Question

If A and B are two events such that

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{P}\left(A\right)=\frac{1}{3},\mathrm{P}\left(B\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{5}{12},\mathrm{then}\mathrm{find}\mathrm{P}\left(A|B\right)\mathrm{and}\mathrm{P}\left(B|A\right). \\ \left(\mathrm{ii}\right)\mathrm{P}\left(A\right)=\frac{6}{11},\mathrm{P}\left(B\right)=\frac{5}{11}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{7}{11},\mathrm{then}\mathrm{find}\mathrm{P}\left(A\cap B\right),\mathrm{P}\left(A|B\right)\mathrm{and}\mathrm{P}\left(B|A\right). \\ \left(\mathrm{iii}\right)\mathrm{P}\left(A\right)=\frac{7}{13},\mathrm{P}\left(B\right)=\frac{9}{13}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{4}{13},\mathrm{then}\mathrm{find}\mathrm{P}\left(\overline{A}|B\right). \\ \left(\mathrm{iv}\right)\mathrm{P}\left(A\right)=\frac{1}{2},\mathrm{P}\left(B\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{1}{4},\mathrm{then}\mathrm{find}\mathrm{P}\left(A|B\right),\mathrm{P}\left(B|A\right),\mathrm{P}\left(\overline{A}|B\right)\mathrm{and}\mathrm{P}\left(\overline{A}|\overline{\mathit{B}}\right). \end{gathered}$$
[NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{1}{3},\mathrm{P}\left(B\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{5}{12} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cup B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{3}+\frac{1}{4}-\frac{5}{12} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{2}{12} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{1}{6} \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left({\displaystyle \frac{1}{6}}\right)}{\left({\displaystyle \frac{1}{4}}\right)}=\frac{4}{6}=\frac{2}{3}\mathrm{and} \\ \mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left({\displaystyle \frac{1}{6}}\right)}{\left({\displaystyle \frac{1}{3}}\right)}=\frac{3}{6}=\frac{1}{2} \\ \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{6}{11},\mathrm{P}\left(B\right)=\frac{5}{11}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{7}{11} \\ \mathrm{As},\mathrm{P}\left(A\cup B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cup B\right) \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{6+5-7}{11} \\ \Rightarrow \mathrm{P}\left(A\cap B\right)=\frac{4}{11} \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left({\displaystyle \frac{4}{11}}\right)}{\left({\displaystyle \frac{5}{11}}\right)}=\frac{4}{5}\mathrm{and} \\ \mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left({\displaystyle \frac{4}{11}}\right)}{\left({\displaystyle \frac{6}{11}}\right)}=\frac{4}{6}=\frac{2}{3} \\ \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ \mathrm{P}\left(A\right)=\frac{7}{13},\mathrm{P}\left(B\right)=\frac{9}{13}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{4}{13} \\ \mathrm{As},\mathrm{P}\left(\overline{A}\cap \mathrm{B}\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{9}{13}-\frac{4}{13} \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{9-4}{13} \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{5}{13} \\ \mathrm{Now}, \\ \mathrm{P}\left(\overline{A}|B\right)=\frac{\mathrm{P}\left(\overline{A}\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left({\displaystyle \frac{5}{13}}\right)}{\left({\displaystyle \frac{9}{13}}\right)}=\frac{5}{9} \\ \left(\mathrm{iv}\right)\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\cap B\right)=\frac{1}{4} \\ \mathrm{Also},\mathrm{P}\left(\overline{B}\right)=1-\mathrm{P}\left(B\right)=1-\frac{1}{3}=\frac{2}{3} \\ \mathrm{As},\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}{4} \\ =\frac{6+4-3}{12} \\ \Rightarrow \mathrm{P}\left(A\cup B\right)=\frac{7}{12} \\ \mathrm{Also},\mathrm{P}\left(\overline{A}\cap B\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right) \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{1}{3}-\frac{1}{4} \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{4-3}{12} \\ \Rightarrow \mathrm{P}\left(\overline{A}\cap B\right)=\frac{1}{12} \\ \mathrm{And},\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{7}{12} \\ =\frac{5}{12} \\ \mathrm{Now}, \\ \mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{1}{3}}\right)}=\frac{3}{4}, \\ \mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left({\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{1}{2}}\right)}=\frac{2}{4}=\frac{1}{2}, \\ \mathrm{P}\left(\overline{A}|B\right)=\frac{\mathrm{P}\left(\overline{A}\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left({\displaystyle \frac{1}{12}}\right)}{\left({\displaystyle \frac{1}{3}}\right)}=\frac{3}{12}=\frac{1}{4}\mathrm{and} \\ \mathrm{P}\left(\overline{A}|\overline{B}\right)=\frac{\mathrm{P}\left(\overline{A}\cap \overline{B}\right)}{\mathrm{P}\left(\overline{B}\right)}=\frac{\left({\displaystyle \frac{5}{12}}\right)}{\left({\displaystyle \frac{2}{3}}\right)}=\frac{15}{24}=\frac{5}{8} \end{gathered}$$