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
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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