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
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Answer:
a. Table completed
b. E(X) = -0.5
c. Miguel loses $0.5 each time he plays
Step-by-step 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) = [tex]\frac{n(A)}{n(S)} =\frac{1}{6}[/tex]
p(B) = [tex]\frac{n(B)}{n(S)} =\frac{5}{6}[/tex]
Then, the table would be as follows:
[tex]X_{i}[/tex] 2 -1
[tex]p(X_{i})[/tex] [tex]\frac{1}{6}[/tex] [tex]\frac{5}{6}[/tex]
b. E(X) = [tex]X_{1}p(X_{1})+X_{2}p(X_{2})[/tex]
[tex]=2(\frac{1}{6} )+(-1)(\frac{5}{6} )[/tex]
[tex]=\frac{2}{6} -\frac{5}{6}[/tex]
[tex]=\frac{2-5}{6}[/tex]
[tex]=-\frac{3}{6}[/tex]
[tex]=-\frac{1}{2}[/tex]
c. Based on the result (b), Miguel loses $0.5 each time he plays
To check all the events (6), we label the chips. Suppose one chip with 1 is labeled R1 and the other B1 (as if they were red and blue). Now, lets take all combinations; for the first chip, we have 4 choices and for the 2nd chip we have 3 remaining choices. Thus there are 12 combinations. Since we dont care about the order, there are only 6 combinations since for example R1, 3 is the same as 3, R1 for us.
The combinations are: (R1, B1), (R1, 3), (R1, 5), (B1, 3), (B1, 5), (3,5)
We have that in 1 out of the 6 events, Miguel wins 2$ 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.