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Question

If A and B are two events such that $\left(\mathrm{i}\right)\mathrm{P}\left(A\right)=\frac{1}{3},\mathrm{P}\left(B\right)=\frac{1}{4}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{5}{12},\mathrm{then}\mathrm{find}\mathrm{P}\left(A|B\right)\mathrm{and}\mathrm{P}\left(B|A\right).\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{P}\left(A\right)=\frac{6}{11},\mathrm{P}\left(B\right)=\frac{5}{11}\mathrm{and}\mathrm{P}\left(A\cup B\right)=\frac{7}{11},\mathrm{then}\mathrm{find}\mathrm{P}\left(A\cap B\right),\mathrm{P}\left(A|B\right)\mathrm{and}\mathrm{P}\left(B|A\right).\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\mathrm{P}\left(A\right)=\frac{7}{13},\mathrm{P}\left(B\right)=\frac{9}{13}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{4}{13},\mathrm{then}\mathrm{find}\mathrm{P}\left(\overline{)A}|B\right).\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\mathrm{P}\left(A\right)=\frac{1}{2},\mathrm{P}\left(B\right)=\frac{1}{3}\mathrm{and}\mathrm{P}\left(A\cap B\right)=\frac{1}{4},\mathrm{then}\mathrm{find}\mathrm{P}\left(A|B\right),\mathrm{P}\left(B|A\right),\mathrm{P}\left(\overline{)A}|B\right)\mathrm{and}\mathrm{P}\left(\overline{)A}|\overline{)\mathit{B}}\right).$ [NCERT EXEMPLAR]

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Solution

$\left(\mathrm{i}\right)\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{1}{3}+\frac{1}{4}-\frac{5}{12}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{2}{12}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left(\frac{1}{6}\right)}{\left(\frac{1}{4}\right)}=\frac{4}{6}=\frac{2}{3}\mathrm{and}\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left(\frac{1}{6}\right)}{\left(\frac{1}{3}\right)}=\frac{3}{6}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\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)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{6+5-7}{11}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cap B\right)=\frac{4}{11}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left(\frac{4}{11}\right)}{\left(\frac{5}{11}\right)}=\frac{4}{5}\mathrm{and}\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left(\frac{4}{11}\right)}{\left(\frac{6}{11}\right)}=\frac{4}{6}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As},\mathrm{P}\left(\overline{)A}\cap \mathrm{B}\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{9}{13}-\frac{4}{13}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{9-4}{13}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{5}{13}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(\overline{)A}|B\right)=\frac{\mathrm{P}\left(\overline{)A}\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left(\frac{5}{13}\right)}{\left(\frac{9}{13}\right)}=\frac{5}{9}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\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}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{P}\left(\overline{)B}\right)=1-\mathrm{P}\left(B\right)=1-\frac{1}{3}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\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)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{6+4-3}{12}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(A\cup B\right)=\frac{7}{12}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{P}\left(\overline{)A}\cap B\right)=\mathrm{P}\left(B\right)-\mathrm{P}\left(A\cap B\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{1}{3}-\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{4-3}{12}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}\left(\overline{)A}\cap B\right)=\frac{1}{12}\phantom{\rule{0ex}{0ex}}\mathrm{And},\mathrm{P}\left(\overline{)A}\cap \overline{)B}\right)=\mathrm{P}\left(\overline{)A\cup B}\right)\phantom{\rule{0ex}{0ex}}=1-\mathrm{P}\left(A\cup B\right)\phantom{\rule{0ex}{0ex}}=1-\frac{7}{12}\phantom{\rule{0ex}{0ex}}=\frac{5}{12}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(A|B\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{3}\right)}=\frac{3}{4},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\cap B\right)}{\mathrm{P}\left(A\right)}=\frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{2}\right)}=\frac{2}{4}=\frac{1}{2},\phantom{\rule{0ex}{0ex}}\mathrm{P}\left(\overline{)A}|B\right)=\frac{\mathrm{P}\left(\overline{)A}\cap B\right)}{\mathrm{P}\left(B\right)}=\frac{\left(\frac{1}{12}\right)}{\left(\frac{1}{3}\right)}=\frac{3}{12}=\frac{1}{4}\mathrm{and}\phantom{\rule{0ex}{0ex}}\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(\frac{5}{12}\right)}{\left(\frac{2}{3}\right)}=\frac{15}{24}=\frac{5}{8}$

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