For x = 10, it would have to be two nickels.
For 1st draw there are 15 total coins and 6 nickels, so probability of getting a nickel is 6/15 = 2/5
For 2nd draw there are 14 total coins and 5 nickels, so probability of getting a nickel is 5/14
Probability of both events occurring is 2/5 * 5/14 = 1/7
For x = 11, it would have to be 1 dime and 1 penny
Which could come in order: dime then penny or penny then dime
For 1st scenario dime then penny,
For 1st draw there are 15 total coins and 4 dimes, so probability of getting a dime is 4/15
For 2nd draw there are 14 total coins and 5 pennies, so probability of getting a penny is 5/14
Probability of both events occurring is 4/15 * 5/14 = 2/21
For 2nd scenario penny then dime.
For 1st draw there are 15 total coins and 5 pennies, so probability of getting a penny is 5/15 = 1/3
For 2nd draw there are 14 total coins and 4 dimes, so probability of getting a dime is 4/14 = 2/7
Probability of both events occurring is 1/3 * 2/7 = 2/21
Probability of either event occurring is 2/21 + 2/21 = 4/21
These are your possibilities for two picks:
X P(x) x*P(x)
Penny, Penny 2 4/42 = 0.095 0.19
Nickel, Nickel 10 1/7 = 0.143 1.43
Dime, Dime 20 2/35 = 0.057 1.14
Penny, Nickel 6 2/7 = 0.286 1.716
Penny, Dime 11 4/21 = 0.190 2.09
Nickel, Dime 15 8/35 = 0.229 3.435
E(X) = sum x*P(x) = 9.971
