Question

Mark the correct alternative in the following question:

If A and B are two events such that P(A) = $\frac{4}{5}$, and $\mathrm{P}\left(A\cap B\right)=\frac{7}{10}$, then P(B|A) =

$\left(\mathrm{a}\right)\frac{1}{10}\left(\mathrm{b}\right)\frac{1}{8}\left(\mathrm{c}\right)\frac{7}{8}\left(\mathrm{d}\right)\frac{17}{20}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{4}{5}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{7}{10} \\ \mathrm{Now}, \\ \mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)} \\ =\frac{\left({\displaystyle \frac{7}{10}}\right)}{\left({\displaystyle \frac{4}{5}}\right)} \\ =\frac{7\times 5}{10\times 4} \\ =\frac{7}{8} \end{gathered}$$

Hence, the correct alternative is option (c).