cap E _3\right)+P\left(E _1 \cap E _2 \cap E _3\right) \\ = & \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} \\ = & 1/4 \end{aligned} $$
Now, (a) $P\left(X_1^{c} \mid X\right)$
$$ =P \frac{X _1^{c} \cap X}{P(X)}=\frac{P\left(\bar{E} _1 \cap E _2 \cap E _3\right)}{P(X)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8} $$
(b) $P$ (exactly two engines of the ship are functioning) $=\frac{P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(E _1 \cap \bar{E} _2 \cap E _3\right)+P\left(\bar{E} _1 \cap E _2 \cap E _3\right)}{P(E_1 \cap E_2 \cap \bar{E}_3 \cup E_1 \cap \bar{E}_2 \cap E_3 \cup \bar{E}_1 \cap E_2 \cap E_3)}$
$=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{7}{8}$
(c) $P \frac{X}{X _2}=\frac{P\left(X \cap X _2\right)}{P\left(X _2\right)}$
$=\frac{P\left(\text { ship is operating with } E _2 \text { function }\right)}{P\left(X _2\right)}$
$$ \begin{aligned} & =\frac{P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(\bar{E} _1 \cap E _2 \cap E _3\right)+P\left(E _1 \cap E _2 \cap E _3\right)}{P\left(E _2\right)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{5}{8} \\ & \text { (d) } P\left(X \mid X_1\right)=\frac{P\left(X \cap X_1\right)}{P\left(X_1\right)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{1 / 2} \\ & =\frac{7}{16} \end{aligned} $$
BETA
