document.write( "Question 1091034: Consider a game where there is a bag filled with x marbles, one of which is red.
\n" ); document.write( "Initially a contestant is given 5 picks from the bag and then the picked marble is placed back in the bag.  If the contestant
\n" ); document.write( "at any point draws the red marble from the bag they get 5 more picks from the bag.
\n" ); document.write( " What is the probability of getting only 5 picks from the bag?
\n" ); document.write( " What is the probability of getting exactly 15 picks from the bag?
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Algebra.Com's Answer #705485 by mathmate(429)\"\" \"About 
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Question:
\n" ); document.write( "Consider a game where there is a bag filled with x marbles, one of which is red.
\n" ); document.write( "Initially a contestant is given 5 picks from the bag and then the picked marble is placed back in the bag. If the contestant
\n" ); document.write( "at any point draws the red marble from the bag they get 5 more picks from the bag.
\n" ); document.write( " What is the probability of getting only 5 picks from the bag?
\n" ); document.write( " What is the probability of getting exactly 15 picks from the bag?
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\n" ); document.write( "Solution:
\n" ); document.write( "Picking WITH replacement,
\n" ); document.write( "P(red)=1/x
\n" ); document.write( "P(not red)=(x-1)/x\r
\n" ); document.write( "\n" ); document.write( "(a) 5 picks from the bag
\n" ); document.write( "That means the person had not picked the red marble, not even once.
\n" ); document.write( "Using the multiplication rule,
\n" ); document.write( "P(no red) = P(not red)^5 = [(x-1)/x]^5\r
\n" ); document.write( "\n" ); document.write( "(b) exactly 15 picks
\n" ); document.write( "Case A: The person picked two red marbles from the first five picks, and no red subsequently.
\n" ); document.write( "P(A)=C(5,2)*(1/x)^2*((x-1)/x)^3 * ((x-1)/x)^10
\n" ); document.write( "=10(x-1)^13/x^15
\n" ); document.write( "Case B: The person picked one red marble from the first five, one red from the next five, and no red subsequently
\n" ); document.write( "P(B)=C(5,1)*(1/x)^4 * C(5,1)*(1/x)^4 * ((x-1)/x)^5
\n" ); document.write( "=25(x-1)^13/x^15\r
\n" ); document.write( "\n" ); document.write( "Total
\n" ); document.write( "P(15)=P(A)+P(B)=35(x-1)^13/x^15.
\n" ); document.write( "Check: probability equals about 9% when 6<=x<=10.
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