Question

An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C, and D. Check whether the following assignments of probability are permissible.

P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34

Answer 1

When A, B, C, D are the possible exclusive and exhaustive events the P(A) + P(B) + P(C) + P(D) = 1.

P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34

Now P(A) + P(B) + P(C) + P(D) = 1

0.22 + 0.38 + 0.16 + 0.34 = 1.10 = ≠1

∴ The assignment of probability is not permissible.