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# A bag contains 12 marbles: 5 of the marbles are red, 3 are g

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A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]  22 Nov 2017, 19:07
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Question Stats:

26% (01:32) correct 73% (01:27) wrong based on 26 sessions
A bag contains 12 marbles: 5 of the marbles are red, 3 are green, and the rest are blue.

 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
[Reveal] Spoiler: OA

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Sandy
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]  24 Nov 2017, 06:21
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qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement

Therefore B>A
GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1820 [1] , given: 397

Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]  29 Dec 2017, 13:25
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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.
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]  12 Mar 2018, 01:34
qty A = (5/12 * 4/11 * 3/10) => Without replacement
qty B = 5/12* 4/12* 3/12) => with Replacement

Therefore B>A
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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g [#permalink]  12 Mar 2018, 17:41
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12 marbles:
Red: 5
Green: 3
Blue: 4
A: The probability of consecutively choosing two red marbles and a green marble without replacement.
The probability of choosing first red marble is 5/12, because there is not any replacement now 4 red marbles are remained and totally 11 marbles. Thus the probability of choosing the second red marble is 4/11. The probability of choosing a green marble in these 10 remaining marbles is 3/10. So P(A) = 5/12 * 4/11 * 3/10

B: The probability of consecutively choosing a red and two blue marbles with replacement
As there are 5 red marbles, the probability of choosing a red marble between 12 marbles is 5/12. The probability of choosing first blue marble between 12 remained marbles(because there is the replacement, we put back drawn marble in last step) is 4/12. The probability of choosing second blue marble between 12 remaining marbles and 4 remaining blue marbles is 4/12. So P(B) = 5/12 * 4/12* 4/12
B is a little bigger.

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Re: A bag contains 12 marbles: 5 of the marbles are red, 3 are g   [#permalink] 12 Mar 2018, 17:41
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