Events A, B, C are mutually exclusive events such that P(A) = (3x + 1)/3, P(B) = (x - 1)/4, P(C) = (1 - 2x)/4

• Probability

Events A, B, C are mutually exclusive events such that P(A) = (3x + 1)/3, P(B) = (x - 1)/4, P(C) = (1 - 2x)/4. The set of possible values of x are in the interval

1. [1/3, 1/2]
2. [1/3, 13/3]
3. [0, 1]
4. [1/3, 2/3]

Answer

P(A) = (3x + 1)/3

P(B) = (x - 1)/4

P(C) = (1 - 2x)/4

These are mutually exclusive events.

-1 ≤ 3x ≤ 2,  -3 ≤ x ≤ 1,  -1 ≤ 2x ≤ 1

-1/3 ≤ x ≤ 2/3,  -2 ≤ x ≤ 1,  -1/2 ≤ x ≤ 1/2

Also, 0 ≤ (3x + 1)/3 + (x - 1)/4 + (1 - 2x)/4 ≤ 1

1/3 ≤ x ≤ 13/3

max{-1/3, -3, -1/2, 1/3} ≤ x ≤  min{2/3, 1/2, 1, 13/3}

1/3 ≤ x ≤ 1/2

The correct option is A.