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# hAlPsSsS

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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.

Mar 6, 2020

#1
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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

Mar 7, 2020
#2
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:C

Guest Mar 7, 2020
#5
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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

Mar 8, 2020
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@geno3141 your 2nd probability is wrong, should be 5/66 *4= 20/66=10/33, instead of 2/33

Mar 8, 2020