I have a bag that contains $6$ nickels and $5$ pennies. I draw coins from the bag at random, without replacement. Find the probability that after drawing $4$ coins, I have removed at most $2$ pennies from the bag.

Guest Mar 6, 2020

#1**+1 **

To have removed at most 2 pennies out of 4 coins, you could have either selected 4 nickels, 3 nickels and 1 penny, or 2 nickels and 2 pennies.

There is only one way to select 4 nickels: 6/11 x 5/10 x 4/9 x 3/8 = 1/22

There are four ways to select 3 nickels and 1 penny; each way is: 6/11 x 5/10 x 4/9 x 5/8 = 1/66

4 x 1/66 = 2/33

There are six ways to select 2 nickels and 2 pennies; each way is: 6/11 x 5/10 x 5/9 x 4/8 = 5/66

6 x 5/66 = 5/11

Adding these probabilities: 1/22 + 2/33 + 5/11 = 37/66

geno3141 Mar 7, 2020

#5**0 **

For this problem, we can use complementary counting. We can count the ways to find 4 pennies and 3 pennies and 1 nickel and subtract it from the total.

For 4 pennies, it is 5/11*4/10*3/9*2/8 which equals 1/66.

For 3 pennies and one nickel, it is 5/11*4/10*3/9*6/8 which equals 1/22. There are 4 ways to do this, so the probability is 4/22.

When we add the probabilities, 1/66+4/22=13/66. We subtract this from 66/66 to get **53/66**

Guest Mar 8, 2020