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If A, B, C are any three events such that the probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P(A \cap B) = \frac{1}{6}, P(B \cap C) = \frac{1}{4}, P(A \cap C) = \frac{1}{8}, P(A \cup B \cup C) = \frac{9}{10}, P(A \cap B \cap C) = \frac{1}{15}, then find P(A), P(B) and P(C)?
Answer:
P(A) =
P(B) =
P(C) =
[Note: Enter numerator and denominator in simplified form.]