A friend who works in a big city owns two cars, one small and one large. Three-quarters of the time he drives the small car to work, and one-quarter of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.8. If he takes the large car, he is on time to work with probability 0.6. Given that he was at work on time on a particular morning, what is the probability that he drove the small car

Answer:

0.8 = 80% probability that he drove the small car

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Being at work on time.

Event B: Driving the small car.

Probability of being at work on time:

0.8 of 0.75(small car).

0.6 of 0.25(large car). So

[tex]P(A) = 0.8*0.75 + 0.6*0.25 = 0.75[/tex]

At work on time, using the small car:

0.8 of 0.75

So

[tex]P(A \cap B) = 0.8*0.75 = 0.6[/tex]

What is the probability that he drove the small car

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.6}{0.75} = 0.8[/tex]

0.8 = 80% probability that he drove the small car

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