I have a bag with 5 pennies and 6 nickels. I draw coins out one at a time at random. What is the probability that after 4 draws I have removed no more than 2 pennies from the bag?

Guest Jan 25, 2020

#2**+1 **

I have a bag with 5 pennies and 6 nickels. I draw coins out one at a time at random. What is the probability that after 4 draws I have removed no more than 2 pennies from the bag?

So not 3 and not 4 pennies

P(pppn) = 5/11 * 4/10 * 3/9 * 6/8 = \(\frac{5*4*3*6}{11*10*9*8}=\frac{1*1*1*1}{11*2*1}=\frac{1}{22}\)

But the n can be in any of 4 places so that is 4/22

P(pppp)= \(\frac{5*4*3*2}{11*10*9*8}=\frac{120}{7920}=\frac{1}{66}\)

P(all pennies or 3 pennies) = \(\frac{4}{22}+\frac{1}{66}=\frac{13}{66}\)

P(2 or less pennies) = \(1-\frac{13}{66}=\frac{53}{66}\)

Needs to be checked I could have made a careless mistake.

Melody Jan 25, 2020