Question

Suppose that Matthew can choose to get home from work by car or bus.

Answer 1

Answer:

B-91%

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: arriving home after 7 p.m.

Event B: getting home by bus.

When he chooses to get home by bus, he arrives home after 7 p.m. 25 percent of the time.

This means that $P(A|B) = 0.25$

Because the bus is cheaper, he uses the bus 70 percent of the time.

This means that $P(B) = 0.7$

Probability of getting home after 7 p.m.

70% of the time he uses bus, and by bus, he arrives arrives home after 7 p.m. 25 percent of the time.

100 - 70 = 30% of the time he uses the car, and by car, he arrives home after 7 p.m. 6 percent of the time.

So

$P(A) = 0.7*0.25 + 0.3*0.06 = 0.193$

What is the approximate probability that Matthew chose to get home from work by bus, given that he arrived home after 7 p.m.?

$P(B|A) = \frac{0.7*0.25}{0.193} = 0.9067$

Rouding up, 91%.

So the correct answer is:

B-91%