A) \[\left( \frac{5}{12},\frac{3}{4} \right)\]
B) \[\left( \frac{2}{3},\frac{3}{4} \right)\]
C) \[\left( \frac{1}{3},\frac{3}{4} \right)\]
D) none of these
Correct Answer: A
Solution :
Given, A and B are two events. Odds against A are 2 to 1, odds in favour of \[A\cup B\]are 3 to 1 also \[x\le P(B)\le y\] i.e., \[P(A)=\left( 1-\frac{2}{3} \right)=\frac{1}{3}\] and \[P(A\cup B)=\frac{3}{4}\] \[\therefore \]\[P(A\cup B)=P(A)P(B)-P(A\cap B)\] \[\Rightarrow \] \[\frac{3}{4}=\frac{1}{3}+P(B)-P(A\cap B)\] \[\Rightarrow \] \[P(A\cap B)=P(B)-\frac{5}{12}\] \[\Rightarrow \] \[P(B)\ge \frac{5}{12}\] Again, \[P(B)=\frac{5}{12}+P(A\cap B)\] \[\Rightarrow \] \[P(B)\le \frac{5}{12}+P(A)\] \[\Rightarrow \] \[P(B)\le \frac{3}{4}\] Hence, \[x\le P(B)\le y\] \[\therefore \] \[\frac{5}{12}\le P(B)\le \frac{3}{4}.\]
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