If A and B are two events such that P(A∩B)=0.1,

Question

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

(1) \(\frac{5}{3}\)

(2) \(\frac{4}{3}\)

(3) \(\frac{9}{4}\)

(4) \(\frac{7}{4}\)
 

Answer 1

Correct option is (3) \(\frac{9}{4}\) 

\(12 \mathrm{x}^{2}-7 \mathrm{x}+1=0\)

\(\mathrm{x}=\frac{1}{3}, \frac{1}{4}\)

Let \(\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1}{3} \ \&\ \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{1}{4}\)

\(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{1}{3}\ \&\ \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{1}{4}\)

\(\Rightarrow \mathrm{P}(\mathrm{B})=0.3\)

\(\&\ \mathrm{P}(\mathrm{A})=0.4\)

\(\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})\)

= 0.3 + 0.4 - 0.1 = 0.6

Now \(\frac{\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}=\frac{\mathrm{P}(\overline{\mathrm{A} \cap \mathrm{B}})}{\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})}\)

\(=\frac{1-P(A \cap B)}{1-P(A \cup B)}=\frac{1-0.1}{1-0.6}=\frac{9}{4}\)