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A bag contains x blue chips and y red chips. If the probabil

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huda
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A bag contains x blue chips and y red chips. If the probabil [#permalink] New post   02 Feb 2020, 08:04
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A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is \(\frac{3}{7}\), then \(\frac{x}{y}\) =

A. \(\frac{7}{11}\)

B. \(\frac{3}{4}\)

C. \(\frac{7}{4}\)

D. \(\frac{4}{3}\)

E. \(\frac{11}{7}\)
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Re: A bag contains x blue chips and y red chips. If the probabil [#permalink] New post   02 Feb 2020, 08:14
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huda wrote:
A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is \(\frac{3}{7}\), then \(\frac{x}{y}\) =

A. \(\frac{7}{11}\)

B. \(\frac{3}{4}\)

C. \(\frac{7}{4}\)

D. \(\frac{4}{3}\)

E. \(\frac{11}{7}\)


We can quickly solve this question by testing values that satisfy the given information

GIVEN: The probability of selecting a red chip at random is 3/7
So, it COULD be the case that there are 3 red chips, and a TOTAL of 7 chips
If there are 7 chips and 3 are red, then the other 4 chips must be blue
In other words, y = 3 and x = 4

This means x/y = 4/3

Answer: D

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Re: A bag contains x blue chips and y red chips. If the probabil [#permalink] New post   01 Feb 2021, 23:22
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Blue chips = y
Red chips = x
Total chips = (x + y)

Prob. of selecting a red chip = y / (x + y) = 3 /7
i.e. 7y = 3x + 3y

Solve for x/y as 4/3, so option D
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Re: A bag contains x blue chips and y red chips. If the probabil [#permalink] New post   06 Mar 2021, 04:07
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P(Blue chip) = 3/7. Therefore y = 3 and total chips = 7

Therefore there are 4 red chips = x

x/y = 4/3


Option D

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Re: A bag contains x blue chips and y red chips. If the probabil [#permalink] New post   11 Jul 2024, 15:38
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Re: A bag contains x blue chips and y red chips. If the probabil [#permalink]
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