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Math 183
Fall 2006
Quiz 1 Solutions
1a. If we look at the wheel, we see that
P [Z = 1] = P [Z = 2] = P [Z = 3] =
So
E[Z] = 1 ·
1b. It is a theorem was proved in class that V ar[Z] = E[Z^2 ] − (E[Z])^2. We have
E[Z^2 ] = 1^2 ·
+ 2^2 ·
+ 3^2 ·
So
V ar[Z] =
1c. There are two ways to do this problem. Since E[X + Y ] = E[X] + E[Y ], we can compute these values separately, and then add them together. We can also compute this value directly from the definition, which I do here; starting from the top of the wheel and moving clockwise:
E[X + Y ] = ((−3) + (−4))
1d. Because X and Z are independent, we have E[X × Z] = E[X] × E[Z]. So
we compute the value of E[X].
E[X] = (−3) ·
So
E[X × Z] = E[X]E[Z] =
1e. One random variable is dependent on another if the value of the first can always be determined from the value of the second. In the given wheel, if Z = 1, Y could be (−4), 1 or 4, so Y can not be dependent on Z. Also, if Z = 1, X could be (−3), 2 or 5, so X is not dependent on Z. Similarly, if X = −3, there are three possible values of Y and three possible values for Z, so neither Y nor Z depends on X. However, if Y = −4, we must have X = −3 and Z = 1. A similar statement can be made for any value of Y –the value of Y always determines the value of X and Z. So X and Z are both dependent on Y.
1f. A pair of random variables is independent if knowing the value of one of them does not affect the probabilities of the other. Since X and Z are dependent on Y , the pairs (X, Y ), (Y, Z) cannot be independent. To check the pair (X, Z), we first recall that P [Z = 1] = P [Z = 2] = P [Z = 3] = 13. Then we check that if X = −3, we have P [Z = 1] = P [Z = 2] = P [Z = 3] = 13. The same is true if X = 2 or if X = 5, so the pair (X, Z) is independent.
1g. From the definition of conditional probability we have
P [Y = 4 | Z = 1] =
P [Y = 4 and Z = 1] P [Z = 1]
=
1 12 1 3 =
1h. From the definition of conditional expectation we have
E[Y | Z = 1] = (−4) · P [Y = − 4 | Z = 1] + 1 · P [Y = 1 | Z = 1] + 4 · P [Y = 4 | Z = 1]
= (−4) ·
- In order to select a red ball from urn B, we could select a red ball from A and then a red ball from B, or we could select a white ball from A and then a red
