# Math Help - 12 Cards are delath from 52 playing cards

1. ## 12 Cards are delath from 52 playing cards

Suppose that a hand of 12 cards is dealt from a normal deck of 52 playing cards. What is the probability that of the 12 cards, 3 will be diamonds, 3 will be clubs, 3 will be spades, and 3 will be hearts?

This is the answer I came up with... 0.07692

Please let me know if this is correct.

Thank you.

2. Originally Posted by desire150
Suppose that a hand of 12 cards is dealt from a normal deck of 52 playing cards. What is the probability that of the 12 cards, 3 will be diamonds, 3 will be clubs, 3 will be spades, and 3 will be hearts?

This is the answer I came up with... 0.07692

Please let me know if this is correct.

Thank you.
Show us how you got that. Then we can answer.

3. Here is how I came up with the answer...

12/52= 0.2307
0.2352/3=0.07692

4. Originally Posted by desire150
Here is how I came up with the answer...

12/52= 0.2307
0.2352/3=0.07692
That is, unfortunately, not even close to understanding this answer.
Do you know about combinations? $_N\mathcal{C}_k=\frac{N!}{k!(N-k)!}$.

Here the number of ways to get three hearts: $_{13}\mathcal{C}_3=\frac{13!}{3!(10)!}$.

Do that for each of the other three suits.

5. Dear Plato,
As you can see, my knowledge is extremely basic.

Here is the result using the formula

13
___
3(10) = 0.433 probability for obtaining 3 diamonds, 3 hearts, 3 clubs, 3 spades.

Thank you.

6. Originally Posted by desire150
Dear Plato,
As you can see, my knowledge is extremely basic. Here is the result using the formula

13
___
3(10) = 0.433 probability for obtaining 3 diamonds, 3 hearts, 3 clubs, 3 spades.
That being the case, has it occurred to you that you are not ready to even understand to solution?
You apparently do not even know what $(13!)$ means.
Now don’t come back at saying that we should tell how.
The function of this forum is not to teach.
We can only help if you do have the basic ideas.

7. Hello, desire150!

A hand of 12 cards is dealt from a normal deck of 52 playing cards.
What is the probability that there are: $3 \diamondsuit s,\;3 \clubsuit s,\; 3 \spadesuit s,\,\text{and }3 \heartsuit s\:?$

$\begin{array}{ccc}\text{Number of ways to get 3 }\diamondsuit s &=& {13\choose3} \\ \\[-4mm]
\text{Number of ways to get 3 }\clubsuit s &=& {13\choose3} \\ \\[-4mm]
\text{Number of ways to get 3 } \spadesuit s &=& {13\choose3} \\ \\[-4mm]
\text{Number of ways to get 3 }\heartsuit s &=& {13\choose3}\end{array}$

Hence, there are: . ${13\choose3}^4$ ways to get 3 of each suit.

There are: . ${52\choose12}$ possible 12-card hands.

Therefore: . $P(\text{3 of each suit}) \;=\;\frac{{13\choose3}^4}{{52\choose12}}$

8. Soroban,
Thank you for explaining it. After doing research and plugging in the numbers into the formula.. this is what I get.

0.231 probability that 3 spades, hears, clubs and diamonds will be chosen.