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) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D)= 0.12

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) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12

Now P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1

0.15 + 0.30 + 0.43 + 0.12 = 1

∴ The assignment of probability is permissible