Question

A jar contains 2 pennies, 6 nickels and 4 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.

a. Find the probability X = 11.

b. Find the expected value of X

Answer 1

1 penny =1 cent

1 dime =10 cents

1 nickel = 5 cents

Let N shows the event that a nickel is selected, D shows the event that a dime is selected and P shows the event that a penny is selected. Since order of coins is not important so possible outcomes are:

S = { NN, PP, DD, NP, ND, PD}

The values of above outcomes in cent are:

NN = 10, PP = 2 , DD = 20, NP = 6, ND = 15, PD = 11

The value of X in each case is 10, 2, 20, 6, 15, 11 respectively.

The total number of coins in the jar is: 2+6+4 = 12

Number of ways of selecting 2 coins out of 12 is C(12, 2) = 66

Now

P(X = 10) = P(NN) = C(6,2) / 66= 15 / 66

P(X =2) = P(PP) = C(2,2) / 66= 1 / 66

P(X =20) = P(DD) = C(4,2) / 66= 6 / 66

P(X=6) = P(NP) = [C(6,1)*C(2,1)] / 66 = 12 / 66

P(X=15) = P(ND) = [C(6,1)*C(4,1)] / 66 = 24 / 66

P(X=11) = P(PD) = [C(2,1)*C(4,1)] / 66 = 8 / 66

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So,

P(X=11) = 8 / 66 = 4 / 33

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Following table shows the calculations for expected value:

X	P(X=x)	xP(X=x)
10	0.22727273	2.27272727
2	0.01515152	0.03030303
20	0.09090909	1.81818182
6	0.18181818	1.09090909
15	0.36363636	5.45454545
11	0.12121212	1.33333333
Total	1	12

So,