Question

For three events A B and C if P (exactly one of A or B occurs) = P (exactly one of B or C occurs) = P (exactly one of C or A occurs) = 1/4 and P (all the three events occur simultaneously) = 1/16, then the probability that at least one of the events occurs is ​

Answer 1

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For three events A, B and C, P(Exactly one of A or B occurs)=P(Exactly one of B or C occurs)=P(Exactly one of C or A occurs)=1/4

hope it helps you

Answer 2

$\sf \red{We \: have}$

$\sf{P \: (exactly \: one \: of \: A \: or \: B \: occurs) }$

$\sf{= P( A \cup B) - P (A \cap B)}$

$\sf \: = P (A) + P (B) - 2P( A \cap B)$

$\sf\red{ According \: to \: the \: question,}$

$\boxed{ \boxed{P(A) + P(B) - 2P(A \cap B) = \dfrac{1}{4}}} - - - (1)$

$\boxed{ \boxed{P(B) + P(C) - 2P(B\cap C) = \dfrac{1}{4}}} - - - (2)$

$\boxed{ \boxed{P(C) + P(A) - 2P(C\cap A) = \dfrac{1}{4}}} - - - (3)$

$\sf \red{On \: adding \: Eqs. 1, 2 \: and \: 3, \: we \: get}$

$\sf{2 \bigg[P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) \bigg] = \dfrac{3}{4}}$

$\sf{ \implies P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = \dfrac{3}{8}}$

$\rm{\therefore \: P \: (atleast \: one \: event \: occurs)}$

$\rm= P(A \cup B \cup C)$

$\sf{ = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)}$

$\implies \dfrac{3}{8} + \dfrac{1}{16} = \dfrac{7}{16} \: \: \: \: \: \: \: \: \bigg[ P(A \cap B \cap C) = \dfrac{1}{16} \bigg]$