Question

Two coins are randomly drawn from a coin pouch. If there are 4 dimes, 3 nickels, and a quarter, how many ways can at least 20 cents be drawn out?

Answer 1

Values in cents

1 quarter =25 cents

1 dime =10 cents

1 nickel = 5 cents

Total number of coins in the pouch 1+4+3=8

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

S = { NN, DD, NQ, ND, DQ}

Let X is a random variable shows the total value. The value of X in above cases is 10, 20, 30, 15, 35. Number of ways of selecting 2 coins out of 8 is C(8,2) = 28. Since there are 3 nickels so

P(X = 10) = P(NN) = C(3,2) / 28 = 3 / 28

There are 4 Dimes so

P(X = 20) = P(DD) = C(4,2) / 28 = 6/ 28

Likewise,

P(X = 30) = P(NQ) = [C(3,1)*C(1,1) ] / 28 = 3 / 28

P(X = 15) = P(ND) = [C(3,1)*C(4,1) ] / 28 = 12 / 28

P(X = 35) = P(DQ) = [C(1,1)*C(4,1) ] / 28 = 4 / 28

Following table shows the pdf:

X	P(X=x)
10	3/28
15	12/28
20	6/28
30	3/28
35	4/28

Since we need to find the number of ways so only numerator of probability expression will be added. The number of ways at least 20 cents be drawn out is

6 + 3+ 4 = 13

Answer: 13