Of ten twenty-dollar bills, two are counterfeit. Six bills are chosen at random. What is the probability that counterfeit bills are chosen?
This is what I did:
C(8,4) * C(2,2) = 70....but the answer needs to be 1/3.
Can anybody explain how to solve this, and also explain why mine is wrong?
Thank you in advance.
This is what I did:
C(8,4) * C(2,2) = 70....but the answer needs to be 1/3.
Can anybody explain how to solve this, and also explain why mine is wrong?
Thank you in advance.
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Number of ways to choose 6 bills from 10 bills = C(10,6) = 210
Number of ways 2 counterfeit bills can occur within the 6 chosen bills = C(8,8) * C(2,2) = 70
You were right until this part. But to calculate the probability you need to divide 70/210 = 1/3.
Number of ways 2 counterfeit bills can occur within the 6 chosen bills = C(8,8) * C(2,2) = 70
You were right until this part. But to calculate the probability you need to divide 70/210 = 1/3.