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A box contains b blue balls and r red balls. If 4 blue balls and 3 red
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29 Sep 2021, 04:21
29 Sep 2021, 04:21
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A box contains b blue balls and r red balls. If 4 blue balls and 3 red balls are added to the box, and if one ball is obtained at random from the box, then the probability that a blue ball is obtained can be represented by
A. \(\frac{b}{r}\)
B. \(\frac{b}{b+r}\)
C. \(\frac{b+4}{r+3}\)
D. \(\frac{b+4}{b+r+4}\)
E. \(\frac{b+4}{b+r+7}\)
A. \(\frac{b}{r}\)
B. \(\frac{b}{b+r}\)
C. \(\frac{b+4}{r+3}\)
D. \(\frac{b+4}{b+r+4}\)
E. \(\frac{b+4}{b+r+7}\)
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Re: A box contains b blue balls and r red balls. If 4 blue balls and 3 red
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29 Sep 2021, 05:18
29 Sep 2021, 05:18
1
Carcass wrote:
A box contains b blue balls and r red balls. If 4 blue balls and 3 red balls are added to the box, and if one ball is obtained at random from the box, then the probability that a blue ball is obtained can be represented by
A. \(\frac{b}{r}\)
B. \(\frac{b}{b+r}\)
C. \(\frac{b+4}{r+3}\)
D. \(\frac{b+4}{b+r+4}\)
E. \(\frac{b+4}{b+r+7}\)
A. \(\frac{b}{r}\)
B. \(\frac{b}{b+r}\)
C. \(\frac{b+4}{r+3}\)
D. \(\frac{b+4}{b+r+4}\)
E. \(\frac{b+4}{b+r+7}\)
P(ball is blue) = (total number of blue balls)/(total number of all balls)
total number of blue balls
We start with b blue balls, and then 4 more blue balls are added.
Total number of blue balls = b + 4
total number of all balls
We start with (b + r) balls, and then 7 more balls are added (4 blue balls and 3 red balls)
Total number of ALL balls = b + r + 7
We get:
P(ball is blue) = (b + 4)/(b + r + 7)
Answer: E
gmatclubot
Re: A box contains b blue balls and r red balls. If 4 blue balls and 3 red [#permalink]
29 Sep 2021, 05:18
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