The probability that a person goes to college by car is $\frac{1}{5}$; by bus $\frac{2}{5}$ and by train is…
The probability that a person goes to college by car is $\frac{1}{5}$; by bus $\frac{2}{5}$ and by train is $\frac{3}{5}$ respectively. The probabilities that he reaches the college late if he takes car, bus, train are $\frac{2}{7}, \frac{4}{7}$ and $\frac{1}{7}$ respectively. If he reaches the college in time, the probability that he traveled by car is
$\frac{6}{29}$
$\frac{24}{29}$
$\frac{5}{29}$
$\frac{23}{29}$
Solution
Let $\mathrm{C}=$ The event that person goes by car
$\mathrm{B}=$ The event that person goes by bus
$\mathrm{T}=$ The event that person goes by train
and $L=$ The event that person reach as college late
Now, $\mathrm{P}(\mathrm{C})=\frac{1}{5}, \mathrm{P}(\mathrm{B})=\frac{2}{5}, \mathrm{P}(\mathrm{T})=\frac{3}{5}$
$\begin{aligned} & \mathrm{P}\left(\frac{\mathrm{L}}{\mathrm{C}}\right)=\frac{2}{7}, \mathrm{P}\left(\frac{\mathrm{L}}{\mathrm{B}}\right)=\frac{4}{7}, \mathrm{P}\left(\frac{\mathrm{L}}{\mathrm{T}}\right)=\frac{1}{7} \\ & \text { and } \mathrm{P}\left(\frac{\mathrm{L}^{\prime}}{\mathrm{C}}\right)=\frac{5}{7}, \mathrm{P}\left(\frac{\mathrm{L}^{\prime}}{\mathrm{B}}\right)=\frac{3}{7}, \mathrm{P}\left(\frac{\mathrm{L}^{\prime}}{\mathrm{T}}\right)=\frac{6}{7}\end{aligned}$
So, $P\left(\frac{C}{L^{\prime}}\right)=\frac{P\left(\frac{L^{\prime}}{C}\right) P(C)}{P\left(\frac{L^{\prime}}{C}\right) P(C)+P\left(\frac{L^{\prime}}{B}\right) P(B)+P\left(\frac{L^{\prime}}{T}\right) P(T)}$
$=\frac{\frac{5}{7} \times \frac{1}{5}}{\frac{5}{7} \times \frac{1}{5}+\frac{3}{7} \times \frac{2}{5}+\frac{6}{7} \times \frac{3}{5}}=\frac{5}{5+6+18}=\frac{5}{29}$