A committee of 100 Democrats is formed with 22 members from the East, 38 members from the Midwest, and 40 members from the West. If a random
sample of 3 members from this committee is selected, what is the probability to the nearest percent that all 3 are from the East?
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
22/100 * 21/99 * 20/98 = ?
22/100*21/99*20/98=.0095238095 or.95%
To find the probability that all 3 members are from the East, we need to calculate the ratio of the number of ways to select 3 members from the East out of the total number of ways to choose 3 members from the committee.
To calculate the number of ways to select 3 members from the East, we can use the combination formula, which is calculated as:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of members from the East (which is 22 in this case), and r is the number of members we want to select (which is 3).
So, C(22, 3) = 22! / (3!(22-3)!) = 22! / (3!19!)
Now, let's calculate the total number of ways to choose 3 members from the committee. Since we have a total of 100 members in the committee, we can use the combination formula again:
C(100, 3) = 100! / (3!(100-3)!) = 100! / (3!97!)
Now, to find the probability, we need to divide the number of ways to select 3 members from the East by the total number of ways to choose 3 members from the committee:
Probability = C(22, 3) / C(100, 3)
Probability = (22! / (3!19!)) / (100! / (3!97!))
Simplifying this expression will result in the probability of selecting 3 members from the East out of the entire committee.