From a standard deck of 52 cards, a five card hand will be drawn. what is the probability that the hand will have a pair of threes?
I assume you mean a pair of threes and no matter (meaning you cannot have a three of a kind in threes).
There are $\displaystyle {{52}\choose 5}$ ways of choosing five cards.
Now we need to find the number of ways of choosing a pair of threes, since there 4 suits and we chose 2 there are $\displaystyle {4\choose 2}$ ways of choosing the suits for the threes. Now with the threes choosen there are 48 possible cards to choose from and from those another 44 cards to chose from and 40 cards for the last card so that the last three cards are not threes and are not the same with each other. There are are $\displaystyle \frac{48\cdot 44\cdot 40}{3!}$ ways of choosing them because $\displaystyle 3!$ is the number of premutations we can form with them so we need to divide by it to remove this case.
Thus, the probability is,
$\displaystyle \boxed{ \frac{ {4\choose 2} \cdot 48\cdot 44\cdot 40}{3! \cdot {{52}\choose 5}} }$
Hello, dclift!
I will assume you mean exactly one pair of threes.From a standard deck of 52 cards, a five card hand will be drawn.
What is the probability that the hand will have a pair of threes?
There are: .$\displaystyle {52\choose5} \:=\:2,598,960$ possible five-card hands.
To pick two 3's from the available four 3's, there are: .$\displaystyle {4\choose2} \,= \,6$ choices.
The remaining three cards must be chosen from the 48 others cards (non-3's).
. . There are: .$\displaystyle {48\choose3} \:=\:17,296$ ways.
Hence, there are: .$\displaystyle 6 \times 17,296 \:=\:103,776$ hands with a pair of 3's.
Therefore: .$\displaystyle P(\text{pair of 3's}) \:=\:\frac{103,776}{2,698,960} \:=\:\frac{2162}{54,145} \:\approx\:4\%$