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04 Oct 2019, 16:53
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (02:08) correct
28%
(02:13)
wrong
based on 199
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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
48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!)
Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question PS Block 2 question pack will win an $85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.
To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct
The OA and official explanation will be held until the 48 hour window has lapsed.
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
48 Hour Window To Win An $85 EMPOWERgmat Tuition Credit (1 Month Free!)
Share your explanation! The GMAT Club member with the most verified Kudos in total on the 5 question PS Block 2 question pack will win an $85 EMPOWERgmat tuition credit, which will entitle the winner to a full month of complete access to the EMPOWERgmat system. Even if you're not sure about your answer or your rationale, share your explanation to help boost your learning and earn a chance to win.
To be eligible, your explanation must be submitted within the 48 hour window after this post was created and should explain your reasoning why the answer you chose is correct
The OA and official explanation will be held until the 48 hour window has lapsed.
04 Oct 2019, 19:36
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EMPOWERgmatRichC
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
04 Oct 2019, 22:15
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EMPOWERgmatRichC
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.
