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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)  <img alt="=\frac{1}{4}" src="https://learn.careers360.com/latex-image/?%5Cdpi%7B100%7D%20%3D%5Cfrac%7B1%7D%7B4%7

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)  =\frac{1}{4}   and P(All the three events occur
simultaneously)  =\frac{1}{16} Then the probability that at least one of the events occurs, is :  
Option: 1 \frac{7}{16}

Option: 2 \frac{7}{64}

Option: 3 \frac{3}{16}

Option: 4 \frac{7}{32}
 

Answers (1)

best_answer

P (exactly one of A or B) =P(exactly B or C)

=P (exactly one of A or C) = \frac{1}{4}

P(A) + P (B) - 2P (A\capB)=\frac{1}{4}

P(B) +P (C) - 2P (B\capC)=\frac{1}{4}

P(C)+P(A)-2P (C\capA)=\frac{1}{4}

P (A) + P (B) -2 P (A\capB) - P(B\capC)-P(C\capA) =\frac{3}{8}

P(A\capB\capC)=\frac{1}{16}

P(A\capB\capC)=\frac{3}{8}+\frac{1}{16}=\frac{4}{16}

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vishal kumar

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