Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $-$ k), the probability that exactly one of B and C occurs is (1 $-$ 2k), the probability that exactly one of C and A occurs is (1 $-$ k) and the probability of all A, B and C occur simultaneously is k2, where 0 < k < 1. Then the probability that at least one of A, B and C occur is :
Solution
$P(\overline A \cap B) + P(A \cap \overline B ) = 1 - k$<br><br>$P(\overline A \cap C) + P(A \cap \overline C ) = 1 - 2k$<br><br>$P(\overline B \cap C) + P(B \cap \overline C ) = 1 - k$<br><br>$P(A \cap B \cap C) = {k^2}$<br><br>$P(A) + P(B) - 2P(A \cap B) = 1 - k$ .....(i)<br><br>$P(B) + P(C) - 2P(B \cap C) = 1 - k$ ..... (ii)<br><br>$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$ ..... (iii)<br><br>$(i) + (ii) + (iii)$<br><br>$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}$$<br><br>So,<br><br>$P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}$<br><br>$P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}$<br><br>$= {{2{{(k - 1)}^2} + 1} \over 2}$<br><br>$P(A \cup B \cup C) > {1 \over 2}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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