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A box contains r red chips and b blue chips only, where r > 0 and b >
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30 Sep 2021, 09:00
30 Sep 2021, 09:00
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A box contains \(r\) red chips and \(b\) blue chips only, where \(r > 0\) and \(b > 0\). If two balls are randomly selected without replacement, what is the probability that the two balls are different colors?
A) \(\frac{br}{(b+r)(b+r-1)}\)
B) \(\frac{2(b+r)}{(b+r)(b+r-1)}\)
C) \(\frac{br}{(b+r)}\)
D) \(\frac{2br}{(b+r)(b+r-1)}\)
E) \(\frac{2br}{(b+r)(b+r)}\)
A) \(\frac{br}{(b+r)(b+r-1)}\)
B) \(\frac{2(b+r)}{(b+r)(b+r-1)}\)
C) \(\frac{br}{(b+r)}\)
D) \(\frac{2br}{(b+r)(b+r-1)}\)
E) \(\frac{2br}{(b+r)(b+r)}\)
Re: A box contains r red chips and b blue chips only, where r > 0 and b >
[#permalink]
30 Sep 2021, 19:12
30 Sep 2021, 19:12
D
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Posted from my mobile device
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: A box contains r red chips and b blue chips only, where r > 0 and b >
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01 Oct 2021, 05:50
01 Oct 2021, 05:50
1
GreenlightTestPrep wrote:
A box contains \(r\) red chips and \(b\) blue chips only, where \(r > 0\) and \(b > 0\). If two balls are randomly selected without replacement, what is the probability that the two balls are different colors?
A) \(\frac{br}{(b+r)(b+r-1)}\)
B) \(\frac{2(b+r)}{(b+r)(b+r-1)}\)
C) \(\frac{br}{(b+r)}\)
D) \(\frac{2br}{(b+r)(b+r-1)}\)
E) \(\frac{2br}{(b+r)(b+r)}\)
A) \(\frac{br}{(b+r)(b+r-1)}\)
B) \(\frac{2(b+r)}{(b+r)(b+r-1)}\)
C) \(\frac{br}{(b+r)}\)
D) \(\frac{2br}{(b+r)(b+r-1)}\)
E) \(\frac{2br}{(b+r)(b+r)}\)
Approach #1: Probability rules
P(different colors) = P(red 1st AND blue 2nd OR blue 1st AND red 2nd
= P(red 1st AND blue 2nd) + P(blue 1st AND red 2nd)
= [P(red 1st) x P(blue 2nd)] + [P(blue 1st) x P(red 2nd)]
= [\(\frac{r}{r+b}\) x \(\frac{b}{r+b-1}\)] + [\(\frac{b}{r+b}\) x \(\frac{r}{r+b-1}\)]
= \(\frac{br}{(b+r)(b+r−1)}\) + \(\frac{br}{(b+r)(b+r−1)}\)
= \(\frac{2br}{(b+r)(b+r−1)}\)
Answer: D
Approach #2: Testing values
Notice that, if \(r=1\) and \(b=1\), then we are guaranteed to have two different colors.
In other words, \(r=1\) and \(b=1\), then P(different colors) = \(1\)
So now we'll plug \(r=1\) and \(b=1\) into each answer choice to see which one equals \(1\)....
A) \(\frac{(1)(1)}{(1+1)(1+1-1)}=\frac{1}{2}\). NO GOOD.
B) \(\frac{2(1+1)}{(1+1)(1+1-1)}=2\). NO GOOD.
C) \(\frac{(1)(1)}{(1+1)}=\frac{1}{2}\). NO GOOD.
D) \(\frac{2(1)(1)}{(1+1)(1+1-1)}=1\). BINGO!!!
E) \(\frac{2(1)(1)}{(1+1)(1+1)}=\frac{1}{2}\). NO GOOD.
Answer: D
