Question

you are planning a three - day trip to seattle, washington, in october. use the fact that on each day, it could either be sunny or rainy, and that each day is equally likely to be sunny or rainy to answer the following question.
what is the probability that it rains on exactly one day?
(type an integer or a decimal. do not round.)

Explanation:

Step1: Determine the total number of possible outcomes

Each day has 2 possibilities (sunny or rainy). For a 3 - day trip, by the multiplication principle, the total number of possible outcomes is \(n = 2\times2\times2=2^{3}=8\).

Step2: Use the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successful trials, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\) is the number of combinations.
Here, \(n = 3\) (number of days), \(k = 1\) (number of rainy days), and \(p=\frac{1}{2}\) (probability of rain on a single day).
First, calculate the combination \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=\frac{3\times2!}{2!}=3\).
Then, \(p^{k}\times(1 - p)^{n - k}=(\frac{1}{2})^{1}\times(1-\frac{1}{2})^{3 - 1}=(\frac{1}{2})\times(\frac{1}{2})^{2}=\frac{1}{8}\).
So, \(P(X = 1)=C(3,1)\times\frac{1}{8}\).
Since \(C(3,1) = 3\), then \(P(X = 1)=3\times\frac{1}{8}=\frac{3}{8}=0.375\).

Answer:

\(0.375\)