A person goes to office by car, scooter, bus, train the probability of which being$\frac{1}{7},\frac{3}{7},\frac{2}{7} \mathrm{and} \frac{1}{7}$ respectively probability that he reaches office late, if he takes car, scooter, bus or train is  $\frac{2}{9},\frac{1}{9},\frac{4}{9} \mathrm{and} \frac{1}{9}$ respectively given that he reached office in in time, if the probability  that he travelled  by car is  $\frac{1}{P}$ then P is _____

Probability that a  person goes to office by car =$P\left(C\right)=\frac{1}{7}$

Probability that a  person goes to office by scooter =$P\left(S\right)=\frac{3}{7}$

Probability that a  person goes to office by bus=$P\left(C\right)=\frac{2}{7}$

Probability that a  person goes to office by train =$P\left(T\right)=\frac{1}{7}$

L= probability that he reaches office by late

I= probability that he reaches office in time

$$\begin{gathered} P(L/C)=\frac{2}{9},P(L/S)=\frac{1}{9},P(L/B)=\frac{4}{9},P(L/T)=\frac{1}{9}, \\ \Rightarrow P(I/C)=\frac{7}{9},P(I/S)=\frac{8}{9},P(I/B)=\frac{5}{9},P(I/T)=\frac{8}{9}, \end{gathered}$$

Using Baye’s Theorem
$$\begin{gathered} \mathrm{P}(\mathrm{C}/\mathrm{I})=\frac{\mathrm{P}\left(\mathrm{C}\right)\mathrm{P}(\mathrm{I}/\mathrm{C})}{\mathrm{P}\left(\mathrm{C}\right)\mathrm{P}(\mathrm{I}/\mathrm{C})+\mathrm{P}\left(\mathrm{S}\right)\mathrm{P}(\mathrm{I}/\mathrm{S})+\mathrm{P}\left(\mathrm{B}\right)\mathrm{P}(\mathrm{I}/\mathrm{B})+\mathrm{P}\left(\mathrm{T}\right)\mathrm{P}(\mathrm{I}/\mathrm{T})} \\ \mathrm{P}(\mathrm{C}/\mathrm{I})=\frac{\frac{7}{9}\times \frac{1}{7}}{\frac{7}{9}\times \frac{1}{7}+\frac{8}{9}\times \frac{3}{7}+\frac{5}{9}\times \frac{2}{7}+\frac{8}{9}\times \frac{1}{7}}=\frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7} \end{gathered}$$p = 7