Probability 4 Question 5

5. 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 the 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: 5. (b, d)

Solution:

  1. PLAN It is based on law of total 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, $P\left(E _1\right)=\frac{1}{2}, P\left(E _2\right)=\frac{1}{4}, P\left(E _3\right)=\frac{1}{4}$

$$ \begin{aligned} \therefore \quad P(X)= & P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(E _1 \cap \bar{E} _2 \cap E _3\right) \\ & +P\left(\bar{E} _1 \cap E _2 \cap E _3\right)+P\left(E _1 \cap E _2 \cap E _3\right) \\ = & \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} \\ = & 1 / 4 \end{aligned} $$

Now, (a) $P\left(X _1^{c} / X\right)$

$$ =P \frac{X _1^{c} \cap X}{P(X)}=\frac{P\left(\bar{E} _1 \cap E _2 \cap E _3\right)}{P(X)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{1}{8} $$

(b) $P$ (exactly two engines of the ship are functioning) $=\frac{P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(E _1 \cap \bar{E} _2 \cap E _3\right)+P\left(\bar{E} _1 \cap E _2 \cap E _3\right)}{P(X)}$

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

(c) $P \frac{X}{X _2}=\frac{P\left(X \cap X _2\right)}{P\left(X _2\right)}$

$=\frac{P\left(\text { ship is operating with } E _2 \text { function }\right)}{P\left(X _2\right)}$

$$ \begin{aligned} & =\frac{P\left(E _1 \cap E _2 \cap \bar{E} _3\right)+P\left(\bar{E} _1 \cap E _2 \cap E _3\right)+P\left(E _1 \cap E _2 \cap E _3\right)}{P\left(E _2\right)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}}=\frac{5}{8} \\ & \text { (d) } P\left(X / X _1\right)=\frac{P\left(X \cap X _1\right)}{P\left(X _1\right)}=\frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4}+\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{1 / 2} \\ & =\frac{7}{16} \end{aligned} $$



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