Question 1139021
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<pre>
After 1-st transfer,


    With the probability  {{{5/(5+9)}}} = {{{5/14}}}  you will get (7+1 blue, 3 green) = (8 blue,3 green) in the bag   

    (call it "case A". It is the case when a blue jelly was transferred);



    with the probability  {{{9/(5+9)}}} = {{{9/14}}}  you will get (7 blue, 3+1 green) = (7 blue,4 green) in the bag   

    (call it "case B". It is the case when a green jelly was transferred).



After the selection from the bag, 


    the probability to get a green jelly from the bag is  {{{3/(8+3)}}} = {{{3/11}}}  in case A,  and

    the probability to get a green jelly from the bag is  {{{4/(4+7)}}} = {{{4/11}}}  in case B.



Thus the probability to get a green jelly from the bag is 


    P(green) = {{{P(A)*(3/11)}}} + {{{P(B)*(4/11)}}} = {{{(5/14)*(3/11)}}} +  {{{(9/14)*(4/11)}}} = {{{(5*3 + 9*4)/(14*11)}}} = {{{51/(14*11)}}}.


Now use the formula for conditional probability to get the answer to the problem's question


    P(green at the first transfer) = {{{P(green)/P(case_B)}}} = {{{(51/(14*11))/((9/14))}}} = {{{51/(9*11)}}} = {{{51/99}}} = {{{17/33}}}.    <U>ANSWER</U>
</pre>

Completed and solved.