Combinations (Grade 12 Data Class)

[Sophiaserna]

1. Use the poker hand tutorials as a guide to help you answer this question. What is the probability of getting 4 of a kind in 5 card poker?





2.Suppose you have 6 men and 8 women in a room.
A) How many ways can you select a 6 person jury if there are no restrictions?





B) What is the probability that a randomly selected jury will have exactly 3 women?





C) What is the probability that a randomly selected jury will have no men?






3. Suppose you have a checkers board. (8 squares wide and 8 squares long) and you
have 16 checkers.
A) How many ways can the checkers be placed on the board if there are no restrictions?


B) Now, suppose you must place 2 checkers in each row. How many ways can this be accomplished?

 

[Prof]

Please read the rules for this forum. You need to do some work on a problem and submit it. We will check your work and help you continue if you are stuck.

 

[pka]

1. Use the poker hand tutorials as a guide to help you answer this question. What is the probability of getting 4 of a kind in 5 card poker?

2.Suppose you have 6 men and 8 women in a room.
A) How many ways can you select a 6 person jury if there are no restrictions?

B) What is the probability that a randomly selected jury will have exactly 3 women?

C) What is the probability that a randomly selected jury will have no men?

3. Suppose you have a checkers board. (8 squares wide and 8 squares long) and you
have 16 checkers.
A) How many ways can the checkers be placed on the board if there are no restrictions?

B) Now, suppose you must place 2 checkers in each row. How many ways can this be accomplished?

You indicate that you may be in a secondary school level. That can be special in terms of our help. However you still should try to do these and post your results. That said, we do not have access to your tutorials. Thus we can only give general directions.
For #1. Because there are thirteen ranks so there are \(13\) ways to have four of a kind. How ways are there to have a fifth card?
There are \(\mathcal{C}_5^{52}=\dbinom{52}{5}=\dfrac{52!}{(47!)(5!)}\), combination of \(52\text{ choose }5\) possible five card hands.
You now post an answer to #1