A) 1/2
B) ¼
C) 0
D) None of these
Correct Answer: B
Solution :
Given, \[P({{A}^{c}})=0.3\Rightarrow P(A)=0.7\] and \[P(B)=0.4\Rightarrow P({{B}^{c}})=0.6\] Also, \[P(A\cap {{B}^{c}})=0.5\] (given) Now, \[P(A\cup {{B}^{c}})=P(A)+P({{B}^{c}})-P(A\cap {{B}^{c}})\] \[=0.7+0.6-0.5\] \[=0.8\] Again, \[P(B/A\cup {{B}^{c}})=\frac{P[B\cap (A\cup {{B}^{c}})]}{P(A\cup {{B}^{c}})}\] \[=\frac{P[(B\cap A)\cup (B\cap {{B}^{c}})]}{0.8}\] \[=\frac{P[(B\cup A)\cup \phi ]}{0.8}=\frac{P(B\cap A)}{0.8}\] \[=\frac{1}{0.8}[P(A)-P(A\cap {{B}^{c}})]\] \[=\frac{0.7-0.5}{0.8}=\frac{0.2}{0.8}=\frac{1}{4}\]
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