Probability 4 Question 17
17. A person goes to office either by car, scooter, bus or train probability of which being $\frac{1}{7}, \frac{3}{7}, \frac{2}{7}$ and $\frac{1}{7}$, respectively. Probability that he reaches offices late, if he takes car, scooter, bus or train is $\frac{2}{9}, \frac{1}{9}, \frac{4}{9}$ and $\frac{1}{9}$, respectively. Given that he reached office in time, then what is the probability that he travelled by a car?
(2005, 2M)
Show Answer
Answer:
Correct Answer: 17. $\frac{1}{7}$
Solution:
- As, the statement shows problem is to be related to Baye’s law.
Let $C, S, B, T$ be the events when person is going by car, scooter, bus or train, respectively.
$\therefore \quad P(C)=\frac{1}{7}, P(S)=\frac{3}{7}, P(B)=\frac{2}{7}, P(T)=\frac{1}{7}$
Again, $L$ be the event of the person reaching office late.
$\therefore \bar{L}$ be the event of the person reaching office in time.
Then, $P \frac{\bar{L}}{C}=\frac{7}{9}, P \frac{\bar{L}}{S}=\frac{8}{9}, P \frac{\bar{L}}{B}=\frac{5}{9}$
and $\quad P \frac{\bar{L}}{T}=\frac{8}{9}$
$$ \begin{aligned} & \therefore \quad P \frac{C}{L}= \frac{P \frac{\bar{L}}{C} \cdot P(C)}{P \frac{\bar{L}}{C} \cdot P(C)+P \frac{\bar{L}}{S} \cdot P(S)+P \frac{\bar{L}}{B} \cdot P(B)} \\ &+P \frac{\bar{L}}{T} \cdot P(T) \\ &= \frac{\frac{7}{9} \times \frac{1}{7}}{\frac{7}{9} \times \frac{1}{7}+\frac{8}{9} \times \frac{3}{7}+\frac{5}{9} \times \frac{2}{7}+\frac{8}{9} \times \frac{1}{7}}=\frac{1}{7} \end{aligned} $$
