1. ## Pack of Cards

3 cards are given from a pack of 52 playing cards. What is the probability that exactly one is a king?

(4 kings in a pack)

I did:

4/52 * 48/51 * 47/50 * 4

= 0.2...

Is this correct?

2. Originally Posted by Aquafina
3 cards are given from a pack of 52 playing cards. What is the probability that exactly one is a king? I did:

4/52 * 48/51 * 47/50 * 4
= 0.2... Is this correct?
I don't see what is being done above.

$\displaystyle \frac{_4C_1\cdot _{48}C_2}{_{52}C_3}$
That gives the same answer and makes it clear how it works.

3. Hello, Aquafina!

3 cards are drawn from a pack of 52 playing cards.
What is the probability that exactly one is a King?

There are: 4 Kings and 48 Others.

Three cards are drawn: X. Y, Z.

The King can be in any of $\displaystyle {\color{red}3}$ positions.
. . (King-Other-Other), (Other-King-Other), (Other-Other-King)

Suppose it is: (King-Other-Other)
. . The probability is: .$\displaystyle \frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50} \;=\;{\color{red}\frac{376}{5525}}$

Therefore: .$\displaystyle P(\text{one King}) \;=\;3\cdot\frac{376}{5525} \;=\;\boxed{\frac{1128}{5525}}$

4. Originally Posted by Soroban
Hello, Aquafina!

There are: 4 Kings and 48 Others.

Three cards are drawn: X. Y, Z.

The King can be in any of $\displaystyle {\color{red}3}$ positions.
. . (King-Other-Other), (Other-King-Other), (Other-Other-King)

Suppose it is: (King-Other-Other)
. . The probability is: .$\displaystyle \frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50} \;=\;{\color{red}\frac{376}{5525}}$

Therefore: .$\displaystyle P(\text{one King}) \;=\;3\cdot\frac{376}{5525} \;=\;\boxed{\frac{1128}{5525}}$

Thank you

Originally Posted by Plato
I don't see what is being done above.

$\displaystyle \frac{_4C_1\cdot _{48}C_2}{_{52}C_3}$