A department store is holding a drawing to give away free shopping sprees. There are 10 customers who have entered the drawing: 6 live in the town of Gaston, 2 live in pike, and 2 live in wells. Two winners will be selected at random. What is the probability that both winners live in Gaston? Write your answer as a fraction in simplest form
Simplify (6/10)(5/9)
To find the probability that both winners live in Gaston, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of possible outcomes. Since there are 10 customers in total, and 2 winners will be selected, we can use the combination formula to determine the number of possible outcomes. The formula for combinations is:
nCr = n! / (r!(n-r)!)
Here, n represents the total number of customers (10), and r represents the number of winners selected (2).
nCr = 10! / (2!(10-2)!) = 10! / (2!8!) = (10 * 9 * 8!) / (2!8!) = (10 * 9) / 2! = 45
There are 45 possible outcomes.
Next, let's find the number of favorable outcomes, which is the number of ways we can choose both winners from the 6 customers who live in Gaston. Again, we can use the combination formula:
nCr = n! / (r!(n-r)!)
Here, n represents the total number of customers from Gaston (6), and r represents the number of winners selected (2).
nCr = 6! / (2!(6-2)!) = 6! / (2!4!) = (6 * 5 * 4!) / (2!4!) = (6 * 5) / 2! = 15
There are 15 favorable outcomes.
Finally, we can determine the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Favorable outcomes / Total outcomes = 15 / 45
To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is 15:
Probability = 15 / 45 = (15 ÷ 15) / (45 ÷ 15) = 1/3
Therefore, the probability that both winners live in Gaston is 1/3.