SOLUTION: If you have a bag of marbles; 7 red, 5 green, and 8 blue, what's the probability of getting at least 2 red marbles in 3 draws from the bag, with replacement? Thanks...this is all

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Question 271350: If you have a bag of marbles; 7 red, 5 green, and 8 blue, what's the probability of getting at least 2 red marbles in 3 draws from the bag, with replacement? Thanks...this is all new to me!
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
If you have a bag of marbles; 7 red, 5 green, and 8 blue, what's the probability of getting at least 2 red marbles in 3 draws from the bag, with replacement?
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Binomial Problem:
n = 3 ; p = 7/20
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P(2<= x <=3) = 3C2(7/20)^2(13/20) + 3C3(7/20)^3(13/20)^0
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= 0.2389 + 0.0429
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= 0.2818
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Cheers,
Stan H.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Here you are looking for the probability of success for a number of trials where the probability of success is constant for each independent trial. The probability is constant because you are replacing the drawn marble each time. The probability on any given trial is the number of red marbles divided by the total number of marbles. You'll have to compute that first. For now, I'm going to call that number . You also need to know the probability of NOT drawing a red marble on any given trial. Two ways to get this: Either count the total number of marbles that aren't red and divide by the total number of marbles or calculate . We'll call this second number .

The probability of exactly successes out of trials is given by:

where

is the number of ways to select things from a collection of things and is equal to:

But you don't want the probability of exactly 2 red out of 3 draws, you want at least 2 red out of 3 draws. So you need the probability of exactly 2 red out of 3 draws PLUS the probability of exactly 3 red out of 3 draws. So, substitute the values you calculated for and into:

and

Then add the two results.

John