[(8 combination 3)(7 combination 2)(5 combination 0)] divided by (20 combination 5)
= .0758513932
= .0758513932
First, we need to calculate the total number of ways to choose 5 people from the total 8 from town A and 7 from town B. This is a combination problem, expressed as "8 choose 3" for town A and "7 choose 2" for town B. The number of ways to choose 3 people from town A and 2 people from town B can be calculated as follows:
Number of ways = (8 choose 3) * (7 choose 2)
For the combination formula "n choose r," where n represents the total items to choose from and r represents the number of items to choose, it can be calculated as:
n! / (r! * (n-r)!)
For the first part, (8 choose 3), the calculation is:
8! / (3! * (8-3)!) = 56
For the second part, (7 choose 2), the calculation is:
7! / (2! * (7-2)!) = 21
Therefore, the number of ways to choose 3 people from town A and 2 people from town B is:
Number of ways = 56 * 21 = 1176
Now, we need to calculate the total number of ways to choose 5 people from the total number of attendees. There are 8 people from town A, 7 people from town B, and 5 people from town C, totaling 20 people. Therefore, this can be calculated as:
Total number of ways = 20 choose 5
Using the combination formula, the calculation is:
20! / (5! * (20-5)!) = 15,504
Finally, to find the probability, we divide the number of desired outcomes (1176) by the total number of possible outcomes (15,504):
Probability = Number of ways / Total number of ways = 1176 / 15,504 = 0.0758 (rounded to four decimal places)
So, the probability that 3 people are from town A and 2 people are from town B in the randomly selected council of 5 people is approximately 0.0758.