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What is the probability that a five-card poker hand con- tains the 2 diamonds and 3 spades?

My Approach/Attempt-: N(contains 2 diamonds and 3 spades)=$\binom{4}{1} * \binom{13}{2} * \binom{4}{1} *\binom{13}{3}$

$\binom{4}{1}$ $\Rightarrow$ For selecting one suite which is diamond.

$\binom{13}{2}$ $\Rightarrow$ For selecting 2 cards from diamond.

$\binom{4}{1}$ $\Rightarrow$ For selecting one suite which is Spade.

$\binom{13}{3}$ $\Rightarrow$ For selecting 3 cards from spade.

$P\left ( E \right )=\left ( \binom{4}{1}*\binom{13}{2}*\binom{4}{1}*\binom{13}{3} \right )/\binom{52}{5}$ Am i correct?

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  • $\begingroup$ You don't need either of the $\binom{4}{1}$. Once you remove those, your answer is correct, assuming the question is asking for the probability of your hand having two diamonds and three spades. However, it might be asking for the probability that your hand contains the card "2 of diamonds" and the card "3 of spades," which is different. $\endgroup$ – angryavian Sep 28 '16 at 5:52
  • $\begingroup$ can you please explain when we are supposed to select the 4 suite and chose it ,like $\binom{4}{1}$ @ang $\endgroup$ – virat Sep 28 '16 at 5:55
  • $\begingroup$ @angryavian ,Yes the question was "2 of diamonds" and the card "3 of spades," i modified it to "2 diamonds and 3 spades" $\endgroup$ – virat Sep 28 '16 at 5:56
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The total number of ways of a 5-card poker hand is: C( 52, 5 ).

The number of ways of the hand of cards which have to include 2 of diamonds and 3 of spades is:

C(1,1)C(1,1)C( 50, 3) = C( 50, 3 ). C(1,1) indicates that there only one way to get 2 of diamonds or 3 of spades. After we have these two in a hand there 50 cards left to hand the rest of the 5-card hand(3 cards left):C(50,3).

So the probability is: C(50, 3)/C(52, 5) = 0.00754

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