gmatter0913 wrote:
I solved this question as follows and I know I am wrong. The problem is: I don't know why am I wrong?
First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4
Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4
It looks like I have done a permutation on the numerator. Could you help me with the error in the above?
First card can be picked in 12C1 ways
Second card can be picked in 10C1 ways
Third card can be picked in 8C1 ways
Fourth card can be picked in 6C1 ways
All possibilities to pick 4 cards out of 12 = 12C4
Probability to have atleast one pair = 1 - prob to have no pairs = 1 - 12c1*10c1*8c1*6c1/12c4
It looks like I have done a permutation on the numerator. Could you help me with the error in the above?
Your denominator is correct 12C4=495. The problem is the numerator: the probability of picking 4 different values among 6 possible values: 6C4=15. But because each of the four cards can be black or red, each of the four slots can have 2 colors , so we have to multiply 15 by 2*2*2*2=16 (each of the four cards can be B or R).
Hope I've explained myself well.
So now P=\frac{495-15*16}{495}=\frac{17}{33}.