SOLUTION: A box has 5 blue and 9 green jelly beans. A bag has 7 blue and 3 green jelly beans. A jelly bean is selected at random from the box and placed in the bag. Then a jelly bean is sele

Algebra ->  Probability-and-statistics -> SOLUTION: A box has 5 blue and 9 green jelly beans. A bag has 7 blue and 3 green jelly beans. A jelly bean is selected at random from the box and placed in the bag. Then a jelly bean is sele      Log On


   



Question 1139065: A box has 5 blue and 9 green jelly beans. A bag has 7 blue and 3 green jelly beans. A jelly bean is selected at random from the box and placed in the bag. Then a jelly bean is selected at random from the bag. If a green jelly bean is selected from the bag, what is the probability that the transferred jelly bean was green? (Round your answer to three decimal places.)
Answer by ikleyn(52848) About Me  (Show Source):
You can put this solution on YOUR website!
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After 1-st transfer,


    With the probability  5%2F%285%2B9%29 = 5%2F14  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%2F%285%2B9%29 = 9%2F14  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%2F%288%2B3%29 = 3%2F11  in case A,  and

    the probability to get a green jelly from the bag is  4%2F%284%2B7%29 = 4%2F11  in case B.



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


    P(green) = P%28A%29%2A%283%2F11%29 + P%28B%29%2A%284%2F11%29 = %285%2F14%29%2A%283%2F11%29 +  %289%2F14%29%2A%284%2F11%29 = %285%2A3+%2B+9%2A4%29%2F%2814%2A11%29 = 51%2F%2814%2A11%29.


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


    P(green at the first transfer) = P%28green%29%2FP%28case_B%29 = %2851%2F%2814%2A11%29%29%2F%28%289%2F14%29%29 = 51%2F%289%2A11%29 = 51%2F99 = 17%2F33.    ANSWER

Completed and solved.