Probability
Probability
nta_pyq_2025_jan
Grade 12
Question:
$A$ 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 $\dfrac{P(\bar{A}\cup\bar{B})}{P(\bar{A}\cap\bar{B})}$ is:
$\dfrac{4}{3}$
$\dfrac{7}{4}$
$\dfrac{5}{3}$
$\dfrac{9}{4}$
Step-by-Step Solution
Key Concept: Sum of roots $=\tfrac{7}{12}$ and product $=\tfrac{1}{12}.$ Product $P(A|B)P(B|A)=\dfrac{P(A\cap B)^{2}}{P(A)P(B)}$ gives $P(A)P(B)$; sum gives $P(A)+P(B).$ Then use De Morgan: $P(\bar A\cup\bar B)=1-P(A\cap B)$ and $P(\bar A\cap\bar B)=1-P(A\cup B).$
Sum $=\dfrac{7}{12}$, product $=\dfrac{1}{12}.$
Product: $\dfrac{P(A\cap B)^{2}}{P(A)P(B)}=\dfrac{1}{12}\Rightarrow P(A)P(B)=12(0.1)^{2}=0.12.$
Sum: $P(A\cap B)\!\left[\dfrac{1}{P(A)}+\dfrac{1}{P(B)}\right]=\dfrac{7}{12}\Rightarrow 0.1\cdot\dfrac{P(A)+P(B)}{0.12}=\dfrac{7}{12}\Rightarrow P(A)+P(B)=0.7.$
$P(A\cup B)=0.7-0.1=0.6.$ $P(\bar A\cup\bar B)=1-P(A\cap B)=0.9.$ $P(\bar A\cap\bar B)=1-P(A\cup B)=0.4.$
Ratio $=\dfrac{0.9}{0.4}=\dfrac{9}{4}.$
Correct Answer: 4