Since there are 4 questions on the image and HomeworkLib policy tells us to answer only one question per request I am answering the 5th question, so please rate accordingly
5) a)
We have 3 chips of $1, 2 for $2 and 2 for $5
And two chips are drawn without replacement.
So the expected value of the game is the all the scenarios combined expected value
If X is the game then
For notation let
E[i,i] = Expected value of drawing $i coin on first draw and $j coin on 2nd draw
Then
E[1,1] = Expected value of drawing $1 coins on both draws = 3/7*1*2/6*1 = 6/42 = 1/7
Therefore
E(X) = E[1,1] + E[1,3] + E[1,5] + E[3,1] + E[3,3] + E[3,5] + E[5,1] + E[5,3] + E[5,5]
E(X) = 1/7 + 3/7*2/6*3 + 3/7*2/6*5 + 2/7*3/6*3 + 2/7*1/6*9 + 2/7*2/6*15+ 2/7*3/6*5 + 2/7*2/6*15 + 2/7*1/6*25
E(X) = 1/7 + 3/7 + 5/7 + 3/7 + 3/7 + 10/7+ 5/7 + 10/7 + 25/21
E(X) = (3 + 9 + 15+ 9+ 9 + 30 + 15 + 30 + 25)/21 = $6.90
b) If the house charges $6.9 then the house makes $0 on average
Thus for the house to make $2 , they should charge $8.9 for the game
5. A bowl contains 7 chips, which cannot be distinguished by a sense of touch alone....
1.8. Expectation of a Random Variable 67 1.8.8. A bowl contains 10 chips, of which 8 are marked $2 each and 2 are $5 each. Let a person choose, at random and without replacement, three chips from this bowl. If the person is to receive the sum of the resulting amounts, find his expectation. Let f(z) = 2z, o < 1, zero elsewhere, be the pdf of X. (a) Compute E(1/X). (b) Find the odf and the pdf of Y...