if 5 card are dealt from 52 cards, determine the probabilty of getting 3 red cards and 2 black face cards.

The answer is 26C3(6C2)/52C5, but I got 52C5(1/2)^5(6/52)^2, how do you know when to use which?

Thank you!

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- Aug 11th 2007, 09:33 PMwbyWhen to use which?
if 5 card are dealt from 52 cards, determine the probabilty of getting 3 red cards and 2 black face cards.

The answer is 26C3(6C2)/52C5, but I got 52C5(1/2)^5(6/52)^2, how do you know when to use which?

Thank you!

- Aug 12th 2007, 03:26 AMSoroban
Hello, wby!

Quote:

If 5 cards are dealt from 52 cards,

determine the probabilty of getting 3 red cards and 2 black face cards.

The answer is: .$\displaystyle \frac{\left(_{26}C_3\right)\left(_6C_2\right)}{_{5 2}C_5}$

but I got: .$\displaystyle \left(_{52}C_5\right)\left(\frac{1}{2}\right)^5\le ft(\frac{6}{52}\right)^2$ .**??**

How do you know when to use which?

That $\displaystyle _{52}C_5$ is the number of possible 5-card hands.

. . It belongs in the*denominator.*

When they state "Five cards are dealt",

. . we generally assume it is__without__replacement.

So the probability of getting a certain color is__not__$\displaystyle \frac{1}{2}$ each time,

. . and the probability of getting a black face card is__not__$\displaystyle \frac{6}{52}$ each time.

Their answer is correct.

You want to draw 3 red cards from the available 26 red cards.

. . There are: .$\displaystyle _{26}C_3$ ways.

You want to draw 2 black face cards from the available 6 black face cards.

. . There are: .$\displaystyle _6C_2$ ways.

. . Hence, there are: .$\displaystyle \left(_{26}C_3\right)\left(_6C_2\right)$ ways to draw 3 red cards and 2 black face cards.

And that is over $\displaystyle _{52}C_5$, the number of possible 5-card hands.