Question

Mark the correct alternative in the following question:

$$\begin{gathered} \mathrm{If}A\mathrm{and}B\mathrm{are}\mathrm{two}\mathrm{events}\mathrm{such}\mathrm{that}\mathrm{P}\left(A\right)=\frac{1}{2},\mathrm{P}\left(B\right)=\frac{1}{3},\mathrm{P}\left(A|B\right)=\frac{1}{4},\mathrm{then}\mathrm{P}\left(\overline{A}\cap \overline{B}\right)\mathrm{equals} \\ \left(\mathrm{a}\right)\frac{1}{12}\left(\mathrm{b}\right)\frac{3}{4}\left(\mathrm{c}\right)\frac{1}{4}\left(\mathrm{d}\right)\frac{3}{16} \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{1}{2},\mathrm{P}\left(B\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left(A|B\right)=\frac{1}{4} \\ \mathrm{As},\mathrm{P}\left(A|B\right)=\frac{1}{4} \\ \Rightarrow \frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{1}{4} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{4}\times \mathrm{P}\left(B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{4}\times \frac{1}{3} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{12} \\ \mathrm{Also},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ =\frac{1}{2}+\frac{1}{3}-\frac{1}{12} \\ =\frac{6+4-1}{12} \\ =\frac{9}{12} \\ =\frac{3}{4} \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}\cap \overline{B}\right)=\mathrm{P}\left(\overline{A\cup B}\right) \\ =1-\mathrm{P}\left(A\cup B\right) \\ =1-\frac{3}{4} \\ =\frac{4-3}{4} \\ =\frac{1}{4} \end{gathered}$$

Hence, the correct alternative is option (c).