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I just need an explaination to a question I saw in a statistics book. I understand the concepts but I don't really understand what the question is asking. This question asks '' Find the probability of being dealt at random and without replacement a thirteen card bridge hand consisting of a)thirteen cards of the same suit b) 2 clubs, 3 diamonds, 5 hearts and 3 spades. A thorough explaination will assist me.

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Part a) is easy: There is a total of $52\choose 13$ hands that can be dealt. Out of these hands, there are $4$ of the same suit, hence the answer is $4\over{52\choose13}$.

For b) the number of "good" possibilities is counted by first selecting 2 out of 13 clubs, then 3 out of 13 diamonds, then 5 out of 13 hearts, and finally 3 out of 13 spades. Thus, the answer is ${{13\choose2}{13\choose3}{13\choose5}{13\choose3}}\over{52\choose13}$.

Do we have to worry about mixing the suits to accommodate for the fact that these cards might be dealt in different orders of suits (e.g. in order "a spade, a heart, a club, ...") No! Since we count unordered hands with $52\choose13$ in the denominator, we must do the same in the numerator - which means that we may assume any specific order we wish, as if after dealing we sort our hand by suits.

Edit: Adapted solution to fixed typo in the problem statement.

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    Hagen, I think its suppose to be 13 cards consiting of 2 clubs, 3 diamonds 5 hearts and 3 spades. It was a typo but i get it now..2012-09-03
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For a: There are four hands that are all the same suit. There are $52 \choose 13$ total hands.

For b: 2+3+5+7=17

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    Although providing solutions to a) and b) e$x$plains the question, all the OP asked for was an explanation of the question which would simply be to find the probability of getting all thirteen cards in the same suit (i.e. all the clubs, hearts, diamonds or spades) and in b) that exactly 2 cards are clubs, 3 are diamonds, 5 are hearts and 3 are spades.2012-09-03