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A bag contains 6 blue marbles and 4 red marbles. Two marbles are sele
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16 Sep 2022, 00:42
16 Sep 2022, 00:42
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Question Stats:
94% (01:10) correct
5% (01:35) wrong based on 36 sessions
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A bag contains 6 blue marbles and 4 red marbles. Two marbles are selected at random, one after the other (the first marble is not replaced in the bag after it is drawn).
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
probability of drawing 2 blue marbles |
probability of drawing 2 red marbles |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
Re: A bag contains 6 blue marbles and 4 red marbles. Two marbles are sele
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16 Sep 2022, 05:11
16 Sep 2022, 05:11
1
2 blue- 6/10+5/9=0.33
2 red- 4/10+3/9=0.1333
so quantity A is greater
2 red- 4/10+3/9=0.1333
so quantity A is greater
A bag contains 6 blue marbles and 4 red marbles. Two marbles are sele
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24 Sep 2022, 05:20
24 Sep 2022, 05:20
1
Approach-1 (not recommended for this problem with such a large sample space but there are other problems where this APPROACH is the only efficient one of course they will have small sample space)
Sample Space
{
b1 b2
b1 b3
b1 b4
b1 b5
b1 b6
b2 b3
b2 b4
b2 b5
b2 b6
b3 b4
b3 b5
b3 b6
b4 b5
b5 b6
Total 15 instances here And
b1 r1
b1 r2
b1 r3
b1 r4
same for b2 b3 b4
b2 {r1....r4}
b3 {r1....r4}
b4 {r1....r4}
b5 {r1....r4}
b6 {r1....r4}
Total of 24 instances here
r1 {r2..r4}
r2 r3
r2 r4
r3 r4
Total 6 instances here
Total Favourable Outcome = 15
Total Possible Outcome = (15+24+6) = 45
Quant A: 15/45 = .333
Quant B: Do this process you will get .133
Approach-2:
Quant A : 6/10+5/9=.333
Quant B : 4/10+3/9=.133
Approach-3:
Quant A : 1-{1-(6/10)*(5/9)}=(1-.667)=.333
Quant B : using the same process would be .133.
Sample Space
{
b1 b2
b1 b3
b1 b4
b1 b5
b1 b6
b2 b3
b2 b4
b2 b5
b2 b6
b3 b4
b3 b5
b3 b6
b4 b5
b5 b6
Total 15 instances here And
b1 r1
b1 r2
b1 r3
b1 r4
same for b2 b3 b4
b2 {r1....r4}
b3 {r1....r4}
b4 {r1....r4}
b5 {r1....r4}
b6 {r1....r4}
Total of 24 instances here
r1 {r2..r4}
r2 r3
r2 r4
r3 r4
Total 6 instances here
Total Favourable Outcome = 15
Total Possible Outcome = (15+24+6) = 45
Quant A: 15/45 = .333
Quant B: Do this process you will get .133
Approach-2:
Quant A : 6/10+5/9=.333
Quant B : 4/10+3/9=.133
Approach-3:
Quant A : 1-{1-(6/10)*(5/9)}=(1-.667)=.333
Quant B : using the same process would be .133.
