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Urn, urn, urn (Posted on 2007-02-20) Difficulty: 3 of 5
Before you are three urns. The first two each contain 4 white and 6 black balls. The third has 3 white and 6 black.

Take one ball from the first urn and add it to the second with out looking at it. Stir it in, then take one ball from the second and add it to the third without looking at it.

If you pick a ball from the third urn, what is the probability it will be white?

What is the least number of balls you can put in the urns (at least one black and one white in each) to make the probability at the end equal to exactly 1/3?

No Solution Yet Submitted by Jer    
Rating: 5.0000 (1 votes)

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Solution Comment 5 of 5 |
    Let there be a white and b black balls in urn 1; c white and d black in urn 2; and e white and f black in urn 3.

The probability of WWW is
(a * (c+1) * (e+1))/((a+b)*(c+d+1)*(e+f+1))
The probability of WBW is
(a * d * e)/((a+b)*(c+d+1)*(e+f+1))
The probability of BWW is
(b * c * (e+1))/((a+b)*(c+d+1)*(e+f+1))
The probability of BBW is
(b * (d+1) * e)/((a+b)*(c+d+1)*(e+f+1))

Therefore the probability of drawing white out of urn 3 is:
[(a*(c+1)*(e+1))+(a*d*e)+(b*c*(e+1))+(b*(d+1)*e)]/
((a+b)*(c+d+1)*(e+f+1))

Let the denomonator equal three times the numerator and reduce the algebra to get:
2*(a+b)*(ce+c+e+de)+2a-b = (a+b)*(cf+df+d+f)

Divide out (a+b):
2*(ce+c+e+de)+[(2a-b)/(a+b)]=cf+df+d+f

There are no integer solutions for a = b = 1 so number of balls will be minimized with a,b = 1,2 leaving us with:
2*(ce+c+e+de)=cf+df+d+f

More algebra gives:
2*(c+d+1)*e + 2c-d = f*(c+d+1)

Divide out (c+d+1):
2e + [(2c-d)/(c+d+1)] = f

No integer solutions for c=d=1 so number of balls will be minimized when c,d = 1,2 leaving us with 2e = f giving:
a,b,c,d,e,f = 1,2,1,2,1,2 or 9 total balls.



  Posted by Eric on 2007-02-20 22:30:41
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