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?
Best Answer
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