Question 700038
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 {{{1/3}}} of the times.
(In fact, each of the three urns will be chosen {{{1/3}}} 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 {{{1/6}}}.
After that, the probability of drawing one of the 3 green balls from the 5 balls left in the urn is {{{3/5}}}.
So, the probability of drawing a white ball followed by a green ball from the first urn is {{{(1/6)(3/5)=1/10}}}.
Similarly, the probability of drawing a green ball followed by a white ball from the first urn is {{{(3/6)(1/5)=1/10}}}.
The probability of drawing a white ball and a green ball (in any order) from the first urn is {{{1/10+1/10=1/5}}}.
 
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
{{{(2/4)(1/3)+(1/4)(2/3)=1/6+1/6=2/6=1/3}}}
 
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
{{{(4/12)(3/11)+(3/12)(4/11)=1/11+1/11=2/11}}}
 
If the drawing of two balls was done {{{3*(3*5*11)=3*165}}},
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(1/5)=3*5*11(2/5)=3*11=highlight(33)}}} 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(1/3)=3*5*11(1/3)=5*11=highlight(55)}}} 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(2/11)=3*5*11(2/11)=3*5*2=highlight(30)}}} times.
 
In all, a white ball and a green ball (in any order) would be drawn (from an urn chosen at random)
{{{33+55+30=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
{{{30/118=highlight(15/59)}}} of the times it would have been using the third urn.