At the first tri-city meeting, there are 8 people from town A, 7 people from town B, and 5 people from town C. If a council consisting of 5 people is randomly selected, find the probability that 3 are from town A and 2 are from town B.
To quote one of our very good math and science tutors: “You will find here at Jiskha that long series of questions, posted with no evidence of effort or thought by the person posting, will not be answered. We will gladly respond to your future questions in which your thoughts are included.”
To find the probability that 3 people are from town A and 2 people are from town B, we need to consider two factors:
1. The total number of ways to choose 5 people from the total number of attendees.
2. The number of ways to choose 3 people from town A and 2 people from town B.
Let's calculate each factor step by step:
1. Total number of ways to choose 5 people from the total attendees:
To calculate this, we'll use the concept of combinations. The formula for combinations is given by C(n, r) = n! / (r!(n-r)!), where n represents the total number of people, and r represents the number of people to be chosen.
In this case, the total number of attendees is 8 + 7 + 5 = 20, and we need to select 5 people.
Therefore, the total number of ways to choose 5 people from 20 attendees is C(20, 5) = 20! / (5!(20-5)!) = 15504.
2. Number of ways to choose 3 people from town A and 2 people from town B:
We can calculate this using combinations as well.
There are 8 people from town A, and we need to choose 3 of them (let's call it C1).
Similarly, there are 7 people from town B, and we need to choose 2 of them (let's call it C2).
Therefore, the number of ways to choose 3 people from town A and 2 people from town B is C(8, 3) * C(7, 2) = (8! / (3!(8-3)!)) * (7! / (2!(7-2)!)) = 56 * 21 = 1176.
Now, we can calculate the probability:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
= (Number of ways to choose 3 people from town A and 2 people from town B) / (Total number of ways to choose 5 people from the total attendees)
= 1176 / 15504
≈ 0.0756 (rounded to four decimal places)
Therefore, the probability that 3 people are from town A and 2 people are from town B is approximately 0.0756.