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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
Solution
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
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