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EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
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A standard set of billiard balls includes 16 balls: 15 numbered balls  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 73% (02:04) correct 27% (02:17) wrong based on 41 sessions

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EMPOWERgmat PS Series:
Pack 2, Question 3

A standard set of billiard balls includes 16 balls: 15 numbered balls – with each of the numbers 1 to 15, inclusive, appearing on one ball each – and 1 white ball which is not numbered. If an additional ball from another standard set of billiard balls is accidentally included with an existing standard set of 16 balls, then what is the probability that the extra ball is NOT numbered with a multiple of 2 or 3?

A. 1/8
B. 1/4
C. 2/5
D. 3/8
E. 9/16

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Re: A standard set of billiard balls includes 16 balls: 15 numbered balls  [#permalink]

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EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Pack 2, Question 3

A standard set of billiard balls includes 16 balls: 15 numbered balls – with each of the numbers 1 to 15, inclusive, appearing on one ball each – and 1 white ball which is not numbered. If an additional ball from another standard set of billiard balls is accidentally included with an existing standard set of 16 balls, then what is the probability that the extra ball is NOT numbered with a multiple of 2 or 3?

A. 1/8
B. 1/4
C. 2/5
D. 3/8
E. 9/16

There are 16 balls, out of which 15 are numbered 1 to 15..

Now, restrictions are..
(1) Not a multiple of 2, so 1 to 14 has 14/2 or 7 numbers.
(2) Not a multiple of 3, so 1 to 15 has 15/3 or 5 numbers.
(3) Common multiples of 2 and 3 or LCM(2, 3) or 6 are TWO - 6 and 12

Total restrictions = 7+5-2=10

Probability of NOT being a multiple of 2 and 3 = $$1-\frac{10}{16}=1-\frac{5}{8}=\frac{3}{8}$$

D
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A standard set of billiard balls includes 16 balls: 15 numbered balls  [#permalink]

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EMPOWERgmatRichC wrote:
EMPOWERgmat PS Series:
Pack 2, Question 3

A standard set of billiard balls includes 16 balls: 15 numbered balls – with each of the numbers 1 to 15, inclusive, appearing on one ball each – and 1 white ball which is not numbered. If an additional ball from another standard set of billiard balls is accidentally included with an existing standard set of 16 balls, then what is the probability that the extra ball is NOT numbered with a multiple of 2 or 3?

A. 1/8
B. 1/4
C. 2/5
D. 3/8
E. 9/16

Total number of balls numbered in multiples of 2 = 7(2,4,6,8,10,12,14)
Total number of balls numbered in multiples of 3 = 5(3,6,9,12,15)

Since 6 and 12 are repeating, total number of balls with multiples of 3 = 5 - 2 = 3

Now,
Required Probability = 1 - (Prob. of balls with multiples of 2 + Prob. of balls with multiples of 3)
= 1 - $$(\frac{7}{16} + \frac{3}{16})$$
= 1 - $$\frac{10}{16}$$
= 1 - $$\frac{5}{8}$$
= $$\frac{3}{8}$$

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EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15946
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: A standard set of billiard balls includes 16 balls: 15 numbered balls  [#permalink]

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OFFICIAL EXPLANATION

Hi All,

We're told that a standard set of billiard balls includes 16 balls: 15 numbered balls – with each of the numbers 1 to 15, inclusive, appearing on one ball each – and 1 white ball which is not numbered. If an additional ball from another standard set of billiard balls is accidentally included with an existing standard set of 16 balls, then what is the probability that the extra ball is NOT numbered with a multiple of 2 or 3. While this question might be a bit 'wordy', the 'math' behind this Probability question isn't that difficult - and by simply 'mapping out' what is described, we can get to the correct answer without too much work.

We have 16 balls: fifteen are numbered 1-15, inclusive and one is un-numbered. To answer the question, we have to determine what fraction of those balls are NOT numbered with a multiple of 2 or a multiple of 3. Listing out the options is a fairly easy way to find what we're looking for.

Multiples of 2: 2, 4, 6, 8, 10, 12, 14
Multiples of 3: 3, 6, 9, 12, 15

Notice how the values 6 and 12 appear in BOTH lists; we should not count those numbers twice though (just once each). At this point, we can either calculate the fraction of balls in this combined list... and subtract that fraction from the number 1... or we can list out the balls that fit what we're looking for...

The list of balls that fits what we are looking for is: 1, 5, 7, 11, 13 and the un-numbered ball --> 6 total balls out of 16 total --> 6/16 = 3/8

GMAT assassins aren't born, they're made,
Rich
_________________ Re: A standard set of billiard balls includes 16 balls: 15 numbered balls   [#permalink] 07 Oct 2019, 19:40
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