Question

probability he come by train bus scooter or by other mean of tranport 3/10,1/5,1/10,2/5. probability he will late 1/4,1/3,1/2 if he come by train bus scooter respectly but if hec come by oter mean of transport then hewill not be late when he arrive he is late what is probability he has come by train

Answer 1

$$\begin{gathered} \mathrm{Let}{\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3\mathrm{and}{\mathrm{E}}_4\mathrm{be}\mathrm{the}\mathrm{events}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{train},\mathrm{bus},\mathrm{scooter}\mathrm{and}\mathrm{other}\mathrm{means}\mathrm{of}\mathrm{transport}\mathrm{respectively}. \\ \mathrm{Then},\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{3}{10},\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{1}{5},\mathrm{P}\left({\mathrm{E}}_3\right)=\frac{1}{10},\mathrm{and}\mathrm{P}\left({\mathrm{E}}_4\right)=\frac{2}{5} \\ \mathrm{Let}\mathrm{E}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late}.\mathrm{Then}, \\ \mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_1\right)=\mathrm{Probability}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{train}=\frac{1}{4} \\ \mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_2\right)=\mathrm{Probability}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{bus}=\frac{1}{3} \\ \mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_3\right)=\mathrm{Probability}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{scooter}=\frac{1}{2} \\ \mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_4\right)=\mathrm{Probability}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{other}\mathrm{means}\mathrm{of}\mathrm{transport}=0 \\ \left[\mathbf{since}\mathbf{}\mathbf{he}\mathbf{}\mathbf{is}\mathbf{}\mathbf{not}\mathbf{}\mathbf{late}\mathbf{}\mathbf{if}\mathbf{}\mathbf{he}\mathbf{}\mathbf{comes}\mathbf{}\mathbf{by}\mathbf{}\mathbf{other}\mathbf{}\mathbf{means}\mathbf{}\mathbf{of}\mathbf{}\mathbf{transport}\right] \\ \mathrm{P}\left({\mathrm{E}}_1/\mathrm{E}\right)=\mathrm{Probability}\mathrm{that}\mathrm{he}\mathrm{comes}\mathrm{by}\mathrm{train},\mathrm{given}\mathrm{that}\mathrm{he}\mathrm{is}\mathrm{late} \\ =\frac{\mathrm{P}\left({\mathrm{E}}_1\right).\mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_1\right)}{\mathrm{P}\left({\mathrm{E}}_1\right).\mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right).\mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_2\right)+\mathrm{P}\left({\mathrm{E}}_3\right).\mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_3\right)+\mathrm{P}\left({\mathrm{E}}_4\right).\mathrm{P}\left(\mathrm{E}/{\mathrm{E}}_4\right)}\left[\mathbf{by}\mathbf{}\mathbf{Bayes}\mathbf{}\mathbf{theorem}\right] \\ =\frac{\left({\displaystyle \frac{3}{10}}\times {\displaystyle \frac{1}{4}}\right)}{\left({\displaystyle \frac{3}{10}}\times {\displaystyle \frac{1}{4}}\right)+\left({\displaystyle \frac{1}{5}}\times {\displaystyle \frac{1}{3}}\right)+\left({\displaystyle \frac{1}{10}}\times {\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{2}{5}}\times 0\right)} \\ =\frac{{\displaystyle \frac{3}{40}}}{{\displaystyle \frac{3}{40}}+{\displaystyle \frac{1}{15}}+{\displaystyle \frac{1}{20}}+0} \\ =\frac{{\displaystyle \frac{3}{40}}}{{\displaystyle \frac{9+8+6}{120}}}=\frac{3}{40}\times \frac{120}{23}=\frac{9}{23} \end{gathered}$$