Question 1117797
you have:


4 ten dollar certificates
5 twenty dollar certificates
3 fifty dollar certificates.


that's a total of 12 certificates, 5 of which are twenty dollar certificates and 7 of which are not twenty dollar certificates.


you are drawing 2 times.


you can draw:


0 twenty dollar certificates on the first 2 draws.
1 twenty dollar certificate on the first draw and 0 twenty dollar certificates on the second draw.
0 twenty dollar certificates on the first draw and 1 twenty dollar certificate on the second draw.
2 twenty dollar certificates on the first 2 draws.


no other options are possible.


your probabilities will be as follows:


0 twenty dollar certificates on the first 2 draws.


the probability will be 7/12 * 6/11 = 42/132.


1 twenty dollar certificate on the first draw and 0 twenty dollar certificates on the second draw.


the probability will be 5/12 * 7/11 = 35/132.


0 twenty dollar certificates on the first draw and 1 twenty dollar certificate on the second draw.


the probability will be 7/12 * 5/11 = 35/132.


2 twenty dollar certificates on the first 2 draws.


the probability will be 5/12 * 4/11 = 20/132.


your total probabilities will be (42 + 35 + 35 + 20) / 132 = 132/132 = 1.


since the sum of all probabilities in the probability distribution must be equal to 1, then this is as it should be.