SOLUTION: A bag contains two red balls, one green ball, and six yellow balls. Balls are drawn from the bag one at a time at random without replacement until a yellow ball has been drawn.

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Question 1125214: A bag contains two red balls, one green ball, and six yellow balls. Balls are drawn from the bag one at a time at random without replacement until a yellow ball has been drawn.

Let random variable R represent the number of red balls drawn. Complete the probability distribution function for R in the table below, putting the values of R in ascending order.

Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
Bag: 2 Red (R), 1 Green (G), 6 Yellow (Y)
Total is 9 balls.

Possible outcomes          R             Probability

Y                          0             (6/9) = 336/504
GY                         0             (1/9)*(6/8) = 6/72 = 42/504
GRY                        1             (1/9)*(2/8)*(6/7) = 12/504
RY                         1  
RGY                        1
RRY                        2
RRGY                       2            (2/9)*(1/8)*(1/7)*(6/6) = 2/504 
RGRY                       2
GRRY                       2


For the first row, a Y is selected ending the trial. This happens with probability 6/9 because there are 6 Y's out of 9 total.

For the 2nd row, a G is selected then a Y, ending the trial. This happens with probability (1/9)*(6/8), 1 G out of 9 for 1st pick then 6 out of the remaining 8 are Y.

For the 3rd row, G,R,Y: (1/9)*(2/8)*(6/7)

I filled in some of the table, you should be able to follow the pattern to fill in the rest (or another tutor will complete it for you, as often happens). I left the denominator as 504 as that is common for all of the calculations, as the numerators should tally up to 504 when you have correctly completed the table (the sum of probabilities of all possible outcomes is 1).

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