find the number of possible 5-card hands taken from a standard 52-card deck that contain 2 aces and 3 cards th?
3 cards that are not aces.
- JaredLv 710 years agoFavorite Answer
There are FOUR Aces.
You MUST choose 3 of the 4. You can do this as follows:
(4 c 3) ......."four choose three"
(which you should immediately know equals 4)
Now how many cards are left to choose from?
NO aces, therefore there are:
52 - 4 = 48 cards left.
Therefore you just choose 2 from the rest (this one you will NOT know immediately...of course it's big):
(48 c 2) = 1128
Therefore the number of hands (distinct hands) should be:
4 * 1128 = 4512
This makes the probability of this type of hand to be:
Total number of hands:
(52 c 5) = 2,598,960
p = 4512 / 2,598,960 = .001736 = .1736%
WOW, I did that totally wrong, it should be 2 aces and 3 non-aces...doesn't really change much here's the correct one quickly:
(4 c 2) * (48 c 3) = 6 * 17,296 = 103,776
Of course this should be an "easier" hand to get (rather than 3 aces like I did above) and sure enough there are MORE hands that satisfy this property.
The probability of getting THIS hand would be:
103,776 / 2,598,960 = .03993 = 4% That's almost 1 in 20.
- 10 years ago
how did you get that number? that might be helpful to the person who asked that way they are able to do these problems on their own.
- RichLv 710 years ago