Five cards are drawn from a pack of 52 cards. What is the probability of getting exactly 3 Diamonds and 2 Jacks?

Question

The probability of getting exactly 3 Diamonds and 2 Jacks in a 5-card hand drawn from a standard deck of 52 cards is calculated as follows:

There are 13 diamonds in a deck of cards and 4 Jacks. The number of ways to choose 3 diamonds out of 13 is “13 choose 3” or (13 C 3) = 286. The number of ways to choose 2 Jacks out of 4 is “4 choose 2” or (4 C 2) = 6.

The total number of ways to choose any 5 cards out of a deck of 52 is “52 choose 5” or (52 C 5) = 2,598,960.

So the probability of getting exactly 3 Diamonds and 2 Jacks is (286 * 6) / 2,598,960 = 0.00066.

I am getting this but this doesn't match with the answer.

Answer 1

Hint

For the numerator, to the case I where you have $3$ non-Jack$\diamond$ and $2$ non-$\diamond$ jacks, add the case II where you have
Jack $\diamond$ + $2$ other $\diamond$ + $1$ other Jack + $1$ non-$\diamond$ non-Jack

${\dbinom{12}3\dbinom32 + \dbinom11\dbinom{12}2\dbinom31\dbinom{36}1\over\dbinom{52}5}$

Answer 2

Let S be the sample space and E be the universal space. We have that $|E|={52\choose 5}=2598960.$

S Type I: Elements of $S$ which have $3$ diamonds with no jack, $2$ jacks which are non-diamond. The number of S Type I elements is ${13-1\choose 3}{4-1\choose 2}={12\choose 3}{3\choose 2}=660$.

S Type II: Elements of $S$ which have $3$ diamonds with jack, $1$ other jack, $1$ other non-jack and non-diamond. The number of S Type II elements is ${13-1\choose 2}{4-1\choose 1}{52-13-3\choose 1}={12\choose 2}{3\choose 1}{36\choose 1}=66\times 108=7128$.

Hence, $S=660+7128=7788$ and the probabilty of getting exactly $3$ diamonds and $2$ jacks in a $5$-card hand is $p=\frac{|S|}{|E|}=\frac{7788}{2598960}\approx 0.00299658$.