1 of the 4 suits ---> C(4,1) = 4
5 from one of those suits = C(13,5) ---> 1287
5 from the whole deck ---> C(52,5) = 2598960
so prob = 4(1287)/2598960
= 33/16660 = .00198
Five cards are drawn at random without replacement from and ordinary deck of 52. What is the probability all 5 cards are from the same suit?
4 answers
I'm confused as to how you're getting these numbers... Could you possible break it down a little farther please? I don't understand how you get from C(52,5)=2598960...Thank you so much for helping!
Have you learned the difference between permutations and combinations?
in combinations the order does not matter.
(if I hold 5 cards in my hand, the order in which they were dealt to me does not matter)
C(13,5) means ..
how many groups of 5 can I form from 13
It is short form for 13!/(5!8!)
= 13x12x11x10x...x2x1/(8x7x6x..x2x1)(5x4x3x2x1))
= 13x12x11x10x9/(5x4x3x2x1) = 154440/120 = 1287
On your calculator there is a button labeled
nCr
so enter
13
2nd F
nCr
5
=
You should get 1287
in combinations the order does not matter.
(if I hold 5 cards in my hand, the order in which they were dealt to me does not matter)
C(13,5) means ..
how many groups of 5 can I form from 13
It is short form for 13!/(5!8!)
= 13x12x11x10x...x2x1/(8x7x6x..x2x1)(5x4x3x2x1))
= 13x12x11x10x9/(5x4x3x2x1) = 154440/120 = 1287
On your calculator there is a button labeled
nCr
so enter
13
2nd F
nCr
5
=
You should get 1287
253/9996