P and Q are considering to apply for a job. The probability that P applies for the job is (1)/(4), the probability that P applies for the job given that Q applies for the job is (1)/(2), and the probability that Q applies for the job given that P applies for the job is (1)/(3). Then the probability that P does not apply for the job given that Q does not apply for the job is A) (4)/(5) B) (5)/(6) C) (7)/(8) D) (11)/(12)

Official answer: A) (4)/(5). Explanation: Let P and Q denote the events that P and Q apply for the job, respectively. Given: - P(P) = (1)/(4) - P(P Q) = (1)/(2) - P(Q P) = (1)/(3) From conditional probability: P(P ∩ Q) = P(Q P) · P(P) = (1)/(3) × (1)/(4) = (1)/(12) Using P(P Q) = (P(P ∩ Q))/(P(Q)): P(Q) = (P(P ∩ Q))/(P(P Q)) = (1/12)/(1/2) = (1)/(6) Now, calculate the probability of the union: P(P ∪ Q) = P(P) + P(Q) - P(P ∩ Q) = (1)/(4) + (1)/(6) - (1)/(12) = (3 + 2 - 1)/(12) = (4)/(12) = (1)/(3) We need to find P(Pc Qc): P(Pc Qc) = (P(Pc ∩ Qc))/(P(Qc)) = (1 - P(P ∪ Q))/(1 - P(Q)) = 1 - (1)/(3)1 - (1)/(6) = (2)/(3)(5)/(6) = (2)/(3) × (6)/(5) = (4)/(5) Therefore, option A is the correct answer.

Probability and Statistics

GATE CSE 2017 Set 2 · Q26

MCQ2 marksMedium

Time00:00

P and Q are considering to apply for a job. The probability that P applies for the job is $\frac{1}{4}$ , the probability that P applies for the job given that Q applies for the job is $\frac{1}{2}$ , and the probability that Q applies for the job given that P applies for the job is $\frac{1}{3}$ . Then the probability that P does not apply for the job given that Q does not apply for the job is

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Solution

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Official answer: A) (4)/(5)

Probability and Statistics

Let $P$ and $Q$ denote the events that P and Q apply for the job, respectively.

Given:

$P(P) = \frac{1}{4}$
$P(P \mid Q) = \frac{1}{2}$
$P(Q \mid P) = \frac{1}{3}$

From conditional probability:

P(P \cap Q) = P(Q \mid P) \cdot P(P) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}

Using $P(P \mid Q) = \frac{P(P \cap Q)}{P(Q)}$ :

P(Q) = \frac{P(P \cap Q)}{P(P \mid Q)} = \frac{1/12}{1/2} = \frac{1}{6}

Now, calculate the probability of the union:

P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12} = \frac{3 + 2 - 1}{12} = \frac{4}{12} = \frac{1}{3}

We need to find $P(P^c \mid Q^c)$ :

P(P^c \mid Q^c) = \frac{P(P^c \cap Q^c)}{P(Q^c)} = \frac{1 - P(P \cup Q)}{1 - P(Q)} = \frac{1 - \frac{1}{3}}{1 - \frac{1}{6}} = \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{2}{3} \times \frac{6}{5} = \frac{4}{5}

Therefore, option A is the correct answer.

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This Probability and Statistics previous year question is Q26 of the GATE CSE 2017 Set 2 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.