Question

A man takes either a bus or the subway to work with probabilities .3 and \(.7,\) respectively. When he takes the bus, he is late \(30 \%\) of the days. When he takes the subway, he is late \(20 \%\) of the days. If the man is late for work on a particular day, what is the probability that he took the bus?

Short answer

Answer: The probability that he took the bus is approximately 0.3913 or 39.13%.

Step-by-step solution

Define the events A and B

Let A be the event that the man takes the bus and B be the event that he is late.

Find the probability of the events

We are given the probabilities as follows: \(P(A) = 0.3\) (the probability of taking the bus) \(P(A') = 0.7\) (the probability of taking the subway) \(P(B | A) = 0.3\) (the probability of being late given that he took the bus) \(P(B | A') = 0.2\) (the probability of being late given that he took the subway)

Apply Bayes' theorem

According to Bayes' theorem, the probability that he took the bus, given that he is late, is: \(P(A | B) = \frac{P(B | A) * P(A)}{P(B | A) * P(A) + P(B | A') * P(A')}\)

Calculate the probability

Now, we substitute the given probabilities to find the required probability: \(P(A | B) = \frac{(0.3)(0.3)}{(0.3)(0.3) + (0.2)(0.7)}\) \(P(A | B) = \frac{0.09}{0.09 + 0.14}\) \(P(A | B) = \frac{0.09}{0.23}\) \(P(A | B) \approx 0.3913\) So, if the man is late for work on a particular day, the probability that he took the bus is approximately \(0.3913\) or \(39.13 \%\).