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प्रश्न
An experiment has four possible outcomes A, B, C and D, that are mutually exclusive. Explain why the following assignments of probabilities are not permissible:
P(A) = `9/120`, P(B) = `45/120`, P(C) = `27/120`, P(D) = `46/120`
उत्तर
P(S) = P(A ∪ B ∪ C ∪ D)
= `9/120 + 45/120 + 27/120 + 46/120`
= `127/120 ≠ 1`
This violates the condition that P(S) = 1.
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