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April 16th, 2017, 04:06 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 541 Thanks: 77  Probability Questions With 10 Marbles
There are 6 blue marbles and 4 red marbles. On each turn, two players pick one marble. This is done for five turns without replacement until all the marbles have been picked. 1. What is the probability that each person picks 2 red marbles? 2. What is the probability that one of the turns will have both people pick their second red marbles? 
April 16th, 2017, 05:10 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,303 Thanks: 666 
The number of ways you can picks 2 reds (and thus 3 blues) is given by $\displaystyle \binom{4}{2}\binom{6}{3}$ The total number of ways of selecting 5 marbiles is $\displaystyle \binom{10}{5}$ Thus $\displaystyle P[\text{2 reds}] = \dfrac{\binom{4}{2}\binom{6}{3}}{\binom{10}{5}}= \dfrac{10}{21}$ (2) is much more complicated. 
April 17th, 2017, 02:37 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 541 Thanks: 77 
I know the second question is more complicated. Given that a person picked two reds, there are 10 orders they could have been picked: 1 and 2 1 and 3 1 and 4 1 and 5 2 and 3 2 and 4 2 and 5 3 and 4 3 and 5 4 and 5 There are 10^2 = 100 possibilities for two people given that they each picked two reds. I can manually list the possibilities and see which ones have both people pick their second red on the same turn. Then I can multiply this by the 10/21 probability that each person picked two reds to get my answer. 
April 17th, 2017, 02:42 PM  #4 
Senior Member Joined: Aug 2012 Posts: 1,438 Thanks: 353 
I have no idea. I've lost my marbles.

April 17th, 2017, 06:18 PM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 541 Thanks: 77 
I found both people picked their second red marble on the same turn 30 out of 100 times. Therefore the answer is (10/21)(30/100) = 1/7.

April 17th, 2017, 09:22 PM  #6  
Senior Member Joined: Sep 2015 From: CA Posts: 1,303 Thanks: 666  Quote:
you end up with $p= \dfrac{\displaystyle \sum_{k=1}^5 k^2}{\displaystyle \binom{10}{6}} = \dfrac{30}{210} = \dfrac 1 7$  
April 18th, 2017, 02:18 PM  #7 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 541 Thanks: 77 
To make the numerator 30, the summation has to stop at k = 4, not k = 5. 1 + 4 + 9 + 16 = 30. Adding 25 would make the numerator 55.


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