A bag contains at least 10 blue marbles and at least 10 red marbles

Question

A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?

(1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.

(2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.

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chetan2u · Answer

A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?

Let there be B blue marbles and R red marbles...
We have to find whether B>R

(1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.
The probability of picking two blue marbles = \(\frac{BC2}{(B+R)C2} \)
The probability of picking one blue marble and one red marble = \(\frac{BC1*RC1}{(B+R)C2}\)
So, \(BC2>BC1*RC1......B(B-1)>2BR......(B-1)>2R\), so surely \(B>R\)
Suff

(2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.
The probability of picking two red marbles = \(\frac{RC2}{(B+R)C2} \)
The probability of picking one blue marble and one red marble = \(\frac{BC1*RC1}{(B+R)C2}\)
So, \(RC2<BC1*RC1......R(R-1)<2BR......(R-1)<2B\),
If R = 20 and B 15...NO
If R=20 and B=30..Yes
Insufficient

A

sanjitscorps18 · Answer

Let the Blue marbles be b+10, where b >= 0
Let the Red marbles be r+10, where r >= 0

(1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.

\(=> ((b+10)C1 * (r+10)C1)/(b+r+20)C2 < (b+10)C2)/(b+r+20)C2\)
\(=> (b+10)C1 * (r+10)C1 < (b+10)C2\)
\(=> 2(b+10)(r+10) < (b+10)(b+9)\)
\(=> 2(r+10) < (b+9)\)
\(=> 2r + 20 < b + 9\)
\(=> b > 2r + 11\)

This suggests that b is definitely greater than r

(2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.

\(=> ((b+10)C1 * (r+10)C1)/(b+r+20)C2 > (r+10)C2)/(b+r+20)C2\)
\(=> (b+10)C1 * (r+10)C1 > (r+10)C2\)
\(=> 2(b+10)(r+10) > (r+10)(r+9)\)
\(=> 2(b+10) > (r+9)\)
\(=> 2b + 20 > r + 9\)
\(=> 2b > r - 11\)
\(=> 2b + 11 > r\)
Here b may or may not be larger than r or may even be equal to r
for e.g.
if b = 0, r could be 10
if b = 0, r could be 0
if b = 1, r could be 10
if b = 1, r could be 0

Option A

G-Man · Answer

Let the number of Blue marbles and red marbles be Nb & Nr
Both Nb and Nr > 10

The question is to check whether Nb is > Nr

Statement 1
The probability of getting 2 blue (without replacement) is higher than one red and one blue
Let's assume Nb = Nr and both are equal to 11 to simplify the calculation (If this satisfies statement 1, then we can disprove the problem)
Then P(2 blue) = P1 = 11/22*10/21
P(1 blue & 1 red) = P2 = 11/22 * 11/21

For P1 to be greater than P2, Numerator should be higher. Or the Number of blue balls should be more than red. 
Sufficient

Statement 2
P1 > P2
Take the same approach
P(1 blue & 1 red) = 11/22 * 10/21
P(2 red) = 11/22 * 10/21
P1 will be greater than P2 for the same number of balls or less number of red balls than blue balls.
Insufficient

A

gmatprepper94111 · Answer

Let b = the quantity of blue marbles
Let r = the quantity of red marbles
Therefore, the quantity of total marbles is b+r

Note the critical point that the marbles are sampled  without replacement.

-----------------
What does A tell us?
-----------------

Well the probability of two successive blue marble draws is the product (it's an  AND  probability) of two quantities:
(1) The probability of drawing a blue marble first
(2) The probability of drawing a blue marble again, after having taken a blue marble  out

Well the probability of a blue and then a red is very similar:
(1) The probability of drawing a blue marble first
(2) The probability of drawing a red marble again, after having taken a blue marble  out

So we can say the following \frac{(b)}{(b+r)} * \frac{(b-1)}{(b+r-1)} > \frac{(b)}{(b+r)} * \frac{(r)}{(b+r-1)}
Notice that we can just dispense with the denominators altogether, since they're identical and we know all of these numbers are positive integers.
So we get the following simplified expression: b^2 - b > br
Which then implies b > r+ 1 (note: again, we know b is a positive integer, so we can safely divide both sides by b)

So A is sufficient

-----
Now what does B tell us?
-----

Well, B works pretty much exactly the same way. Applying virtually the same exact logic, we get the following expression: b * r > r(r-1)
This tells us b > r - 1

This is insufficient; b could be either r or r+1.

A bag contains at least 10 blue marbles and at least 10 red marbles

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GMAT Data Sufficiency (DS) Questions

A bag contains at least 10 blue marbles and at least 10 red marbles

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