Let E₁, E₂, E₃ be three mutually exclusive events such that P(E₁)=(2+3 p)/6, P(E₂)=(2-p)/8 and P(E₃)=(1-p)/2. If the maximum and minimum values of p are p₁ and p₂, then (p₁+p₂) is equal to :

Solution

Maths sir · 3 steps

Each probability must lie in [0, 1], and the three together cannot exceed 1. Intersect all of them.

1. Two kinds of condition. Each P(Eᵢ) must satisfy 0 ≤ P(Eᵢ) ≤ 1. The events are mutually exclusive, so P(E₁ ∪ E₂ ∪ E₃) is their sum, and that sum must be at most 1.

   (2+3p)/6+(2-p)/8+(1-p)/2 = (4(2+3p)+3(2-p)+12(1-p))/24 = (26-3p)/24

   Condition	Gives
   0 ≤ (2+3p)/6 ≤ 1	-2/3 ≤ p ≤ 4/3
   0 ≤ (2-p)/8 ≤ 1	-6 ≤ p ≤ 2
   0 ≤ (1-p)/2 ≤ 1	-1 ≤ p ≤ 1
   (26-3p)/24 ≤ 1	26 - 3p ≤ 24, so p ≥ 2/3

2. Intersect. All four must hold at once. The tightest upper bound is p ≤ 1 and the tightest lower bound is p ≥ 2/3, so 2/3 ≤ p ≤ 1.
3. So. p₁ = 1 and p₂ = 2/3, giving p₁ + p₂ = 5/3, which is option (b).

The idea to keep. For mutually exclusive events with probabilities in terms of a parameter, put each one in [0, 1] and also make the total at most 1. The sum condition is usually the one that sets a bound.

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