A certain jar contains only b black marbles, w white marble, and r red marble. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?
(1) r/(b + w) > w/(b + r)
(2) b-w > r
A [/spoiler]
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Marbles Probability
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scoobydooby
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P(r)=r/(r+b+w)
P(w)=w/(r+b+w)
simplifying the stimulus,
Q. P(r)>P(w)?
=>r>w?
1) r/(b + w) > w/(b + r)
=>r/(b + w)-w/(b + r)>0
=>r(b+r)-w(b+w)/{(b+w)(b+r)}>0
=>rb+r^2-wb-w^2>0
=>b(r-w)+r^2-w^2>0
=>b(r-w)+(r+w)(r-w)>0
=>(r-w)(b+r+w)>0
=>(r-w)>0 => r>w sufficient
2) b-w > r
cant say from here whether r>w not sufficient
hence A
=
P(w)=w/(r+b+w)
simplifying the stimulus,
Q. P(r)>P(w)?
=>r>w?
1) r/(b + w) > w/(b + r)
=>r/(b + w)-w/(b + r)>0
=>r(b+r)-w(b+w)/{(b+w)(b+r)}>0
=>rb+r^2-wb-w^2>0
=>b(r-w)+r^2-w^2>0
=>b(r-w)+(r+w)(r-w)>0
=>(r-w)(b+r+w)>0
=>(r-w)>0 => r>w sufficient
2) b-w > r
cant say from here whether r>w not sufficient
hence A
=
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Ian Stewart
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There are quite a few ways to look at this problem, though it's actually based on an idea we encounter all the time in day to day life. I'll suggest two other approaches:ieeyorei wrote:A certain jar contains only b black marbles, w white marble, and r red marble. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?
(1) r/(b + w) > w/(b + r)
(2) b-w > r
A [/spoiler]
If you know the concept of 'odds' from probability (and from real life), then the first statement is clearly sufficient here. If you say that "the odds are 21 to 5 in favor of picking a consonant when you pick a random letter from the alphabet", that is just a ratio of the number of ways to pick a consonant to the number of ways to not pick a consonant. From those odds, we can see that the probability of picking a consonant is 21/26, and the probability of picking a vowel is 5/26.
That's what statement 1 is telling us: the odds of getting a red marble are better than the odds of getting a white marble (the fraction on the left is the ratio of red marbles to non-red marbles, while the fraction on the right is the ratio of white marbles to non-white marbles). If the odds are higher, the probability must be higher.
Alternatively, you could look at this abstractly. We don't need to rewrite the fractions at all. We know that:
r/(b + w) > w/(b + r)
and all of the letters represent positive quantities. We also know that if a fraction has *both* a larger numerator and a smaller denominator than another, it must be larger in value. Looking at the above inequality, it's impossible for w to be greater than r; if it were, then the fraction on the right would have both a larger numerator and a smaller denominator than the fraction on the left. So r must be greater than w (they can't be equal, if that inequality is true), which is all we need to know here.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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cramya
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Simply superb explanation; thanks Ian!
Wish we could make a book out of Stuart's/your explanations for all of us.....
No kidding I still cant believe where u could have missed that 20 in your GMAT.
Regards,
CR
Wish we could make a book out of Stuart's/your explanations for all of us.....
No kidding I still cant believe where u could have missed that 20 in your GMAT.
Regards,
CR
Ian your solution makes perfect sense. I'm just wondering if you can help me clarify this thought:
If R/W > W/R then we know that R > W. But there's a "b" in the fraction, so it's talking about the ratio of red marbles to non-red marbles and white marbles to non-white marbles. However how does ratio of Red vs. Non-red being higher than white vs. non white show us that red > white?
If R/W > W/R then we know that R > W. But there's a "b" in the fraction, so it's talking about the ratio of red marbles to non-red marbles and white marbles to non-white marbles. However how does ratio of Red vs. Non-red being higher than white vs. non white show us that red > white?
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Domnu
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Here's another mathematical solution...
A -
r/(b+w) > w/(b+r)
(b+w)/r < (b+r)/w
1 + (b+w)/r < 1 + (b+r)/w
(b+w+r)/r < (b+w+r)/w
r/(b+w+r) > w/(b+w+r),
so P(r) > P(w). Here, we relied on the fact that a > b -> 1/a < 1/b.
A -
r/(b+w) > w/(b+r)
(b+w)/r < (b+r)/w
1 + (b+w)/r < 1 + (b+r)/w
(b+w+r)/r < (b+w+r)/w
r/(b+w+r) > w/(b+w+r),
so P(r) > P(w). Here, we relied on the fact that a > b -> 1/a < 1/b.
Have you wondered how you could have found such a treasure? -T
Ian,
I am unable to follow your logic below. Can you please provide any examples?
Thanks
I am unable to follow your logic below. Can you please provide any examples?
Thanks
Ian Stewart wrote: Alternatively, you could look at this abstractly. We don't need to rewrite the fractions at all. We know that:
r/(b + w) > w/(b + r)
and all of the letters represent positive quantities. We also know that if a fraction has *both* a larger numerator and a smaller denominator than another, it must be larger in value. Looking at the above inequality, it's impossible for w to be greater than r; if it were, then the fraction on the right would have both a larger numerator and a smaller denominator than the fraction on the left. So r must be greater than w (they can't be equal, if that inequality is true), which is all we need to know here.
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Frankenstein
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Hi,gmat1978 wrote:Ian,
I am unable to follow your logic below. Can you please provide any examples?
Thanks
Lets say r>w. So b+r > b+w
Consider r/(b+w). Numerator r > w.
Denominator (b+w) is less than (b+r).
if denominator is constant, greater the numerator greater the fraction
if numerator is constant, lower the numerator greater the fraction
In this case, the numerator is greater, combined with denominator being lesser makes the fraction larger.
Example: consider b=3
2>1, so(3+2)>(3+1)
Hence, 2/(3+1) > 1/(3+2).
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
