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A lottery game consists of the host removing one ball at a

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A lottery game consists of the host removing one ball at a [#permalink] New post 26 Jun 2020, 00:48
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A lottery game consists of the host removing one ball at a time from an opaque jar. Each ball has one of the digits (0 – 9) written on it, and no two have the same number on them. If the host removes three balls without replacement, what is the probability that the sum of the numbers written on the balls equals 24?

(A) 1/3

(B) 3/10

(C) 1/720

(D) 1/120

(E) 1/40
[Reveal] Spoiler: OA

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Re: A lottery game consists of the host removing one ball at a [#permalink] New post 26 Jun 2020, 06:33
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The total number of potential combinations of 3 balls being drawn (with order mattering):
10*9*8 = 720
(the first draw we have 10 potential balls to draw from, second draw we have 9 leftover balls, and last draw we have 8 to choose from)

The only way to get 24 is to draw these numbers:
9+8+7 =24
However, 9 8 and 7 can all be drawn in any order (ex. getting 7,9,8 is the same as 8,7,9)
There are 6 different orderings of how we can draw those 3 numbers (7,8,9 - 7,9,8 - 8,9,7 - 8,7,9 - 9,7,8 - 9,8,7) (with order mattering)

Thus the ratio of our desired outcome to total outcomes is 6/720 , which simplifies to 1/120 (Answer D)


* Can also not worry about the order in the calculation of total outcomes and divide 720 by 3!. Then we would have only 1 outcome that's desired (since order wouldn't matter) and we would once again get 1/120
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Re: A lottery game consists of the host removing one ball at a   [#permalink] 26 Jun 2020, 06:33
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