Stats Help Please!!

Question

. There are nine $1 dollar bills and one $50 dollars bill in an urn. Let 𝑋 be the total amount that results when two bills are drawn with replacement.

a) Compute the mean of the amount obtained, 𝔼[𝑋].

b) Compute the variance of the amount obtained, Var(𝑋).

c) Compute the mean of the amount obtained, 𝔼[𝑋] when the drawings are without replacement.

d) Compute the variance of the amount obtained, Var(𝑋) when the drawings are without replacement.

Answer 1

If there is replacement, the expected value of each draw is the same, (.1*50) + (.9*1) or $5.90. This is the probability of getting a $50 bill times the value of the 50, and the probability of getting a $1 bill, times the value of the $1 bill. The outcome for the second draw is the same (remember - we replace the bills that are drawn)

Thus the expected value for the sum of two draws, with replacement , is $5.90 plus $5.90 or $11.80

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Without replacement

When calculating without replacement the probability for the outcome of two simultaneous draws must be determined. One can draw a $50 and a $1, or two $1's, but not two $50.

The number of different ways two bills can be drawn from a stack of 10 bills (order does not matter) would be 10C2 or 45

[note: nCr is a shorthand notation for a combination of r picks from n items and can be determined using a scientific calculator with probability functions]

The number of different ways that one can draw a $50 and a $1 is 9 (order does not matter, and we have one $50 bill and 9 $1 bills).

So, the probability of getting $51 is 9/45 or .2 , of course the probability of getting every other combination (i.e. two $1) is .8 so

The expected value is then (.2*51) + (.8*2) or $11.80 (thanks to Bob A for correction)