1. ## Cards...

Hi guys!

I'm working on a probablity question that seems relatively easy, but for some reason has me totally stumped

Four cards are drawn from a deck at random without replacement from a standard deck of 52 cards. What is the probability that all are of different suits?

Any suggestions!!

Thanks a bunch!

2. Originally Posted by strgrl
Hi guys!

I'm working on a probablity question that seems relatively easy, but for some reason has me totally stumped

Four cards are drawn from a deck at random without replacement from a standard deck of 52 cards. What is the probability that all are of different suits?

Any suggestions!!

Thanks a bunch!
$\left( \frac{13}{52} \right) \left( \frac{13}{51} \right) \left( \frac{13}{50} \right) \left( \frac{13}{49} \right)$.

Alternativley:

$(4!) \frac{{13 \choose 1} {13 \choose 1} {13 \choose 1} {13 \choose 1}}{{52 \choose 4}}$.

3. thanks, mr fantastic, for your quick response.

the second solution correlates to the format i've been learning in class and is very similar to what i was toying with on my homework. however, the issue i keep running across with both your solution and mine is that the probability ends up being over 1 (approx. 2.53) in both cases.

how is this possible?

thank you!

4. Originally Posted by strgrl
thanks, mr fantastic, for your quick response.

the second solution correlates to the format i've been learning in class and is very similar to what i was toying with on my homework. however, the issue i keep running across with both your solution and mine is that the probability ends up being over 1 (approx. 2.53) in both cases.

how is this possible?

thank you!
Then you're making a data entry error on your calculator. I get 2197/499800 = 0.0044 (correct to four decimal places).

5. Hello, strgrl!

I'm getting a different answer . . .

Four cards are drawn at random without replacement from a 52-card deck.
What is the probability that all are of different suits?

The first card can be any card: . $\frac{52}{52}$

The second must be not match the first suit: . $\frac{39}{51}$

The third must not match the first two suits: . $\frac{26}{50}$

The fourth must be of the fourth suit: . $\frac{13}{49}$

Therefore: . $P(\text{4 suits}) \:=\:\frac{52}{52}\cdot\frac{39}{51}\cdot\frac{26} {50}\cdot\frac{13}{49} \;=\;\frac{2,197}{20,825}$

6. Originally Posted by Soroban
Hello, strgrl!

I'm getting a different answer . . .

The first card can be any card: . $\frac{52}{52}$

The second must be not match the first suit: . $\frac{39}{51}$

The third must not match the first two suits: . $\frac{26}{50}$

The fourth must be of the fourth suit: . $\frac{13}{49}$

Therefore: . $P(\text{4 suits}) \:=\:\frac{52}{52}\cdot\frac{39}{51}\cdot\frac{26} {50}\cdot\frac{13}{49} \;=\;\frac{2,197}{20,825}$
You're quite right, Soroban. I mucked up. (Now we're even )

7. Thank you soroban and mr fantastic!