SOLUTION: Consider a game where there is a bag filled with x marbles, one of which is red. Initially a contestant is given 5 picks from the bag and then the picked marble is placed back in

Question 1091034: Consider a game where there is a bag filled with x marbles, one of which is red.
Initially a contestant is given 5 picks from the bag and then the picked marble is placed back in the bag.��If the contestant
at any point draws the red marble from the bag they get 5 more picks from the bag.
 What is the probability of getting only 5 picks from the bag?
 What is the probability of getting exactly 15 picks from the bag?
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
Consider a game where there is a bag filled with x marbles, one of which is red.
Initially a contestant is given 5 picks from the bag and then the picked marble is placed back in the bag. If the contestant
at any point draws the red marble from the bag they get 5 more picks from the bag.
 What is the probability of getting only 5 picks from the bag?
 What is the probability of getting exactly 15 picks from the bag?

Solution:
Picking WITH replacement,
P(red)=1/x
P(not red)=(x-1)/x
(a) 5 picks from the bag
That means the person had not picked the red marble, not even once.
Using the multiplication rule,
P(no red) = P(not red)^5 = [(x-1)/x]^5
(b) exactly 15 picks
Case A: The person picked two red marbles from the first five picks, and no red subsequently.
P(A)=C(5,2)*(1/x)^2*((x-1)/x)^3 * ((x-1)/x)^10
=10(x-1)^13/x^15
Case B: The person picked one red marble from the first five, one red from the next five, and no red subsequently
P(B)=C(5,1)*(1/x)^4 * C(5,1)*(1/x)^4 * ((x-1)/x)^5
=25(x-1)^13/x^15
Total
P(15)=P(A)+P(B)=35(x-1)^13/x^15.
Check: probability equals about 9% when 6<=x<=10.

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