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?
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: .$\displaystyle P(\text{one King}) \;=\;3\cdot\frac{376}{5525} \;=\;\boxed{\frac{1128}{5525}}$
Thank you
Sorry I meant what Soroban did, but wrote 4 at the end by mistake instead of 3.
Why is it that the binomial coefficient calculation works?
Is it like:
(Number of ways of choosing 1 king out of 4)*(Number of ways of choosing 2 others out of 58)/(Number of ways of choosing 3 cards out of 52)
Is this conditional probability?