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

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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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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VeritasPrepKarishma and Bunuel - thanks a lot for ur explanations.

+1 from me... again.

keep up the good job.

awesome explanation. I have translated the question to is r>w but did not know how to solve the (1). Now get it clear

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

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

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.

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

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

Posted from my mobile device
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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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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) \(\frac{r}{b+w} > \frac{w}{b+r}\)
\(r(b+r) > w(b+w)\)

From the above we can conclude r > w. SUFFICIENT.

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

INSUFFICIENT.

Answer is A.

In statement 2: if B>R+W, then can't we conclude that B>R because if B is greater than R+W combined, then B should obviously be greater than R? and hence, sufficient to conclude that probability of selecting a red marble will be lesser than probability of selecting a white marble. VeritasKarishma

Yes, you know from Statement 2 that B > R + W, and since these unknowns are positive, you are correct that we can be sure that B > R and B > W. But the question is asking, essentially, do we have more red marbles than white marbles. So it is asking if we can be sure that R > W is true. And there's no way to prove that from either of the inequalities "B > R" or "B > W".

You can conclude that B is greater than R but you cannot compare R and W.
B > R + W means B > R and B > W (all positive integers)
But does it mean R > W? No.
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(1) \(\frac{r}{b+w} > \frac{w}{b+r}\)

Cross multiply :

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

now think about this for a sec.

r and w are number of marbles, so they can't be -ve or non-integers. based on that, if we now \(r(b+r) > w(b+w)\) -- we're adding and multiplying r and w with the same constant "b" and getting that the result with r is the greater one, then the # of red marbles must be greater.

if we take an e.g. to prove ourselves wrong: b=2 , r=3 and w= 5 (we've taken w>r)

then \(3(3+2)\) >( cannot be greater than) \(5(5+2)\), for this inequality to work, r must be > w.

Hence, Probability of r is > probability of w.

A is sufficient.

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

doesn't really say anything specific about the # of marbles, you will get sum/ difference between two types of marbles.

Insufficient.

A is the answer

I just did it like this:

1) the number of "r" compared to everything else is more than the number of "w" compared to everything else: Sufficient

2) b - w >r --> b > w + r --> b - r > w: basically gives us nothing regarding whether or not the counts for "w" or "r" affect the overall understanding of the probabilities or ratios. Insufficient

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