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

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) = $\frac{n(A)}{n(S)} =\frac{1}{6}$

p(B) = $\frac{n(B)}{n(S)} =\frac{5}{6}$

Then, the table would be as follows:

$X_{i}$            2    -1

$p(X_{i})$        $\frac{1}{6}$     $\frac{5}{6}$

b.  E(X) = $X_{1}p(X_{1})+X_{2}p(X_{2})$

$=2(\frac{1}{6} )+(-1)(\frac{5}{6} )$

$=\frac{2}{6} -\frac{5}{6}$

$=\frac{2-5}{6}$

$=-\frac{3}{6}$

$=-\frac{1}{2}$

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

Answer 2

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.