If $A$ and $B$ are two events such that $P(A\cap B)=0.1$, and $P(A\mid B)$ and $P(B\mid A)$ are the roots of the equation $12x^2-7x+1=0$, then the value of $\frac{P(\bar A\cup\bar B)}{P(\bar A\cap\bar B)}$ is :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The roots are $\frac13,\frac14$: $\frac{0.1}{P(B)}\cdot\frac{0.1}{P(A)}=\frac1{12}$ gives $P(A)P(B)=0.12$, and $0.1\left(\frac1{P(A)}+\frac1{P(B)}\right)=\frac7{12}$ gives $P(A)+P(B)=0.7$. So $P(\bar A\cup\bar B)=1-0.1=0.9$ and $P(\bar A\cap\bar B)=1-(0.7-0.1)=0.4$. Ratio $=\frac94$. Answer (4).

Q. If $A$ and $B$ are two events such that $P(A\cap B)=0.1$, and $P(A\mid B)$ and $P(B\mid A)$ are the roots of the equation $12x^2-7x+1=0$, then the value of $\frac{P(\bar A\cup\bar B)}{P(\bar A\cap\bar B)}$ is :

Explanation

Reviewed by Dr. Abubacker

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