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Let \(A\) and \(B\) are two independent events. The probability that both \(A\) and \(B\) happen is \(\frac{1}{12}\) and probability that neither A nor \(B\) happen is \(\frac{1}{2}\). Then
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The correct answer is:
\(P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\)
\(\begin{array}{ll}
\text {Hint : } P(A \cap B)=\frac{1}{12} & \Rightarrow P(A) P(B)=\frac{1}{12} \cdots (1) \\
P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{2} & \Rightarrow P(A \cup B)=1-\frac{1}{2}=\frac{1}{2}
\end{array}\)
\(\begin{aligned} & \Rightarrow P(A)+P(B)-P(A) P(B)=\frac{1}{2} \\ & \Rightarrow P(A)+P(B)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\end{aligned}\)
\(\begin{aligned} & \text { From (1), } \quad P(A)\left(\frac{7}{12}-P(A)\right)=\frac{1}{12} \\ & \Rightarrow P(A)^2-\frac{7}{12} P(A)+\frac{1}{12}=0 \\ & \Rightarrow 12 P(A)^2-7 P(A)+1=0 \\ & \Rightarrow 12 P(A)^2-4 P(A)-3 P(A)+1=0 \\ & \Rightarrow(3 P(A)-1)-(4 P(A)-1)=0\end{aligned}\)
\(\begin{aligned} & \therefore P(A)=\frac{1}{3} \text { or } \quad P(A)=\frac{1}{4} \\ & \therefore \quad P(B)=\frac{1}{12} \times 3 \Rightarrow P(B)=\frac{1}{12} \times 4 \\ & \qquad\qquad=\frac{1}{4} \qquad \qquad \qquad=\frac{1}{3} \\ & \therefore P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\end{aligned}\)
\text {Hint : } P(A \cap B)=\frac{1}{12} & \Rightarrow P(A) P(B)=\frac{1}{12} \cdots (1) \\
P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{2} & \Rightarrow P(A \cup B)=1-\frac{1}{2}=\frac{1}{2}
\end{array}\)
\(\begin{aligned} & \Rightarrow P(A)+P(B)-P(A) P(B)=\frac{1}{2} \\ & \Rightarrow P(A)+P(B)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\end{aligned}\)
\(\begin{aligned} & \text { From (1), } \quad P(A)\left(\frac{7}{12}-P(A)\right)=\frac{1}{12} \\ & \Rightarrow P(A)^2-\frac{7}{12} P(A)+\frac{1}{12}=0 \\ & \Rightarrow 12 P(A)^2-7 P(A)+1=0 \\ & \Rightarrow 12 P(A)^2-4 P(A)-3 P(A)+1=0 \\ & \Rightarrow(3 P(A)-1)-(4 P(A)-1)=0\end{aligned}\)
\(\begin{aligned} & \therefore P(A)=\frac{1}{3} \text { or } \quad P(A)=\frac{1}{4} \\ & \therefore \quad P(B)=\frac{1}{12} \times 3 \Rightarrow P(B)=\frac{1}{12} \times 4 \\ & \qquad\qquad=\frac{1}{4} \qquad \qquad \qquad=\frac{1}{3} \\ & \therefore P(A)=\frac{1}{3}, P(B)=\frac{1}{4}\end{aligned}\)
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