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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?

 Jan 25, 2020
 #1
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The probability is 14/47. cool

 Jan 25, 2020
 #2
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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.

 Jan 25, 2020

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