Let A, B, C be three events. If the probability of occurring exactly one out of A and B is \(\displaystyle\frac{3}{5}\), exactly one out of B and C is \(\displaystyle\frac{1}{5}\), exactly one out of C and A is \(\displaystyle\frac{3}{5}\) and that of occurring of three events is \(\displaystyle\frac{4}{25}\), then the probability of occurring at least one of them is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{43}{50}\)
The probability that exactly one of A and B occurs is \(P(A \cap B') + P(A' \cap B)\):
\( \displaystyle P(A) + P(B) - 2P(A \cap B) = \frac{3}{5} \)
In the same way:
\( \displaystyle P(B) + P(C) - 2P(B \cap C) = \frac{1}{5} \)
\( \displaystyle P(C) + P(A) - 2P(C \cap A) = \frac{3}{5} \)
Adding the three equations:
\( \displaystyle 2\left[P(A) + P(B) + P(C)\right] - 2\left[P(A \cap B) + P(B \cap C) + P(C \cap A)\right] = \frac{7}{5} \)
Dividing by 2:
\( \displaystyle \left[P(A) + P(B) + P(C)\right] - \left[P(A \cap B) + P(B \cap C) + P(C \cap A)\right] = \frac{7}{10} \)
By the addition theorem for three events:
\( \displaystyle P(A \cup B \cup C) = \frac{7}{10} + P(A \cap B \cap C) \)
\( \displaystyle = \frac{7}{10} + \frac{4}{25} = \frac{35}{50} + \frac{8}{50} = \frac{43}{50} \)