Determine whether the following individual events are independent or dependent. Then find the probability of the combined event.
Randomly selecting a four-person committee consisting entirely of Americans from a pool of 18 Americans and 10 Canadians.
The event of selecting an American and the event of selecting an American the next time are
The probability of randomly selecting a four-person committee consisting entirely of Americans from a pool of 18 Americans and 10 Canadians is
(Round to four decimal places as needed.)
Rounding to four decimal places as needed, the final answer is \(\boxed{0.1495}\).
Step 1 :The problem is asking for the probability of selecting a four-person committee consisting entirely of Americans from a pool of 18 Americans and 10 Canadians. This is a combination problem, where order does not matter.
Step 2 :The total number of ways to select a four-person committee from 28 people (18 Americans and 10 Canadians) is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of people and k is the number of people to be selected. In this case, \(n = 28\) and \(k = 4\).
Step 3 :The number of ways to select a four-person committee consisting entirely of Americans from 18 Americans is given by the same combination formula, but with \(n = 18\) and \(k = 4\).
Step 4 :The probability of the event is then given by the ratio of the number of ways to select a four-person committee consisting entirely of Americans to the total number of ways to select a four-person committee.
Step 5 :Calculating the total number of ways to select a four-person committee from 28 people, we get \(C(28, 4) = 20475\).
Step 6 :Calculating the number of ways to select a four-person committee consisting entirely of Americans from 18 Americans, we get \(C(18, 4) = 3060\).
Step 7 :The probability of randomly selecting a four-person committee consisting entirely of Americans from a pool of 18 Americans and 10 Canadians is then given by \(\frac{3060}{20475} = 0.14945054945054945\).
Step 8 :Rounding to four decimal places as needed, the final answer is \(\boxed{0.1495}\).