Question

A town keeps weather records. From these records, it has been determined that it rains on an average of 12% of the days each year. If 30 days are selected at random from one year, what is the probability that at most 3 days had rain?

a) 0.7257

b) 0.5987

c) 0.8413

d) 0.1587

Answer 1

To calculate the probability that at most 3 days had rain out of a selection of 30 days, you need to calculate the probabilities of 0, 1, 2, and 3 rainy days using the binomial probability formula. Then, you add these probabilities together to get the final probability.

Step-by-step explanation:

To find the probability that at most 3 days had rain out of a selection of 30 days, we need to calculate the probability of 0, 1, 2, and 3 rainy days and add them together.

The probability of 0 rainy days can be found using the binomial probability formula: P(X=k) = (n C k) * p^k * (1-p)^(n-k), where n is the number of trials (30), k is the number of successes (0), and p is the probability of success (0.12).

The same formula can be used for 1, 2, and 3 rainy days, but with different values of k. Once we have calculated the probabilities for each case, we can add them together to get the final probability.

P(X=0) = (30 C 0) * 0.12^0 * (1-0.12)^(30-0)

P(X=1) = (30 C 1) * 0.12^1 * (1-0.12)^(30-1)

P(X=2) = (30 C 2) * 0.12^2 * (1-0.12)^(30-2)

P(X=3) = (30 C 3) * 0.12^3 * (1-0.12)^(30-3)

Finally, we add the probabilities of 0, 1, 2, and 3 rainy days together to get the probability that at most 3 days had rain.