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

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.