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chetan2u
Abhi077
There are 10 red and blue marbles in a box. If two marbles are randomly taken out of the box, is the probability of picking two red marbles greater than 1/3?

1)The probability of getting both blue marbles is less than 2/3
2) There are more red marbles than blue marbles in the box.

Let the red balls be R, so we are looking for \(\frac{RC2}{10C2}>\frac{1}{3}.....\frac{R(R-1)}{90}>\frac{1}{3}...R(R-1)>30\) So, R>6

1)The probability of getting both blue marbles is less than 2/3
So \(\frac{BC2}{10C2}>\frac{2}{3}.....\frac{B(B-1)}{45}>\frac{2}{3}...B(B-1)<30\) So, B<6
If B=5, R=5..NO R is not >6
If B=2, R=8..YES
Insuff

2) There are more red marbles than blue marbles in the box.
If R=6..No
If R>6..Yes
Insuff

Combined
If R=6..No
If R>6..Yes
Insuff

E
Should second not be
1)The probability of getting both blue marbles is less than 2/3
So \(\frac{BC2}{10C2}>\frac{2}{3}.....\frac{B(B-1)}{90}>\frac{2}{3}...B(B-1)<60\) So, B<8
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globaldesi
chetan2u
Abhi077
There are 10 red and blue marbles in a box. If two marbles are randomly taken out of the box, is the probability of picking two red marbles greater than 1/3?

1)The probability of getting both blue marbles is less than 2/3
2) There are more red marbles than blue marbles in the box.

Let the red balls be R, so we are looking for \(\frac{RC2}{10C2}>\frac{1}{3}.....\frac{R(R-1)}{90}>\frac{1}{3}...R(R-1)>30\) So, R>6

1)The probability of getting both blue marbles is less than 2/3
So \(\frac{BC2}{10C2}>\frac{2}{3}.....\frac{B(B-1)}{45}>\frac{2}{3}...B(B-1)<30\) So, B<6
If B=5, R=5..NO R is not >6
If B=2, R=8..YES
Insuff

2) There are more red marbles than blue marbles in the box.
If R=6..No
If R>6..Yes
Insuff

Combined
If R=6..No
If R>6..Yes
Insuff

E
Should second not be
1)The probability of getting both blue marbles is less than 2/3
So \(\frac{BC2}{10C2}>\frac{2}{3}.....\frac{B(B-1)}{90}>\frac{2}{3}...B(B-1)<60\) So, B<8

Thanks. Yes, it should be B(B-1)/10*9<2/3.....B(B-1)<60..B<9 as 8*7=56
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