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A bag contains x blue chips and y red chips. If the probability of sel

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Joined: 02 Sep 2009
Posts: 49251
A bag contains x blue chips and y red chips. If the probability of sel  [#permalink]

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New post 23 Aug 2018, 04:31
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (00:38) correct 24% (00:39) wrong based on 37 sessions

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Director
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Joined: 20 Feb 2015
Posts: 728
Concentration: Strategy, General Management
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Re: A bag contains x blue chips and y red chips. If the probability of sel  [#permalink]

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New post 23 Aug 2018, 04:39
Bunuel wrote:
A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7



probability = \(\frac{red}{(red + blue)}\)= \(\frac{3}{7}\)
that means blue = 4

\(\frac{blue}{red}\) =\(\frac{4}{3}\)
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Re: A bag contains x blue chips and y red chips. If the probability of sel  [#permalink]

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New post 23 Aug 2018, 11:53
Bunuel wrote:
A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7


Given #Blue chips=x, #Red chips=y
Given \(P(R)=\frac{3}{7}\)
Or, \(\frac{y}{x+y}=\frac{3}{7}\)
Or, 7y=3x+3y
Or, 4y=3x
Or, \(\frac{x}{y}=\frac{4}{3}\)

Ans. (D)
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Director
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Concentration: Accounting, Finance
GPA: 3.68
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A bag contains x blue chips and y red chips. If the probability of sel  [#permalink]

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New post 23 Aug 2018, 14:59
Bunuel wrote:
A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =

A. 7/11
B. 3/4
C. 7/4
D. 4/3
E. 11/7



Probability of Red chips:

\(\frac{y}{x + y}\)

Given ,

Blue: x

Red : y

\(\frac{y}{x + y}\) = \(\frac{3}{7}\)
So, x is 4.

Now we are looking for \(\frac{x}{y}\).

\(\frac{x}{y}\) =\(\frac{4}{3}\)
The best answer is D.
A bag contains x blue chips and y red chips. If the probability of sel &nbs [#permalink] 23 Aug 2018, 14:59
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