Question

Archie has to go to school this morning for an important test, but he woke up late. He can either take the bus or take his unreliable car. If he takes the car, Archie knows from experience that he will make it to school without breaking down with probability 0.2. However, the bus to school runs late 65% of the time. Archie decides to choose between these options by tossing a coin. Suppose that Archie does, in fact, make it to the test on time. What is the probability that he took his car? Round your answer to two decimal places.

Answer 1

Let

M = the event that Archie made it to school on time

C = the event that Archie took his car

B = the event that Archie took the bus

The bus is late 65% of the time (or .65). Therefore, it is on time 1-.65=.35 or 35% of the time, since the bus can only be late or on time.

So, we know from the problem that the probability that Archie makes it to school on time, given that he took the bus is P(M|B) = .35 and the probability that Archie Makes it to school on time, given that he took his car is P(M|C) = .2

If he flips a “fair” coin, to choose between the bus and car, then the probability that he takes the bus

P(B), probability that he takes bus = probability that he takes his car, P(C) = probability of a heads = probability of a tails = .5 or 50%.

We want to find out the probability that he took the car given that he made it to the test on time, or,

P(C|M).

Then,

Baye’s

Rule states:

Now, we just substitute the values for each probability on the right side of the equation to determine

P(C|M):

.35*.5/ .35*.5 + .2*.5 = .6363

Final Answer: 0.64

So the probability that he took the bus is .67.