Question

In my town, it's rainy one third of the days. Given that it is rainy, there will be heavy traffic with probability 1/2, and given that it is not rainy, there will be heavy traffic with probability 1/4. If it's rainy and there is heavy traffic, I arrive late for work with probability 1/2. On the other hand, the probability of being late is reduced to 1/8 if it is not rainy and there is no heavy traffic. In other situations (rainy and no traffic, not rainy and traffic) the probability of being late is 0.25. You pick a random day.
a. What is the probability that it’s not raining and there is heavy traffic and I am not late?
b. What is the probability that I am late?
c. Given that I arrived late at work, what is the probability that it rained that day?

Answer 1

Answer:

a) 0.125 = 12.5% probability that it’s not raining and there is heavy traffic and I am not late.

b) 0.2292 = 22.92% probability that I am late.

c) 0.5454 = 54.54% probability that it rained that day.

Step-by-step explanation:

Conditional Probability

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

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

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.

Question a:

2/3 probability of not raining.

If not raining, 1/4 probability of heavy traffic.

1 - 0.25 = 0.75 = 3/4 probability of not late.

So

$p = \frac{2}{3} \times \frac{1}{4} \times \frac{3}{4} = \frac{2}{16} = 0.125$

0.125 = 12.5% probability that it’s not raining and there is heavy traffic and I am not late.

b. What is the probability that I am late?

0.5 of (1/3)*(1/2) = 1/6(rainy and heavy traffic).

0.25 of (1/3)*(1/2) = 1/6(rainy and no traffic).

1/8 = 0.125 of (2/3)*(3/4) = 1/2(not rainy and no traffic).

0.25 of (2/3)*(1/4) = 1/6(not rainy and traffic). So

$P(A) = 0.5\frac{1}{6} + 0.25\frac{1}{6} + 0.125\frac{3}{6} + 0.25\frac{1}{6} = \frac{0.5 + 0.25 + 3*0.125 + 0.25}{6} = 0.2292$

0.2292 = 22.92% probability that I am late.

c. Given that I arrived late at work, what is the probability that it rained that day?

Event A: Late

Event B: Rained

0.2292 = 22.92% probability that I am late.

This means that $P(A) = 0.2292$

Probability of late and rain:

0.5 of 1/6(rain and heavy traffic).

0.25 of 1/6(rain and no traffic). So

$P(A \cap B) = 0.5\frac{1}{6} + 0.25\frac{1}{6} = \frac{0.5 + 0.25}{6} = \frac{0.75}{6} = 0.125$

Probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.125}{0.2292} = 0.5454$

0.5454 = 54.54% probability that it rained that day.