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A bag contains x blue chips and y red chips. If the probabil
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02 Feb 2020, 08:04
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. \(\frac{11}{7}\)
A. \(\frac{7}{11}\)
B. \(\frac{3}{4}\)
C. \(\frac{7}{4}\)
D. \(\frac{4}{3}\)
E. \(\frac{11}{7}\)
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Joined: 10 Apr 2015
Posts: 6218
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Re: A bag contains x blue chips and y red chips. If the probabil
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02 Feb 2020, 08:14
02 Feb 2020, 08:14
1
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. \(\frac{11}{7}\)
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
Cheers,
Brent
General Discussion
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Status:Founder & Quant Trainer
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Re: A bag contains x blue chips and y red chips. If the probabil
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01 Feb 2021, 23:22
01 Feb 2021, 23:22
1
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
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
