# Thread: Probablity Question? Choosing bills?

Out of 10 $20 bills, 2 are counterfeit, 6 bills are chosen at random. What is the probability that both counterfeit bills are chosen? --my attempt-- C(10,6)/(C(6,2) *C(10,2)) Is this correct? I am not sure. Thanks 2. ## Re: Probablity Question? Choosing bills? Originally Posted by blackZ Out of 10$20 bills, 2 are counterfeit, 6 bills are chosen at random. What is the probability that both counterfeit bills are chosen?
Standard notation is $\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$, n things choosing k.
So the answer will be $\displaystyle \frac{\dbinom{8}{4}}{\dbinom{10}{6}}$.
Out of ten $20 bills, two are counterfeit. Six bills are chosen at random. What is the probability that both counterfeit bills are chosen? There are:$\displaystyle _{10}C_6 \:=\:210$possible outcomes. There are: 2 fake and 8 real bills. We want: 2 fake and 4 real bills. There are: .$\displaystyle (_2C_2)(_8C_4) \:=\:1\cdot70 \:=\:70$ways.$\displaystyle P(\text{2 fake, 4 real}) \:=\:\frac{70}{210} \:=\:\frac{1}{3}\$