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There are 18 balls in a jar. You take out 3 blue balls without putting

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There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 11 Jul 2016, 12:42
1
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

93% (00:53) correct 7% (01:17) wrong based on 104 sessions

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Re: There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 11 Jul 2016, 13:32
E.

3 + 15/5
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Re: There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 11 Jul 2016, 16:45
6 = 3 blue balls + 15 / 5
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Re: There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 09 Jul 2017, 04:44
There are 18 balls in total and 3 are removed without replacement . So there are total of 18-3=15 balls remaining. the probability of drawing the remaining blue balls is 1/5 and lets says there are n balls remaining out of a total of 15 balls. So n/15=1/5 therefore n=3+ 3 which were removed in the beginning =6
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There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 09 Jul 2017, 08:13
Bunuel wrote:
There are 18 balls in a jar. You take out 3 blue balls without putting them back inside, and now the probability of pulling out a blue ball is 1/5. How many blue balls were there in the beginning?

A. 12.
B. 9.
C. 8.
D. 7.
E. 6.

Another method, very quick in this case: test answers. b = blue.

1. Start with C = 8b.
(8b - 3b) = 5 blue
18 - 3 = 15 total (whole)
Blue part: Whole = probability
5:15 = \(\frac{1}{3}\) --> Number of blue balls creates probability higher than prompt, so # of blue is too high.

2. Try E = 6b.
(6b - 3b) = 3 blue
(18 - 3) = 15 total
Blue part: Whole = probability
\(\frac{3}{15}\) = \(\frac{1}{5}\). That works.

Answer E
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Re: There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 15 Jul 2017, 07:15
Bunuel
Can you Please help...
The explanation above are not of much help for me
TIA
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Re: There are 18 balls in a jar. You take out 3 blue balls without putting  [#permalink]

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New post 15 Jul 2017, 07:21
ehsan090 wrote:
Bunuel
Can you Please help...
The explanation above are not of much help for me
TIA


There are 18 balls in a jar. You take out 3 blue balls without putting them back inside, and now the probability of pulling out a blue ball is 1/5. How many blue balls were there in the beginning?

A. 12.
B. 9.
C. 8.
D. 7.
E. 6.

Say there are x blue balls.

If we take out 3 blue balls without putting them back inside, the probability of pulling out a blue ball will be 1/5, so (x - 3)/(18 - 3) = 1/5. Solving gives x = 6.

Answer: E.
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Re: There are 18 balls in a jar. You take out 3 blue balls without putting   [#permalink] 15 Jul 2017, 07:21
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