# Probability counting excercise

• March 8th 2011, 03:27 PM
Sheld
Probability counting excercise
Ten cards are drawn randomly from a deck of 52 cards (13 cards of each of the 4 different suits).

Each selected card is put into a pile that depends on what suit the card is. (4 total piles)

a) What is the probability that the largest pile contains 4 cards, the next largest pile has 3 cards, the next largest has two cards and the smallest has 1 card?

I am not sure how to begin part a. You want to select 4 of one suit, 3 of a different suit, then 2 of a yet again different suit and then a single card of the last suit. If the numbers were tied to specific suits I could see how to do part a, but this more general case is confusing me.

b) What is the probability that two of the piles have 3 cards, one has 4 cards, and one has no cards?

So when you draw ten cards you don't get 1 of the 4 suits. However, like part a, since the numbers aren't tied to specific suits, I am confused on how to proceed.

Any help would be greatly appreciated. Thank you kindly.
• March 8th 2011, 05:01 PM
awkward
a) Do you suppose you can solve the problem if given the suits? Say 4 hearts, 3 clubs, 2 diamonds, and 1 spade?

If so, then just multiply your answer by 4! to allow for the possible arrrangements of the suits.
• March 8th 2011, 06:47 PM
Sheld
so the answer to part a is

$\frac{4!*\begin{pmatrix}\;\;13
&\\\;\;4
\end{pmatrix}*\begin{pmatrix}\;\;13
&\\\;\;3
\end{pmatrix}\begin{pmatrix}\;\;13
&\\\;\;2
\end{pmatrix}\begin{pmatrix}\;\;13
&\\\;\;1
\end{pmatrix}}{\begin{pmatrix}\;\;52
&\\\;\;10
\end{pmatrix}}$

two questions

why isn't the denominator $52*51*50*49*48*47*46*45*44*43$

also in part b do we multiply by $3!$ or still by $4!$ since we still have to consider the pile of zero cards?
• March 8th 2011, 07:23 PM
Soroban
Hello, Sheld!

Here's another approach . . . I'll do part (a).

Quote:

Ten cards are drawn randomly from a deck of 52 cards.
Each selected card is put into a pile that depends on what suit the card is.

(a) What is the probability that the largest pile contains 4 cards, the next largest
has 3 cards, the next largest has 2 cards and the smallest has 1 card?

There are: . $\displaystyle{52\choose10}$ possible 10-card hands.
. . This is the denominator of the probability.

It is not $_{52}P_{10}$ because the order of the cards is not considered.

For the first pile, there are 4 choices for the suit.
. . Then there are ${13\choose4}$ ways to get 4 cards of that suit.

For the second pile, there are 3 choices for the suit.
. . Then there are ${13\choose3}$ ways to get 3 cards of that suit.

For the third pile, there are 2 choices for the suit.
. . Then there are ${13\choose2}$ ways to get 2 cards of that suit.

For the fourth pile, there is 1 choice for the suit.
. . Then there are ${13\choose1}$ ways to get 1 card of that suit.

Hence, the numerator is: . $\displaystyle 4\!\cdot\!{13\choose4}\!\cdot\! 3\cdot\!{13\choose3}\!\cdot\!2\cdot\!{13\choose2}\ !\cdot\!1\!\cdot\!{13\choose1}$

• March 8th 2011, 08:49 PM
Sheld
Ah I understand now. The 10 cards are the 10 cards, it doesnt matter what order they are in.

I also believe I got the answer to part b)

$\frac{4!*\begin{pmatrix}\;\;13
&\\\;\;4
\end{pmatrix}*\begin{pmatrix}\;\;13
&\\\;\;3
\end{pmatrix}\begin{pmatrix}\;\;13
&\\\;\;3
\end{pmatrix}\begin{pmatrix}\;\;13
&\\\;\;0
\end{pmatrix}}{\begin{pmatrix}\;\;52
&\\\;\;10
\end{pmatrix}}$

where
$\begin{pmatrix}\;\;13
&\\\;\;0
\end{pmatrix} = 1$

Thanks a lot!