Question

Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has number 3 and rest is in pic

Answer 1

To check all the events (6), we label the chips. Suppose one chip with 1 is labeled R1 and the other B1 (as if they were red and blue). Now, lets take all combinations; for the first chip, we have 4 choices and for the 2nd chip we have 3 remaining choices. Thus there are 12 combinations. Since we dont care about the order, there are only 6 combinations since for example R1, 3 is the same as 3, R1 for us.

The combinations are: (R1, B1), (R1, 3), (R1, 5), (B1, 3), (B1, 5), (3,5)

We have that in 1 out of the 6 events, Miguel wins 2$ and in five out of the 6 events, he loses one. The expected value of this bet is: 1/6*2+5/6*(-1)=-3/6=-0.5$. In general, the expected value of the bet is the sum of taking the probabilities of the outcome multiplied by the outcome; here, there is a 1/6 probability of getting the same 2 chips and so on. On average, Miguel loses half a dollar every time he plays.

Answer 2

a. Table completed

b. E(X) = -0.5

c. Miguel loses $0.5 each time he plays

Explanation:

a. The total outcomes S = {11, 13, 15, 13, 15, 35}

Let A be the event he wins $2 and B be the even he loses $1.

Then, A ={11} and

B = {13, 15, 13, 15, 35}

p(A) =
`(n(A))/(n(S)) =(1)/(6)`

p(B) =
`(n(B))/(n(S)) =(5)/(6)`

Then, the table would be as follows:

`X_(i)` 2 -1

`p(X_(i))`
`(1)/(6)`
`(5)/(6)`

b. E(X) =
`X_(1)p(X_(1))+X_(2)p(X_(2))`

`=2((1)/(6) )+(-1)((5)/(6) )`

`=(2)/(6) -(5)/(6)`

`=(2-5)/(6)`

`=-(3)/(6)`

`=-(1)/(2)`

c. Based on the result (b), Miguel loses $0.5 each time he plays