marble probability

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marble probability

by dreamv » Tue Feb 21, 2012 1:16 pm
A certain jar contains only b black marbles, w white marbles, and r red marbles. 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
Source: — Data Sufficiency |

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by pemdas » Tue Feb 21, 2012 3:25 pm
prompt: P(red)>P(white)
st(1) r/(b+w)> w/(b+r) implies given b is fixed parameter, the relationship of r/w and w/r will be as following r/w > w/r.
Example b=1,r=2,w=3 ---> 2/(1+3)<3/(1+2), 2/4<1 and 2<3 or applying the same to r and w, r<w
In our case, it's opposite since r/w > w/r and r>w

Sufficient to answer Yes

st(2) b-w>r implies only b>r+w which is again not explicit for our knowing about r and w individual values. Not Sufficient

a
dreamv wrote:A certain jar contains only b black marbles, w white marbles, and r red marbles. 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
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by Anurag@Gurome » Wed Feb 22, 2012 5:10 am
dreamv wrote:A certain jar contains only b black marbles, w white marbles, and r red marbles. 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
(1) r/(b + w)> w/(b + r)
Since both (b + w) and (b + r) are > 0 so we can write r/(b + w)> w/(b + r) as r(b + r) > w(b + w) or rb + r² > wb + w²
(r² - w²) + (rb - wb) > 0
(r - w)(r + w) + b(r - w) > 0
(r - w)[r + w + b] > 0
Since r + w + b > 0, so (r - w) > 0 or r > w; SUFFICIENT.

(2) b - w > r does not imply if r > w or not; NOT sufficient.

The correct answer is A.
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by [email protected] » Sun Mar 18, 2012 10:47 pm
I got the answer as A but through trial and error...

But anurag gave a wonderful explanation....

thank you...
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