Probability of a Hand Consisting of Clubs or 2 Suits: Answers & Questions

[cbarker1]

A deck of 52 cards is mixed well, and 5 cards are dealt.(a) It can be shown that (disregarding the order in which the cards are dealt) there are 2,598,960 possible hands, of which only 1287 are hands consisting entirely of clubs.
What is the probability that a hand will consist entirely of clubs? (Give the answer to six decimal places.)
1287/2598960= 4.951980792 x 10^-4

What is the probability that a hand will consist entirely of a single suit? (Give the answer to six decimal places.)

.001981=1287/2598960*4

(b) It can be shown that exactly 63,206 hands contain only diamonds and hearts, with both suits represented.
What is the probability that a hand consists entirely of diamonds and hearts with both suits represented? (Give the answer to five decimal places.)
63206/2598960=.024320

(c) Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits? (Give the answer to five decimal places.)
6*(63206/2598960)=.145918

I am curious if these are the correct answer.

Thank you

 

[Jameson]

Hi Cbarker1,

I get the same answer for the all clubs hand. Can you walk me through your logic for the diamonds and hearts question? I would like to hear how you approached it. :)

 

[cbarker1]

Using the information above in part a, 2598960 and using the other information, 63206.

63206/2598960

 

[Jameson]

Sorry, I didn't see that this was given to you. I thought you should figure out how to get those numbers.

For example, if we want to have a hand with all diamonds and all hearts, then there are 4 possible situations:

1H 4D, 2H 3D, 3H 2D, 4H 1D

To calculate the number of combos we can do this: $$\binom{13}{1}\binom{13}{4}+\binom{13}{2}\binom{13}{3}+\binom{13}{3}\binom{13}{2}+\binom{13}{4}\binom{13}{1}=63206$$

That's how they got that number.

So yep, all of your answers look good to me. (Yes)