A) \[\frac{3}{16}\]
B) \[\frac{7}{32}\]
C) \[\frac{7}{16}\]
D) \[\frac{7}{64}\]
Correct Answer: C
Solution :
P (exactly one of A or B occurs) \[=P(A)+P(B)-2P(A\cap B)=\frac{1}{4}\] P(Exactly one of B or C occurs) \[=P(B)+P(C)-2P(B\cap C)=\frac{1}{4}\] P(Exactly one of C or A occurs) \[=P(C)+P(A)-2P(C\cap A)=\frac{1}{4}\] Adding all, we get \[2\sum P(A)-2\sum P(A\cap B)=\frac{3}{4}\] \[\therefore \]\[\sum P(A)-\sum P(A\cap B)=\frac{3}{8}\] Now, \[P(A\cap B\cap C)=\frac{1}{16}\](given) \[\therefore \] \[P(A\cup B\cup C)\] \[=\sum P(A)-\sum P(A\cap B)+P(A\cap B\cap C)\] \[=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}\]You need to login to perform this action.
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