Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
What is the probability of drawing AT LEAST one ace, when drawing 13 cards?
When working with a deck that is missing one non-ace. So I am drawing from a deck of 51 cards. And a description would be nice :)
3 Answers
- M3Lv 71 decade agoFavorite Answer
there are only 51 cards which include 4 aces
ways of not drawing an ace = 47C13
ways of drawing any 13 cards = 51C13
so P[ no ace ] = 47C13 / 51C13
and P[ at least one ace ] = 1 - 47C13 / 51C13
= 0.7046
------------
- ?Lv 71 decade ago
The odds of NOT drawing an ace on the first card (47/51) is multiplied by the chances of NOT drawing an ace on the second card (46/50) multiplied by then chances of NOT drawing an ace on the third card (45/49) etc., etc., etc., all the way to the 13th card. (35/39)
The product of all multiplying all 13 of these fractions is: .2953782
That's the chances of NOT drawing an ace. Subtract this from 1 to find the probability OF drawing an ace. 1 - .2953782 = .7046218. That's your answer. 70.46218%
- 1 decade ago
ok so u have 4/51 for one so 51-13 is 38 so go 4/38 then u get 2/19 so that should be the answer
