A) \[0.23\le x\le 0.48\]
B) \[0.23\le x\le 0.47\]
C) \[0.22\le x\le 0.48\]
D) None of these
Correct Answer: A
Solution :
[a] Since \[P(A\cup B\cup C)=P(A)+P(B)+P(C)\] \[-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)\] Or \[P(A\cup B\cup C)=0.3+0.4+0.8-(0.08+0.28)\] \[+P(B\cap C)+0.09\] \[F=1.23-P(B\cap C)\] Or \[P(B\cap C)=1.23-P(A\cup B\cup C)\] But we know that \[0\le P(A\cup B\cup C)\,\,1\] Hence \[0.23\le P(B\cap C)\le 0.48\]You need to login to perform this action.
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