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by topspin360 » Tue Dec 18, 2012 2:30 am
Hello,

I understand how to do the original question but can you please explain how to solve my version of the question (posted underneath the question below).

If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw? -understand how to do this.

***What if the question said: draw 3 times, replacing marbles each time. What is the probability of getting red marbles exactly twice? At least twice?

I know I can't use the counting principal alone (12/30 * 12/30 * 18/30). How do i do it? Can i use combinatorics on the counting principal? so it would be: 3! * (12/30 * 12/30 * 18/30)?

Thanks.
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by Anurag@Gurome » Tue Dec 18, 2012 9:20 am
topspin360 wrote:What if the question said: draw 3 times, replacing marbles each time. What is the probability of getting red marbles exactly twice? At least twice?

I know I can't use the counting principal alone (12/30 * 12/30 * 18/30). How do i do it? Can i use combinatorics on the counting principal? so it would be: 3! * (12/30 * 12/30 * 18/30)?
Yes, you can but the red part is wrong.

Probability of drawing a red marble in exactly two draw = (Probability of drawing a red marble)*(Probability of drawing a red marble)*(Probability of drawing a blue marble) = (12/30)*(12/30)*(18/30) = (2/5)*(2/5)*(3/5) = 12/125

Now this can happen in 3C2 = 3 ways.

Hence, required probability = 3*(12/125) = 36/125

Probability of drawing a red marble at least twice = (Probability of drawing a red marble exactly twice) + (Probability of drawing a red marble all the three times) = 36/125 + (12/30)*(12/30)*(12/30) = 36/125 + 8/125 = 42/125
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by ritind » Wed Dec 19, 2012 10:27 pm
Anurag@Gurome wrote:
topspin360 wrote:What if the question said: draw 3 times, replacing marbles each time. What is the probability of getting red marbles exactly twice? At least twice?

I know I can't use the counting principal alone (12/30 * 12/30 * 18/30). How do i do it? Can i use combinatorics on the counting principal? so it would be: 3! * (12/30 * 12/30 * 18/30)?
Yes, you can but the red part is wrong.

Probability of drawing a red marble in exactly two draw = (Probability of drawing a red marble)*(Probability of drawing a red marble)*(Probability of drawing a blue marble) = (12/30)*(12/30)*(18/30) = (2/5)*(2/5)*(3/5) = 12/125

Now this can happen in 3C2 = 3 ways.

Hence, required probability = 3*(12/125) = 36/125

Probability of drawing a red marble at least twice = (Probability of drawing a red marble exactly twice) + (Probability of drawing a red marble all the three times) = 36/125 + (12/30)*(12/30)*(12/30) = 36/125 + 8/125 = 42/125
44/125
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