# probability

• Mar 22nd 2013, 04:49 PM
amyl95
probability
(i) how many different poker hands can be dealt to a player?
(ii) how many hands include the Ace of spades?
(iii) how many hands do not contain include the ace of spades?
(iv)how many hands include all 4 aces?
(v) how many hands contain exactly 3 aces?

so i know that there are 52 cards in a pack and that a poker hand is 5 cards, so the answer to (i) is 52 choose 5 52C5 , using the calculator i get 2598960.

and i know that there is only one ace of spades in a pack of cards but i'm not quite sure how to work out part (ii) and if i knew the answer to this then obviously i could work out the answer to part (iii)

could anyone give me a bit of help please? thanks.
• Mar 22nd 2013, 05:41 PM
Soroban
Re: probability
Hello, amyl95!

Quote:

(1) How many different poker hands can be dealt to a player?
You are correct!

There are: . ${52\choose5} \:=\:\frac{52!}{5!\,47!} \:=\:2,\!598,\!960$ possible hands.

Quote:

(2) How many hands include the Ace of Spades?
We want the $A\spadesuit$ and four of the 51 other cards.

. . $1\cdot{51\choose4} \:=\:249,\!900$

Quote:

(3) How many hands do not contain the Ace of Spades?
We want 5 of the other 51 cards.

. . ${51\choose5} \:=\:2,\!349,\!060$

Quote:

(4) How many hands include all 4 Aces?
There is one way to have all four Aces.
And we want one of the other 48 cards.

. . $1\cdot{48\choose1} \:=\:48$

Quote:

(5) How many hands contain exactly 3 Aces?

There are ${4\choose3}$ choices for the three Aces.
And there are ${48\choose2}$ ways to choose the other two cards.

. . ${4\choose3}{48\choose2} \:=\:4\cdot 1128 \:=\:4512$
• Mar 23rd 2013, 03:58 AM
amyl95
Re: probability
thanks so much I'm new to probability so I need all the help I can get.