Question

A certain football team has the reputation of being a very good team when the weather is bad. In fact, the probability of their winning when the weather is bad is 0.8, whereas the probability of their winning on a clear day is only 0.4. During the month of November, weather statistics indicate the probability of bad weather to be 0.6. If it is known that the team won a given game in November, what is the probability that the weather was bad on that day?

Answer 1

The probability that the weather was bad on the day the football team won a game in November is 0.75. This is calculated using Bayes' theorem, considering the overall probability of winning and the conditional probabilities of winning given the weather conditions.

Step-by-step explanation:

To find the probability that the weather was bad given that the football team won a game in November, we use Bayes' theorem. The formula for Bayes' theorem in this context is:

P(Bad Weather | Won) = (P(Won | Bad Weather) * P(Bad Weather)) / P(Won)

We know the following:

• P(Won | Bad Weather) = 0.8
• P(Won | Clear Weather) = 0.4
• P(Bad Weather) = 0.6
• P(Clear Weather) = 1 - P(Bad Weather) = 0.4

First, we calculate the total probability of winning, P(Won), which is the sum of the probabilities of winning in bad weather and winning in clear weather, weighted by the probability of each type of weather:

P(Won) = P(Won | Bad Weather) * P(Bad Weather) + P(Won | Clear Weather) * P(Clear Weather)

Substitute in the values:

P(Won) = (0.8 * 0.6) + (0.4 * 0.4)

P(Won) = 0.48 + 0.16

P(Won) = 0.64

Now we can find the probability that the weather was bad given that the team won:

P(Bad Weather | Won) = (0.8 * 0.6) / 0.64

P(Bad Weather | Won) = 0.48 / 0.64

P(Bad Weather | Won) = 0.75

Therefore, the probability that the weather was bad on the day the football team won a game in November is 0.75.