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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

Miguel is playing a game in which a box contains four chips with numbers written on-example-1
User HWD
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2 Answers

5 votes

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.

User HedgeHog
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7.9k points
7 votes

Answer:

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

User Ojreadmore
by
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