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A bag contains 12 marbles: 5 of the marbles are red, 3 are g
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22 Nov 2017, 19:07
22 Nov 2017, 19:07
Expert Reply
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Question Stats:
35% (01:38) correct
64% (01:30) wrong based on 131 sessions
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A bag contains 12 marbles: 5 of the marbles are red, 3 are green, and the rest are blue.
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 |
The probability of consecutively choosing two red marbles and a green marble without replacement |
The probability of consecutively choosing a red and two blue marbles with replacement |
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.
Drill 2
Question: 5
Page: 525
Question: 5
Page: 525
Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g
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24 Nov 2017, 06:21
24 Nov 2017, 06:21
2
qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement
Therefore B>A
qty B = 5/12* 4/12* 3/12) => with Replacement
Therefore B>A
Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g
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29 Dec 2017, 13:25
29 Dec 2017, 13:25
2
Expert Reply
1
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Explanation
Quantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.
The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{4}{11}\), and then a green marble is \(\frac{3}{10}\).
This is an “and” probability problem, so you have to multiply the probability of each event together: \(\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\).
For Quantity B, you do the same thing, but the total stays the same for each draw: \(\frac{5}{12}\times\frac{4}{12}\times\frac{4}{12}=\frac{80}{1728}=\frac{5}{108}\).
Quantity B is greater.
Quantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.
The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{4}{11}\), and then a green marble is \(\frac{3}{10}\).
This is an “and” probability problem, so you have to multiply the probability of each event together: \(\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\).
For Quantity B, you do the same thing, but the total stays the same for each draw: \(\frac{5}{12}\times\frac{4}{12}\times\frac{4}{12}=\frac{80}{1728}=\frac{5}{108}\).
Quantity B is greater.
