Question

A man takes either a bus or the subway to work with probabilities 0.4 and 0.6, respectively. When he takes the bus, he is late 50% of the days. When he takes the subway, he is late 40% of the days. If the man is late for work on a particular day, what is the probability that he took the bus? (Round your answer to four decimal places.)

Answer 1

Answer: 0.4545

============================================

Work Shown:

Define the following events

• B = the man takes the bus
• S = the man takes the subway
• L = the man is late
• L' = the man is not late (ie the man is on time)

Draw out a probability tree. See diagram below.

From the tree diagram, we can figure out that P(L) = 0.44 which is the probability of being late. Note how I considered the two probabilities P(B & L) and P(S & L)

• P(B & L) = probability of taking the bus and being late
• P(S & L) = probability of taking the subway and being late

The events (B & L) and (S & L) are mutually exclusive, so we can add them up. Combining them will have us find the overall probability of being late. For more information, check out the law of total probability.

--------------------

Now use a conditional probability formula to find...

P(B|L) = probability the man takes bus given he is late

P(B|L) = P(B & L)/P(L)

P(B|L) = 0.2/0.44

P(B|L) = 0.45454545454546 approximately

P(B|L) = 0.4545 rounding to four decimal places

note: when I write "P(B|L)", I am using a vertical bar. I'm not using an uppercase letter 'i' or a lowercase letter L.