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