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Questions › GATE › GATE 2021 › Q21
2 marks, −0.667Statistics › Probability and Bayes' theorem › Addition rule, complements and inclusion-exclusionHardabout 3 min

GATE 2021 Question 21: Probability and Bayes' theorem

Let AAA and BBB be two events with probabilities P(A)=34P(A)=\dfrac34P(A)=43​ and P(B)=13P(B)=\dfrac13P(B)=31​; then which of the following options is true?

Tap an option to check your answer.

Answer and worked solution

Answer: (B)

For any two events, the probability of their intersection is bounded as follows, without further assumptions about independence or the specific relationship between AAA and BBB:

Upper bound: P(A∩B)≤min⁡[P(A),P(B)]P(A\cap B) \le \min[P(A),P(B)]P(A∩B)≤min[P(A),P(B)], since the intersection can never be more probable than the less likely of the two events on its own.

min⁡[34,13]=13\min\left[\frac34,\frac13\right] = \frac13min[43​,31​]=31​

Lower bound (Bonferroni inequality): P(A∩B)≥max⁡[0, P(A)+P(B)−1]P(A\cap B) \ge \max[0,\,P(A)+P(B)-1]P(A∩B)≥max[0,P(A)+P(B)−1].

P(A)+P(B)−1=34+13−1=912+412−1212=112P(A)+P(B)-1 = \frac34+\frac13-1 = \frac{9}{12}+\frac{4}{12}-\frac{12}{12} = \frac{1}{12}P(A)+P(B)−1=43​+31​−1=129​+124​−1212​=121​

Since 112>0\dfrac{1}{12}>0121​>0, the lower bound is 112\dfrac{1}{12}121​.

So 112≤P(A∩B)≤13\dfrac{1}{12} \le P(A\cap B) \le \dfrac13121​≤P(A∩B)≤31​, which matches option (B) exactly (written as 13≥P(A∩B)≥112\dfrac13\ge P(A\cap B)\ge\dfrac{1}{12}31​≥P(A∩B)≥121​).

The trap: (D) P(A∩B)≥min⁡[34,13]P(A\cap B) \ge \min\left[\dfrac34,\dfrac13\right]P(A∩B)≥min[43​,31​]

Options (A) and (D) present plausible-sounding but incorrect max/min-based lower bounds that don't match the actual Bonferroni inequality — a student who recalls 'the intersection is bounded by max or min of the two probabilities' loosely, without the precise formula (max[0, P(A)+P(B)-1] for the lower bound, min[P(A),P(B)] for the upper), may pick one of these structurally-similar-looking wrong options.

Plain-text version

Let A and B be two events with probabilities P(A)=3/4 and P(B)=1/3; then which of the following options is true? (A) P(A∩ B) ≥ max [3/4,1/3] (B) 1/3 ≥ P(A∩ B) ≥ 1/12 (C) 3/4 ≥ P(A∩ B) ≥ 1/3 (D) P(A∩ B) ≥ min [3/4,1/3]

Probability and Bayes' theorem has come up 190 times in 7 exams since 2004.

What this question tests

This question is on Addition rule, complements and inclusion-exclusion, part of Probability and Bayes' theorem in Statistics.

The probability of at least one of several events, and of none.

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