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March 22nd, 2012, 02:47 PM  #1 
Newbie Joined: Feb 2012 Posts: 10 Thanks: 0  Probability, marbles
Hi, can someone check this and tell me if I am doing this correctly? I would really appreciate it! Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen. a) Find the probability distribution of X. b) Find the mean of the random variable X ok so for part a i did: P(0B) (6 choose 0) 1*(0.10)^0 (0.3)^6 = 0.000729 P(1B) (6 choose 1) 6* (0.10)^1 (0.3)^5 = .001458 P(2B) (6 choose 2) 15 *(0.10)^2 (0.3)^4 = .001215 P(3B) (6 choose 3) 20 *(0.10)^3 (0.3)^3 = 0.00054 P(4B) (6 choose 4) 15 * (0.10)^4 (0.3)^2 = 0.000135 P(5B) (6 choose 5) 6 * (0.10)^5 (0.3)^1 = 0.00000180 P(6B) (6 choose 6) 1 * (0.10)^6 (0.3)^0 = 0.00000010 Is this right? Or am I doing this completely wrong? Some of the numbers I calculated aren't coming out right. 
March 23rd, 2012, 11:43 AM  #2 
Newbie Joined: Feb 2012 Posts: 10 Thanks: 0  Re: Probability, marbles
I tried it again and I still am getting wrong answers 
March 23rd, 2012, 08:42 PM  #3  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 1  Re: Probability, marbles Quote:
If that's the case, I see that the event of drawing blue marbles are dependent, and the probability of success (i.e. to get blue marbles) is not the same for each draw. In other words, it can't constitute any probability distribution, don't you think???  
March 24th, 2012, 01:55 PM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Probability, marbles Hello, Ataloss! Quote:
There are 8 Blue marbles and 13 Others.  

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