Let E 1 , E 2 , E 3 be three mutually exclusive events such that P E 1 = 2 + 3 p 6 , P E 2 = 2 - p 8 and P E 3 = 1 - p 2 . If the maximum and minimum values of p are p 1 and p 2 then p 1 + p 2 is equal to:

Step-by-step solution

We know that, 0 ≤ P E i ≤ 1 for i = 1 , 2 , 3 0 ≤ P E 1 ≤ 1 ⇒ 0 ≤ 2 + 3 P 6 ≤ 1 ⇒ P ∈ - 2 3 , 4 3   . . . . . . . . i Now for 0 ≤ P E 2 ≤ 1 ⇒ 0 ≤ 2 - P 8 ≤ 1 ⇒ P ∈ - 6 , 2 . . . . . . . . i i Now for 0 ≤ P E 3 ≤ 1 ⇒ 0 ≤ 1 - P 2 ≤ 1 ⇒ P ∈ - 1 , 1 . . . . . . . i i i Also, E 1 and E 2 and E 3 are mutually exclusive P E 1 + P E 2 + P E 3 ≤ 1 ⇒ 0 ≤ 13 12 - P 8