Question

A person has to catch a train. To catch train, from his home he can take a taxi or take rickshaw or walk by foot with respective probabilities 1/2, 1/3, 1/6 Probability of him catching train when he takes rickshaw from his home is half that of when the takes the taxi and probability of catching the train when he walked by foot is 1 th 4 that of when he takes rickshaw. He finally reached the train, the probability he walked by his foot to catch the train, is

Answer 1

1/33

Answer 2

Let $X$, $R$, and $W$ be the events that the person takes a taxi, rickshaw, or walks by foot, respectively. Let $T$ be the event that the person catches the train.

We are given the probabilities of choosing each mode of transport: $P(X) = 1/2$ $P(R) = 1/3$ $P(W) = 1/6$

We are given relationships between the conditional probabilities of catching the train: $P(T|R) = \frac{1}{2} P(T|X)$ $P(T|W) = \frac{1}{4} P(T|R)$

Let $P(T|X) = p$. Then $P(T|R) = \frac{1}{2}p$. And $P(T|W) = \frac{1}{4} \left(\frac{1}{2}p\right) = \frac{1}{8}p$.

We need to find the probability that the person walked by foot given that he reached the train, which is $P(W|T)$. Using Bayes' theorem: $P(W|T) = \frac{P(T|W) P(W)}{P(T)}$

First, we calculate the total probability of catching the train, $P(T)$, using the law of total probability: $P(T) = P(T|X) P(X) + P(T|R) P(R) + P(T|W) P(W)$ Substitute the known values and the expressions in terms of $p$: $P(T) = (p) \left(\frac{1}{2}\right) + \left(\frac{p}{2}\right) \left(\frac{1}{3}\right) + \left(\frac{p}{8}\right) \left(\frac{1}{6}\right)$ $P(T) = \frac{p}{2} + \frac{p}{6} + \frac{p}{48}$

To sum these fractions, find a common denominator, which is 48: $P(T) = \frac{p \cdot 24}{2 \cdot 24} + \frac{p \cdot 8}{6 \cdot 8} + \frac{p}{48}$ $P(T) = \frac{24p}{48} + \frac{8p}{48} + \frac{p}{48}$ $P(T) = \frac{24p + 8p + p}{48} = \frac{33p}{48}$

Now, substitute the values into the Bayes' theorem formula for $P(W|T)$: $P(W|T) = \frac{P(T|W) P(W)}{P(T)}$ $P(W|T) = \frac{\left(\frac{p}{8}\right) \left(\frac{1}{6}\right)}{\frac{33p}{48}}$ $P(W|T) = \frac{\frac{p}{48}}{\frac{33p}{48}}$

The term $p$ cancels out (assuming $p \neq 0$, which must be true since the person caught the train): $P(W|T) = \frac{p/48}{33p/48} = \frac{p}{48} \times \frac{48}{33p} = \frac{1}{33}$

The probability that he walked by his foot to catch the train, given that he finally reached the train, is $1/33$.