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September 19th, 2009, 10:54 AM   #1
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A probability issue.

I've been having an argument about the following problem. Hopefully someone can help me to resolve it!

I have four marbles in a bag; three red, one blue. I want to find the probability of choosing the blue marble in two draws, without replacing the first marble.

I know that on the first draw, I have a 25% chance of drawing blue.
I know that on the second draw, if I draw red on the first go and do not put the red marble back in, I have a 33.3% chance of drawing blue.

I know that if I were to draw two marbles at once, I have a 50% chance of drawing blue.

However, I'm positive that drawing one at a time without replacing yields a higher probability than 50% (to the tune of 58.3%, or 7/12, I think) --- am I wrong?
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September 19th, 2009, 05:50 PM   #2
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Re: A probability issue.

Quote:
Originally Posted by Scalfaro
I've been having an argument about the following problem. Hopefully someone can help me to resolve it!

I have four marbles in a bag; three red, one blue. I want to find the probability of choosing the blue marble in two draws, without replacing the first marble.

I know that on the first draw, I have a 25% chance of drawing blue.
I know that on the second draw, if I draw red on the first go and do not put the red marble back in, I have a 33.3% chance of drawing blue.

I know that if I were to draw two marbles at once, I have a 50% chance of drawing blue.

However, I'm positive that drawing one at a time without replacing yields a higher probability than 50% (to the tune of 58.3%, or 7/12, I think) --- am I wrong?
You are wrong.
As you noted the prob of picking blue on first draw is 1/4.
For the second draw your prob is 3/4 (prob of red on first draw) x 1/3 (prob of blue on second draw) = 1/4.

Therefore the prob of getting a blue is 1/2.
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September 19th, 2009, 06:29 PM   #3
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Re: A probability issue.

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Originally Posted by mathman
You are wrong.
As you noted the prob of picking blue on first draw is 1/4.
For the second draw your prob is 3/4 (prob of red on first draw) x 1/3 (prob of blue on second draw) = 1/4.

Therefore the prob of getting a blue is 1/2.
I don't doubt you, but how did you get from the 1/4 to 1/2?
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