Question

Dill and Bromberg, Problem 1.23. The San Francisco football team plays better in fair weather. They have a 70% chance of winning in good weather but only a 20% chance of winning in bad weather. (a) If they play in the Super Bowl in Wisconsin and the weatherman predicts a 60% chance of snow that day, what is the probability that San Francisco will win. (b) Given that San Francisco lost, what is the probability that the weather was bad?

Answer 1

Answer:

a) 40% probability that San Francisco will win.

b) 80% probability that the weather was bad

Step-by-step explanation:

We have these following probabilities:

60% chance of snow(bad weather).

If there is bad weather, 20% probability of SF winning.

100-60 = 40% chance of good weather.

If there is good weather, 70% probability of SF winning.

(a) If they play in the Super Bowl in Wisconsin and the weatherman predicts a 60% chance of snow that day, what is the probability that San Francisco will win.

20% of 60%, plus 70% of 40%. So

0.2*0.6 + 0.7*0.4 = 0.4

40% probability that San Francisco will win.

(b) Given that San Francisco lost, what is the probability that the weather was bad?

We use the conditional probability formula to solve this question.

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.

In this question:

Event A: SF losing

Event B: Bad weather

P(A)

From a), 40% probability that SF wins. So 100-40 = 60% = 0.6 probability that SF losses.

So $P(A) = 0.6$

P(A and B)

60% chance of snow. If it snows, 80% probability of SF losing. So

$P(A \cap B) = 0.6*0.8 = 0.48$

$P(B|A) = \frac{0.48}{0.60} = 0.8$

80% probability that the weather was bad