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GMAT PREP II (2) PROBABILITY QUESTION

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GMAT PREP II (2) PROBABILITY QUESTION

by pkw209 » Sat Dec 05, 2009 9:20 pm
Hey all-

Just finished GMAT PREP II and can't understand the answer to this question:

Q) A jar contains only B black marbles, W white marbles, and R red marbles. If one marble is to be chosen at random from a 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

If someone could provide a brief explanation, that would be great.

Thanks!
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by viju9162 » Sun Dec 06, 2009 2:55 am
Hi pkw209 ,

Is the answer
E ?

The question asks whether:

R/ B+W+R > W/B+W+R .

From (A), we cant determine as we are not sure about the denominator value

From (B), W can either be greater than R or can be lesser than R

ex: B-W > R

8-5> 1 ... Here W > R

8-2 > 6 ... Here R > W

Combining both A and B, values cant be determined for any of them..

Let me know the official answer.
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by Stuart@KaplanGMAT » Sun Dec 06, 2009 8:53 am
pkw209 wrote:Hey all-

Just finished GMAT PREP II and can't understand the answer to this question:

Q) A jar contains only B black marbles, W white marbles, and R red marbles. If one marble is to be chosen at random from a 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

If someone could provide a brief explanation, that would be great.

Thanks!
I really want to say that (1) is sufficient alone, although every time I try to prove it mathematically I end up with ugly quadratics.

(2) is easier, so let's start there. We can rewrite it as:

b > r + w

which shows that there are more than 50% black marbles, but says nothing at all about the relationship between red and white, so is insufficient alone.

(1) is tougher to deal with algebraically, so let's work theoretically instead.

We know that r, b and w are all non-negative integers.

If b=0, then we get:

r/w > w/r

which, since r and w are non-negative, we can safely rewrite as:

r^2 > w^2

and, since r and w are integers, conclude that r > w.

Since b appears in the denominator of each ratio, increasing the value of b won't change that fundamental relationship. For example, if b = 4, we'd get:

r/(4+w) > w/(r+4)

r(r+4) > w(w+4)

and again, since r and w are non-negative integers, r must be greater than w for that inequality to hold. Increasing the value of b just increases the number in each bracket.
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by viju9162 » Sun Dec 06, 2009 5:58 pm
amazing Stuart! really good explanation.. I didn't think so much while solving the problem..

Thank you very much.

Regards,
Viju
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by nirmalp » Sun Dec 06, 2009 7:12 pm
The mathematical way to solve this one:

1. r/ (b+w) > w / (b+r)

Probability of choosing a red marble is r/ (b+w+r) and a white marbel is w/ (b+w+r)

We know r, b and w are all positive whole numbers. So we need not worry about -ve signs.

Transposing both sides from the given equation:
(b+r) / w > (b+w) / r

Adding 1 to both sides:
[(b+r)/ w] + 1 > [(b+w) / r] + 1

Take 1 inside the fractional part:
(b+ r+ w)/ w > (b + w + r) / r

Transposing back:
r / (b + w + r) > w / (b+ r+ w)

Hence the p(r) > p(w).

Answer: A.
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by pkw209 » Sun Dec 06, 2009 8:38 pm
Thanks, you guys are extremely helpful!

I think I'm more comfortable with the mathematical way.

nirmalp, why did you choose to transpose both sides from the given equation:
(b+r) / w > (b+w) / r

Also, can you please explain the steps below? Confused as to why you added the 1...

Adding 1 to both sides:
[(b+r)/ w] + 1 > [(b+w) / r] + 1

Take 1 inside the fractional part:
(b+ r+ w)/ w > (b + w + r) / r
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by Testluv » Sun Dec 06, 2009 11:45 pm
...but, pkw209, the reasoning approach is quicker, and requires much less work!

1) r / (b + w) > w / (b + r)

In general, how can fraction a be bigger than fraction b? Either fraction a's numerator is larger than that of fraction b or else fraction a 's denominator is smaller than that of fraction b.

Here, the fraction on the left is bigger than the fraction on the right. How can this be? Simple: either the numerator on the left is bigger than the one on the right OR the denominator on the left is smaller than the one on the right.

Therefore:


if the left-hand side numerator is bigger, then:
r > w

and

if the left-hand side denominator is smaller, then:
b + w < b + r
w < r
r > w

In both cases, r > w.

Therefore, the first statement is sufficient by itself.
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by Ian Stewart » Wed Dec 09, 2009 3:25 am
I gave an alternative explanation to this problem in this thread:

www.beatthegmat.com/marbles-probability-t32705.html
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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by restero » Thu Jan 21, 2010 6:45 pm
Answer is A.

Simply,
Inverse the inequality, add one to both side and inverse the inequality again..
u get r/(b+r+w)>w/(b+r+w)
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