SOLUTION: There are three urns:first urn contains 1 white,2red,3green balls,second urn contains 2 white,1red,1green balls and third urn contains 4 white,5red,3green balls. Two balls are dra

Algebra ->  Permutations -> SOLUTION: There are three urns:first urn contains 1 white,2red,3green balls,second urn contains 2 white,1red,1green balls and third urn contains 4 white,5red,3green balls. Two balls are dra      Log On


   

Question 700038: There are three urns:first urn contains 1 white,2red,3green balls,second urn contains 2 white,1red,1green balls and third urn contains 4 white,5red,3green balls.
Two balls are drawn from an urn chosen at random.These are found to be one white and one green.Probability that balls are drawn from third urn is:
(A)15/59, (B)14/59, (C)13/59, (D)12/59
Please explain the procedure also!In book only the correct answer is given 15/59....but i dont get how to slove it,so plz explain the procedure,plz.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
DISCLAIMER:
I do not memorize procedures, and I do not follow procedures unless someone makes me.
What follows is my just my understanding of the problem.
Your teacher may have a procedure to reach the solution more efficiently.

If an urn chosen at random, the third urn will be chosen 1%2F3 of the times.
(In fact, each of the three urns will be chosen 1%2F3 of the times).

If the first urn is chosen (with 6 balls, including 1 white and 3 green),
the probability of drawing the white ball first is 1%2F6.
After that, the probability of drawing one of the 3 green balls from the 5 balls left in the urn is 3%2F5.
So, the probability of drawing a white ball followed by a green ball from the first urn is %281%2F6%29%283%2F5%29=1%2F10.
Similarly, the probability of drawing a green ball followed by a white ball from the first urn is %283%2F6%29%281%2F5%29=1%2F10.
The probability of drawing a white ball and a green ball (in any order) from the first urn is 1%2F10%2B1%2F10=1%2F5.

If the second urn is chosen (with 4 balls, including 2 white and 1 green),
the probability of drawing a white ball and a green ball (in any order) is
%282%2F4%29%281%2F3%29%2B%281%2F4%29%282%2F3%29=1%2F6%2B1%2F6=2%2F6=1%2F3

If the third urn is chosen (with 12 balls, including 4 white and 3 green),
the probability of drawing a white ball and a green ball (in any order) is
%284%2F12%29%283%2F11%29%2B%283%2F12%29%284%2F11%29=1%2F11%2B1%2F11=2%2F11

If the drawing of two balls was done 3%2A%283%2A5%2A11%29=3%2A165,
you would expect the first urn to be chosen 165 times,
drawing one a white ball and a green ball (in any order) from the first urn
165%281%2F5%29=3%2A5%2A11%282%2F5%29=3%2A11=highlight%2833%29 times.
The second urn would be also chosen 165 times,
drawing one a white ball and a green ball (in any order) from the second urn
165%281%2F3%29=3%2A5%2A11%281%2F3%29=5%2A11=highlight%2855%29 times.
The third urn would be also chosen 165 times,
drawing one a white ball and a green ball (in any order) from the third urn
165%282%2F11%29=3%2A5%2A11%282%2F11%29=3%2A5%2A2=highlight%2830%29 times.

In all, a white ball and a green ball (in any order) would be drawn (from an urn chosen at random)
33%2B55%2B30=118 times.
Of those 118 times, 30 times it would be from the third urn,
so if a white ball and a green ball (in any order) had been drawn
30%2F118=highlight%2815%2F59%29 of the times it would have been using the third urn.