Question

A 3-card poker hand is dealt at random from a standard 52-card deck. What is the total number of possible hands? What is the total number of possible hands if the hand contains exactly one heart?

Answer 1

Answer:

What is the total number of possible hands?

$nCx = \frac{n!}{x! (n-x)!}$

For this case n = 52 and x =3 and if we replace we got:

$\frac{52!}{3! (52-3)!}= \frac{52!}{3! 49!}= \frac{52*51*50*49!}{3! 49!}= \frac{52*51*50}{3*2*1}= 22100$

So then we have a total of 22100 ways in order to select 3 cards from a total of 52 cards.

What is the total number of possible hands if the hand contains exactly one heart?

For the possible cases we can do this:

$(13C1) *(39C2)= \frac{13!}{1! 12!} \frac{39!}{2! 37!}= 13 \frac{39*38*37!}{2 37!}= 13 *\frac{39*38}{2}= 13*741 =9633$

Since we have 13 hearts and we want to select 1 and the two remain cards are non hearts. So we have 9633 ways in order to have only one heart in a hand of 3, and the probability would be:

$p = \frac{9633}{22110}= 0.436$

Step-by-step explanation:

What is the total number of possible hands?

For this case we assume that w ehave an standard deck of 52 cards and we want to know on how many ways we can select 3 cards from the total of 52, so for this case since the order no matter we can use the formula for combination given by:

$nCx = \frac{n!}{x! (n-x)!}$

For this case n = 52 and x =3 and if we replace we got:

$\frac{52!}{3! (52-3)!}= \frac{52!}{3! 49!}= \frac{52*51*50*49!}{3! 49!}= \frac{52*51*50}{3*2*1}= 22100$

So then we have a total of 22100 ways in order to select 3 cards from a total of 52 cards.

What is the total number of possible hands if the hand contains exactly one heart?

For this case we need to remember that in a tandard deck we have 13 hearts and we want that in the 3 selected card we have just one heart, and we can find this probability using the definition of empirical probability:

$p =\frac{Possible}{Total}$

Where the total cases on this case are 22100

For the possible cases we can do this:

$(13C1) *(39C2)= \frac{13!}{1! 12!} \frac{39!}{2! 37!}= 13 \frac{39*38*37!}{2 37!}= 13 *\frac{39*38}{2}= 13*741 =9633$

Since we have 13 hearts and we want to select 1 and the two remain cards are non hearts. So we have 9633 ways in order to have only one heart in a hand of 3, and the probability would be:

$p = \frac{9633}{22110}= 0.436$