Ok, just to begin lets enumerate the 12 cards in our pile. I'll use JQK for the values and SDHC for the suits (hopefully obvious)
We've got: JS JD JH JC QS QD QH QC KS KD KH KC
Now, for problem 1:
What is the probability that if we select 3 cards they will all have the same value? Since the suits don't matter we have JJJJQQQQKKKK, the probability will be 3 times the probability of picking a specific value three times. P(selecting 3 J)=P(selecting J 1st)*P(selecting J 2nd)*P(selecting J 3rd)=4/12*3/11*2/10=1/55 so P(selecting 3 of the same)=3*P(selecting 3 J)=3/55
I assume you want the probability of selecting 3 of the same suit. There are obviously 3*(4 choose 3) = 12 ways to do that.
For problem 2:
So we want to select 3 cards and get 1 of each value. Think of it like this: There are 4 ways to pick 1 J, then for each of those 4 ways there are 4 ways to add 1 Q, so that's 4*4=16. Then for each of those 16 ways there are 4 ways to add 1 K so that's 16*4=64.
Someone might come by and give you some stock probability answer but I like doing it the hard way!
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