Bacon goes for the prize

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Tue Feb 21, 2012 4:46 am

At a carnival, there is a prize on randomly drawing a green ball in three fresh attempts, from a bag that contains 4 red, 3 green, and 2 white all identical balls. Bacon goes for the prize. What is the probability that he'd get the prize?
A. 4/27
B. 2/9
C. 1/3
D. 5/9
E. 19/27

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Source: Beat The GMAT — Problem Solving | Tue Feb 21, 2012 4:46 am

MBACenter: GMAT Instructor; Posts: 56; Joined: Thu Jun 30, 2011 12:54 amFollowed by: 4 members; Location: Paris, FRANCE; Thanked: 16 times; GMAT Score:740

by MBACenter » Tue Feb 21, 2012 5:19 am

sanju09 wrote:At a carnival, there is a prize on randomly drawing a green ball in three fresh attempts, from a bag that contains 4 red, 3 green, and 2 white all identical balls. Bacon goes for the prize. What is the probability that he'd get the prize?
A. 4/27
B. 2/9
C. 1/3
D. 5/9
E. 19/27

[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]

Depends. Do you mean the probability that he gets three green balls, or the probability that with three tries he gets at least one green ball? And do the marbles stay out, or does he put them back after each try?

If it's all three and the marbles stay out, then that would be:

(3/9) * (2/8) * (1/7) = (1/3) * (1/4) * (1/7) = 1/84.

If it's at least one of the three and the marbles stay out, then that would be one minus the probability that he DOESN'T pick any greens:

1 - (6/9) * (5/8) * (4/7) = 1 - (1/3) * (5/1) * (1/7) = 1 - 5/21 = 16/21

If it's all three and the marbles do not stay out, then that would be:

(3/9) * (3/9) * (3/9) = (1/3) * (1/3) * (1/3) = 1/27.

If it's at least one of the three and the marbles do not stay out, then that would be one minus the probability that he DOESN'T pick any greens:

1 - (6/9) * (6/9) * (6/9) = 1 - (2/3) * (2/3) * (2/3) = 1 - 8/27

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Feb 22, 2012 12:27 am

MBACenter wrote:
sanju09 wrote:At a carnival, there is a prize on randomly drawing a green ball in one of the three fresh attempts, from a bag that contains 4 red, 3 green, and 2 white all identical balls. Bacon goes for the prize. What is the probability that he'd get the prize?
A. 4/27
B. 2/9
C. 1/3
D. 5/9
E. 19/27

[spoiler]made up by Sanjeev K Saxena for Avenues Abroad[/spoiler]
Depends. Do you mean the probability that he gets three green balls, or the probability that with three tries he gets at least one green ball? And do the marbles stay out, or does he put them back after each try?

If it's all three and the marbles stay out, then that would be:

(3/9) * (2/8) * (1/7) = (1/3) * (1/4) * (1/7) = 1/84.

If it's at least one of the three and the marbles stay out, then that would be one minus the probability that he DOESN'T pick any greens:

1 - (6/9) * (5/8) * (4/7) = 1 - (1/3) * (5/1) * (1/7) = 1 - 5/21 = 16/21

If it's all three and the marbles do not stay out, then that would be:

(3/9) * (3/9) * (3/9) = (1/3) * (1/3) * (1/3) = 1/27.

If it's at least one of the three and the marbles do not stay out, then that would be one minus the probability that he DOESN'T pick any greens:

1 - (6/9) * (6/9) * (6/9) = 1 - (2/3) * (2/3) * (2/3) = 1 - 8/27 I meant this only, please suggest the adequate wordings.

Thanks MBACenter for pointing out the flaw. The term "fresh attempts" should mean that the experiments are carried out with replacement, if I am not wrong.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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