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September 19th, 2009, 10:54 AM  #1 
Newbie Joined: Sep 2009 Posts: 2 Thanks: 0  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? 
September 19th, 2009, 05:50 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,807 Thanks: 717  Re: A probability issue. Quote:
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.  
September 19th, 2009, 06:29 PM  #3  
Newbie Joined: Sep 2009 Posts: 2 Thanks: 0  Re: A probability issue. Quote:
 

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