# in a poker hand consisiting of 5 cards,find the probability of holding a) 3 aces b) 4 hearts & 1 club?

..nid help..tnx

Relevance

a) Since there are 4 aces in a deck, there are several ways to arrange 3 aces:

nCr = n!/((n-r)!*r!)

4C3 = 4!((4-3)!*3!) = 4

Now, there are 48 cards left in the deck if you exclude the fourth ace as a possible card for the remaining 2.

The number of possibilities for those 2 cards is 48C2 = 48!/(46!*2!) = 48*47/2 = 1128

Since there are 52C5 possible poker hands, that is the denominator in the Probability equation

So P(3 aces) = 4C3 * 48C2 / 52C5 = 4*1128/2598960 = 1736 * 10^-3 = 0.1736%

b) There are 13C4 ways to pick 4 hearts. There are 13C1 ways to pick 1 club

P(4 hearts + club) = 13C4 * 13C1 / 52C5 = 9295/2598960

= 3.5764 * 10^-3 = 0.3576 %

• Login to reply the answers
• If you have n objects and chose r of them, the number of combinations is:

n! / ( r! (n-r)! )

this can be written as nCr

There are 52 C 5 = 2598960 total possible hands

there are ( 4 C 3 ) * (48 C 2 ) = 4512 hands with three aces

P( three aces ) = 4512 / 2598960 = 0.001736079

there are (13 C 4 ) * (13 C 1) = 9295 hands with four hearts and one club

P( 4hearts & 1club) = 9295 / 2598960 = 0.003576431

• Login to reply the answers
• Anonymous

1/20, 5/24

• Login to reply the answers
• Anonymous
6 years ago

94/54145

• Login to reply the answers
• Login to reply the answers