A standard set of billiard balls includes 16 balls: 15 numbered balls

Question

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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chetan2u · Accepted Answer

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

unraveled · Answer

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}

Answer D.

EMPOWERgmatRichC · Answer

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

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

shaliny · Answer

here query is that why additional ball is not counted in total balls, that is 16+1=17??

Bunuel · Answer

From a standard set, the extra ball is equally likely to be any of the 16 balls.

Among 1 to 15, the multiples of 2 or 3 are 10 numbers (2, 3, 4, 6, 8, 9, 10, 12, 14, 15), so 5 numbers are not multiples of 2 or 3 (1, 5, 7, 11, 13). Also, the white ball is not numbered.

Favorable outcomes = 5 + 1 = 6 out of 16, so the probability is 6/16 = 3/8.

Answer: D.

Bunuel · Answer

Because the probability is about what the extra ball is before it gets mixed in.

You are selecting 1 ball from another standard set to be the “extra ball.” That set has 16 possible balls, all equally likely. So the denominator is 16, not 17.

The 17 only matters after it’s added, but at that point you are no longer choosing which ball it was.

Hope it's clear.

A standard set of billiard balls includes 16 balls: 15 numbered balls

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A standard set of billiard balls includes 16 balls: 15 numbered balls

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