Probability, can anyone help me?

Question

A man takes either a bus or the subway to go to work with probabilities 0.25 and 0.75, respectively. When he takes the bus, he is late 40% of the days. When he takes the subway, he is late 30% of the days. If he is late for work on a particular day, what is
the probability that he took the bus?

Answer 1

Use the formula for conditional probability: P(A|B) = P(A n B) / P(B), where I'm using the "n" for the intersection symbol.

In this case, A = "he took the bus", and B = "he is late".

P(B) = 0.25*0.4 + 0.75*0.3 = 0.325.

P(A n B) = 0.25*0.4 = 0.1.

Therefore P(A|B) = 0.1 / 0.325 = 0.3077.

Answer 2

Take enough days that we can do this without fractions or decimals. It's 40 days.

10 days he takes the bus, 4 of those days he is late.
30 days he takes the subway, 9 of those days he is late.

So if he is late, it's on one of those 13 days, and the probability is 4/13 that he took the bus. That's about 0.3077.

Answer 3

P(Bus¦ Late)=P(Bus n Late) / P (Late)