JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

The probabilities of three events A, B and C are given by
P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5.
If P(A∪B) = 0.8, P(A∩C) = 0.3, P(A∩B∩C) = 0.2, P(B∩C) = β
and P(A∪B∪C) = α, where 0.85 ≤ α ≤ 0.95, then β lies in the interval :

  1. (a) [0.35, 0.36]
  2. (b) [0.20, 0.25]
  3. (c) [0.25, 0.35]
  4. (d) [0.36, 0.40]
Answer and solution

Answer (c) [0.25, 0.35]

Solution

Maths sir · 4 steps

Get α as a function of β by inclusion–exclusion, then use the range of α.

  1. Missing intersection. From P(A ∪ B) = P(A) + P(B) - P(A ∩ B) we get 0.8 = 0.6 + 0.4 - P(A ∩ B), so P(A ∩ B) = 0.2.

    P(A ∪ B ∪ C) = P(A)+P(B)+P(C) - P(A∩ B) - P(B∩ C) - P(A∩ C) + P(A∩ B∩ C)

  2. Put in numbers. α = 0.6 + 0.4 + 0.5 - 0.2 - β - 0.3 + 0.2 = 1.2 - β.
  3. Use the range. 0.85 ≤ 1.2 - β ≤ 0.95. Subtract 1.2 and multiply by -1, which flips both inequality signs: 0.25 ≤ β ≤ 0.35.
  4. So. β lies in [0.25, 0.35], which is option (c). It also satisfies β ≥ P(A ∩ B ∩ C) = 0.2, so it is consistent.

The idea to keep. When one pairwise intersection is missing, get it from the two-event union first. Then inclusion–exclusion makes the three-event union a simple linear expression in the unknown.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Probability — Basic

Asked about 80 times in JEE Main since 2020. This question is from 2020.