Question

A hat contains eight one-dollar bills and two thousand-dollar bills. A coin is tossed. If it falls heads, Bill gets to draw, at random, two bills from the hat. If the coin falls tails, Bill draws only one bill, unless this first bill is a one-dollar bill; in the latter case, he gets to draw two additional bills. What is the average value of the bills Bill draws?

Answer 1

Let H shows the event that coin shows head and T shows the event that coin shows tail. So we have

P(H) = P(T) = 0.50

Let X is a random variable shows the amount draw.

If Coin shows head:

Since now he draws to two bills so possible draws are 1 1, 1 1000, 1000 1000.

That is X can take values 1+1 = 2, 1+1000=1001 and 1000+1000=2000.

Total number of bills in the hat = 8 +2 = 10

Number of ways of selecting 2 bills out of 10 is C(10,2).

When X=2, means two $1 bill selected. Number of ways of selecting two $1 bills is C(8,2). Therefore

P(X=2 | H) = C(8,2) / C(10, 2) = 0.6222

Likewise

P(X=1001 |H) = [C(8,1)*C(2,1)] / C(10, 2) = 0.3556

P(X=2000 |H) = [C(2,2)] / C(10, 2) = 0.0222

Now,

P(X=2 and H) = P(X=2 | H)*P(H) = 0.3111

P(X=1001 and H) = P(X=1001 | H)*P(H) = 0.1778

P(X=2000 and H) = P(X=2000 | H)*P(H) = 0.0111

If Coin shows tail:

If first bill is not one dollar then X can take value 1000. That is we need to select 1 bill out of 2 thousand bills.  and we have

P(X=1000 | T and not 1 dollar) = C(2,1) / C(10,1) = 0.2

If one dollar bill is selected, then we have 9 bills remaining out of which 7 are one dollar and 2 are thousand bills. Here X can take values 1+1+1 = 3, 1+1+1000=1002 and 1+1000+1000=2001.

Probabilities are:

P(one dollar bill | T) = C(8,1) / C(10,1) = 0.8

P(X = 3 | T and one dollar bill) = C(7,2) / C(9,2) = 0.5833

P(X = 1002 | T and one dollar bill) = [C(7,1)*C(2,1)] / C(9,2) = 0.3889

P(X = 2001 | T and one dollar bill) = [C(2,2)] / C(9,2) = 0.0278

Now,

P(X = 3 and T and one dollar bill) = P(X = 3 | T and one dollar bill)*P(T and one dollar bill) *P(T)= 0.2333

P(X = 1002 and T and one dollar bill) = P(X = 1002 | T and one dollar bill)*P(T and one dollar bill) *P(T)= 0.1556

P(X = 2001 and T and one dollar bill) = P(X = 2001 | T and one dollar bill)*P(T and one dollar bill) *P(T)= 0.0111

P(X=1000 and T and not 1 dollar) = P(X=1000 | T and not 1 dollar) *P(T) = 0.10

Since all values of X are unique so marginal pdf of X will be

X	P(X=x)
2	0.3111
1001	0.1778
2000	0.0111
3	0.2333
1002	0.1556
2001	0.0111
1000	0.1
Total	1

Following table shows the calculations for E(X):

X	P(X=x)	xP(X=x)
2	0.3111	0.6222
1001	0.1778	177.9778
2000	0.0111	22.2
3	0.2333	0.6999
1002	0.1556	155.9112
2001	0.0111	22.2111
1000	0.1	100
Total	1	479.6222

The average value of the bills Bill draws is

Answer: $479.62