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A bag contains 3 orange, 6 red, and 5 blue marbles
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23 Feb 2024, 03:44
23 Feb 2024, 03:44
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Question Stats:
71% (01:40) correct
28% (01:35) wrong based on 42 sessions
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A bag contains 3 orange, 6 red, and 5 blue 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.
Quantity A |
Quantity B |
Chance of drawing marbles in the order: blue, blue, red , orange |
Chance of drawing marbles in the order: red, red, orange, red |
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.
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Re: A bag contains 3 orange, 6 red, and 5 blue marbles
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11 May 2024, 13:58
11 May 2024, 13:58
1
There are 3+6+5=14 total marbles in the bag.
Quantity A
Draw 1: There are 5 blue, so the chance of drawing blue is 5/14. Now there are 4 blue marbles and 13 total marbles.
Draw 2: The chance of drawing blue is 4/13. Now there are 12 total marbles.
Draw 3: The chance of drawing red is 6/12. Now there are 11 total marbles.
Draw 4: The chance of drawing orange is 3/11.
The chance of drawing in that order is \(\frac{5}{14}*\frac{4}{13}*\frac{6}{12}*\frac{3}{11} = \frac{360}{24024}\)
Quantity B
Draw 1: There are 6 red, so the chance of drawing red is 6/14. Now there are 5 red marbles and 13 total marbles.
Draw 2: The chance of drawing red is 5/13. Now there are 4 red marbles and 12 total marbles.
Draw 3: The chance of drawing orange is 3/12. Now there are 11 total marbles.
Draw 4: The chance of drawing red is 4/11.
The chance of drawing in that order is \(\frac{6}{14}*\frac{5}{13}*\frac{3}{12}*\frac{4}{11} = \frac{360}{24024}\)
Quantity A
Draw 1: There are 5 blue, so the chance of drawing blue is 5/14. Now there are 4 blue marbles and 13 total marbles.
Draw 2: The chance of drawing blue is 4/13. Now there are 12 total marbles.
Draw 3: The chance of drawing red is 6/12. Now there are 11 total marbles.
Draw 4: The chance of drawing orange is 3/11.
The chance of drawing in that order is \(\frac{5}{14}*\frac{4}{13}*\frac{6}{12}*\frac{3}{11} = \frac{360}{24024}\)
Quantity B
Draw 1: There are 6 red, so the chance of drawing red is 6/14. Now there are 5 red marbles and 13 total marbles.
Draw 2: The chance of drawing red is 5/13. Now there are 4 red marbles and 12 total marbles.
Draw 3: The chance of drawing orange is 3/12. Now there are 11 total marbles.
Draw 4: The chance of drawing red is 4/11.
The chance of drawing in that order is \(\frac{6}{14}*\frac{5}{13}*\frac{3}{12}*\frac{4}{11} = \frac{360}{24024}\)
Re: A bag contains 3 orange, 6 red, and 5 blue marbles
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15 May 2024, 19:23
15 May 2024, 19:23
1
Did we make the assumption that it was without replacement, or it wold be the same answer?
