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A certain jar contains only b black marbles, w white marbles and r red

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A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 16 Nov 2010, 05:43
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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 then the probability that the marble chosen will be white?


(1) \(\frac{r}{b+w} > \frac{w}{b+r}\)

(2) \(b-w > r\)

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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 16 Nov 2010, 05:52
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anilnandyala 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 then the probability that the marble chosen will be white?

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


The question is \(\frac{R}{R+B+W}>\frac{W}{R+B+W}\) true? Or is \(R>W\) true?

(1) \(\frac{R}{B+W} > \frac{W}{B+R}\) --> \(\frac{R}{B+W} +1> \frac{W}{B+R}+1\) --> \(\frac{R+B+W}{B+W}> \frac{W+B+R}{B+R}\) --> \(\frac{1}{B+W}> \frac{1}{B+R}\) --> \(B+R>B+W\) --> \(R>W\). Sufficient.

OR:
Given: \(\frac{R}{B+W} > \frac{W}{B+R}\) -->

Cross multiply, we can safely do this as \(B+W\) and \(B+R\) are more than zero.

We'll get \(R(B+R)>W(B+W)\) --> \(RB+R^2>WB+W^2\) --> \((R^2-W^2)+(RB-WB)>0\) --> \((R-W)(R+W)+B(R-W)>0\) --> \((R-W)(R+W+B)>0\).

As \(R+W+B>0\), the above inequality to hold true \(R-W\) must also be more than zero, so \(R-W>0\) --> \(R>W\).

(2) \(B-W>R\), not sufficient to determine whether \(R>W\) or not.

Answer: A.
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 16 Nov 2010, 11:40
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anilnandyala wrote:
a certain jar contains only b black marbles, w white marbles & r red marbles. if one marble is to be chosen random from jar is the probability that the marble chosen will be red greater then the probability the marble chosen is white?
a) r/(b+w) > w/(b+r)
b) b-w > r


The probability that red marble is chosen will be greater than the probability that white marble is chosen if there are more red marbles than white marbles.
So the queestion is just: Is r > w

Statement 1: r/(b + w) > w/(b + r)
Cross multiply to get r(b + r) > w(b + w) .... [(b + w) and (b + r) are definitely positive so cross multiplying is not a problem.]
Now, if r > w, (b + r) has to be greater than (b + w)
If r were less than w, then (b + r) < (b + w) and the left side would have been smaller than the right side.
So this implies that r must be greater than w. Sufficient.

Statement 2: b > r + w
But we cant compare r and w so not sufficient.

Answer (A).
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 17 Nov 2010, 22:55
VeritasPrepKarishma and Bunuel - thanks a lot for ur explanations.

+1 from me... again.

keep up the good job.
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 23 Nov 2010, 22:36
Bunuel wrote:
anilnandyala wrote:
a certain jar contains only b black marbles, w white marbles & r red marbles. if one marble is to be chosen random from jar is the probability that the marble chosen will be red greater then the probability the marble chosen is white?
a) r/(b+w) > w/(b+r)
b) b-w > r


(1) \(\frac{R}{B+W} > \frac{W}{B+R}\) --> \(\frac{R}{B+W} +1> \frac{W}{B+R}+1\) --> \(\frac{R+B+W}{B+W}> \frac{W+B+R}{B+R}\) --> \(\frac{1}{B+W}> \frac{1}{B+R}\) --> \(B+R>B+W\) --> \(R>W\). Sufficient.

OR:
Given: \(\frac{R}{B+W} > \frac{W}{B+R}\) -->

Cross multiply, we can safely do this as \(B+W\) and \(B+R\) are more then zero.

Answer: A.


awesome explanation. I have translated the question to is r>w but did not know how to solve the (1). Now get it clear
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 30 Dec 2013, 10:17
Given 8 red marbles and y white marbles.
Number of ways you can pick any two marbles is \((8+Y)C2\).
Ways of picking 2 red marbles is 8*7 (first time you can pick out of 8 red marbles and second time you can pick one of the remaining 7 marbles).
Ways of picking 1 marble of each color. This can happen in 2 ways. 1 way) First pick white and second pick red [y*8 ways] or 2 way) first pick red and second pick white [8*y ways]. So total number of ways to pick two different colors will be \(y*8 + 8*y\) = \(2*8y\)

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

Question is asking for - is probability of picking 2 red marbles > probability of picking different color marbles, is \(\frac{8*7}{(8+Y)C2} > \frac{2*8y}{(8+Y)C2}\)
Therefore, \(y<7/2=3.5\)

(1) y ≤ 8 y can be less than or greater than 3.5 - NS
(2) y ≥ 4 y is always greater than 3.5 - S

So B
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 10 Jan 2014, 05:31
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anilnandyala 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 then the probability that the marble chosen will be white?

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


Is r>w?

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

rb + r^2 ? wb + w^2

(r+w)(r-w) > b(w-r)

Now if r-w>0 then w-r < 0 and inequality holds true.
Other way around if r-w<0, LHS is negative and RHS is positive and inequality does NOT hold true

So only valid scenario is r-w>0 and thus r>w

Sufficient

(B) Not enough Info

Answer is A

Cheers
J :)

PS. Alternatively, first statement can be treated as

r-w / b+r>0

Well b+r is always positive, thus r-w has to be positive too
Then r>w
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 19 Jun 2015, 23:25
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I like this problem because there are at least five different ways to solve it. I'll mention a more conceptual solution since no one has mentioned it yet, but there are some great solutions above as well:

If you know the concept of "odds" that is used in daily life, you can answer this question very quickly. "Odds" are just ratios of good outcomes to bad outcomes, while probabilities are ratios of good outcomes to total outcomes (good+bad). So when we say the odds that something will happen are 2 to 1, that means there's a 2/3 probability it will happen, and a 1/3 probability it will not.

In this question, the fraction r/(b+w) is just the ratio of red marbles to other marbles, so it just represents the odds of picking a red marble. Similarly the fraction w/(b+r) is the ratio of white marbles to other marbles, so it represents the odds of picking a white marble. And if the odds of getting red are better than the odds of getting white, the probability of getting red must be higher than the probability of getting white, so S1 is sufficient.
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 24 Apr 2018, 10:12
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anilnandyala 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 then the probability that the marble chosen will be white?

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


Target question: Is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white?

We can rephrase the target question as...
REPHRASED target question: Is r > w?

Statement 1: r/(b + w) > w/(b + r)
Let's let T = the TOTAL number of marbles in the jar.
This means that b + w + r = T
This also means that b + w = T - r
And it means that b + r = T - w
So, we can take statement 1, r/(b + w) > w/(b + r), and rewrite it as...
r/(T - r) > w/(T - w)
Multiply both sides by (T - r) to get: r > w(T - r)/(T - w)
Multiply both sides by (T - w) to get: r(T - w) > w(T - r)
Expand both sides: rT - rw > wT - rw
Add rw to both sides: rT > wT
Divide both sides by T to get: r > w
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: b - w > r
Add w to both sides to get: b > w + r
All this means is that there are more black marbles than there are white and red marbles combined.
Given this information, there's no way to determine whether or not r is greater than w
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 25 Aug 2019, 09:58
Bunuel wrote:
anilnandyala 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 then the probability that the marble chosen will be white?

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


The question is \(\frac{R}{R+B+W}>\frac{W}{R+B+W}\) true? Or is \(R>W\) true?

(1) \(\frac{R}{B+W} > \frac{W}{B+R}\) --> \(\frac{R}{B+W} +1> \frac{W}{B+R}+1\) --> \(\frac{R+B+W}{B+W}> \frac{W+B+R}{B+R}\) --> \(\frac{1}{B+W}> \frac{1}{B+R}\) --> \(B+R>B+W\) --> \(R>W\). Sufficient.

OR:
Given: \(\frac{R}{B+W} > \frac{W}{B+R}\) -->

Cross multiply, we can safely do this as \(B+W\) and \(B+R\) are more than zero.

We'll get \(R(B+R)>W(B+W)\) --> \(RB+R^2>WB+W^2\) --> \((R^2-W^2)+(RB-WB)>0\) --> \((R-W)(R+W)+B(R-W)>0\) --> \((R-W)(R+W+B)>0\).

As \(R+W+B>0\), the above inequality to hold true \(R-W\) must also be more than zero, so \(R-W>0\) --> \(R>W\).
(2) \(B-W>R\), not sufficient to determine whether \(R>W\) or not.

Answer: A.


I don't understand the highlighted part. Would you please explain it in simple words? Bunuel

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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 25 Aug 2019, 19:42
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RashedVai wrote:
Bunuel wrote:
anilnandyala 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 then the probability that the marble chosen will be white?

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


The question is \(\frac{R}{R+B+W}>\frac{W}{R+B+W}\) true? Or is \(R>W\) true?

(1) \(\frac{R}{B+W} > \frac{W}{B+R}\) --> \(\frac{R}{B+W} +1> \frac{W}{B+R}+1\) --> \(\frac{R+B+W}{B+W}> \frac{W+B+R}{B+R}\) --> \(\frac{1}{B+W}> \frac{1}{B+R}\) --> \(B+R>B+W\) --> \(R>W\). Sufficient.

OR:
Given: \(\frac{R}{B+W} > \frac{W}{B+R}\) -->

Cross multiply, we can safely do this as \(B+W\) and \(B+R\) are more than zero.

We'll get \(R(B+R)>W(B+W)\) --> \(RB+R^2>WB+W^2\) --> \((R^2-W^2)+(RB-WB)>0\) --> \((R-W)(R+W)+B(R-W)>0\) --> \((R-W)(R+W+B)>0\).

As \(R+W+B>0\), the above inequality to hold true \(R-W\) must also be more than zero, so \(R-W>0\) --> \(R>W\).
(2) \(B-W>R\), not sufficient to determine whether \(R>W\) or not.

Answer: A.


I don't understand the highlighted part. Would you please explain it in simple words? Bunuel

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We have \((R-W)(R+W+B)>0\).

The product of two multiples, (R-W) and (R+W+B), to be positive they must have the same sign. Since, the second multiple, R+W+B, is positive, then the first multiple, R-W, mut also be positive: R - W > 0.
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Re: A certain jar contains only b black marbles, w white marbles and r red  [#permalink]

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New post 15 May 2020, 12:51
Bunuel - please could you explain the simplification for (1) again?
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Re: A certain jar contains only b black marbles, w white marbles and r red   [#permalink] 15 May 2020, 12:51

A certain jar contains only b black marbles, w white marbles and r red

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