Question

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




When he chooses to get home by car, he arrives home after 7 p.m. 14 percent of the time.

When he chooses to get home by bus, he arrives home after 7 p.m. 62 percent of the time.

Because the bus is cheaper, he uses the bus 83 percent of the time.



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

Answer 1

Using conditional probability, it is found that there is a 0.9556 = 95.56% probability that Juan chose to get home from work by bus, given that he arrived home after 7 p.m.

Conditional Probability

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

In which

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

• $P(A \cap B)$ is the probability of both A and B happening.

• P(A) is the probability of A happening.

In this problem:

• Event A: Arrived home after 7 p.m.

• Event B: Got home by bus.

The percentages associated with arriving home after 7 p.m. are:

• 14% of 17%(by car).

• 62% of 83%(by bus).

Hence:

$P(A) = 0.14(0.17) + 0.62(0.83) = 0.5384$

The probability of both arriving home after 7 p.m. and using bus is:

$P(A \cap B) = 0.62(0.83)$

Hence, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.62(0.83)}{0.5384} = 0.9556$

0.9556 = 95.56% probability that Juan chose to get home from work by bus, given that he arrived home after 7 p.m.

You can learn more about conditional probability at https://brainly.com/question/14398287