SATHEE: Probability 4 Question 5

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) $P [X_{1}^{c} ∣ X] = 3 / 16$

(b) $P$ [exactly two engines of the ship are functioning] $= \frac{7}{8}$

(d) $P [X ∣ X_{1}] = \frac{7}{16}$

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

(d) Statement I is false; Statement II is true

Show Answer

Answer:

Correct Answer: 5. (b, d)

Solution:

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

$\begin{aligned} ∴ P (X) = & P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) \\ + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3}) \\ = & \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} \\ = & 1 / 4 \end{aligned}$

Now, (a) $P (X_{1}^{c} / X)$

$= P \frac{X_{1}^{c} \cap X}{P (X)} = \frac{P ({\bar{E}}_{1} \cap E_{2} \cap E_{3})}{P (X)} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{1}{8}$

(b) $P$ (exactly two engines of the ship are functioning) $= \frac{P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P (E_{1} \cap {\bar{E}}_{2} \cap E_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3})}{P (X)}$

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

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

$\begin{aligned} = \frac{P (E_{1} \cap E_{2} \cap {\bar{E}}_{3}) + P ({\bar{E}}_{1} \cap E_{2} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3})}{P (E_{2})} \\ = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4}}{\frac{1}{4}} = \frac{5}{8} \\ (d) P (X / X_{1}) = \frac{P (X \cap X_{1})}{P (X_{1})} = \frac{\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4}}{1 / 2} \\ = \frac{7}{16} \end{aligned}$

Probability 4 Question 5

Assertion and Reason

Answer:

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Probability 4 Question 5

Assertion and Reason

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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