45-733 Probability and Statistics I
(3rd Mini AY 1999-2000 Flex-Mode and Flex-Time)

Assignment #4 Answers

**5.1 (a)**

**
Y**_{2}
1 2 3
------------
1 | 1 2 1 | 4
| |
Y_{1} 2 | 2 2 0 | 4
| |
3 | 1 0 0 | 1
| |
---------------
4 4 1 | 9

**(b)** **F(1,0) = P[Y**_{1}£
1, Y_{2}£
0] =

**5.2 (a)**
**
Y**_{2} = Side Bet
-1 1 2 3
--------------
0 | 1 0 0 0 | 1
| |
1 | 0 1 1 1 | 3
Y_{1} | |
2 | 0 2 1 0 | 3
| |
3 | 0 1 0 0 | 1
| |
----------------
1 4 2 1 | 8

**(b)** **F(2,1) = P[Y**_{1} £
2, Y_{2} =
1] =

**5.3 ** The Sample Space has elements. Ranges:

**Y**_{1} = 0, 1, 2, 3 Number of married executives in the sample

**Y**_{2} = 0, 1, 2, 3 Number of never married executives in the sample

To get the joint distribution, **f(y**_{1}, y_{2}), you have to do a series of
"committee selection" problems like those we did in class. For example:

**P(Y**_{1}=2, Y_{2}=0) = f(y_{1}=2, y_{2}=0) =
and so on.

This produces:
**
Y**_{2}
0 1 2 3
--------------
0 | 0 3 6 1 | 10
| |
1 | 4 24 12 0 | 40
Y_{1} | |
2 |12 18 0 0 | 30
| |
3 | 4 0 0 0 | 4
| |
----------------
20 45 18 1 | 84

**5.4 (a)** =
**[k(y**_{2}^{2})/4]|_{0}^{1} = k/4

Hence, **k = 4**

**(b)** **F(y**_{1},y_{2}) =
=

Hence:
**F(y**_{1},y_{2}) =

**(c)**
=

**5.13 (a)** Using results in 5.1

**f**_{1}(y_{1}) =

**(b)** No. Note **f(y**_{1}) =

Which yields: **f(0)** =
,
**f(1)** = ,
**f(2)** =

**5.14 (a)** Using the results in 5.2

**f**_{2}(y_{2}) =

**(b)** **P[Y**_{1}=3|Y_{2}=1] =

**5.15 (a)** Using results from 5.3

**
{ 10/84 y**_{1} = 0
{
{ 40/84 y_{1} = 1
{
f(y_{1}) = { 30/84 y_{1} = 2
{
{ 4/84 y_{1} = 3
{
{ 0 otherwise

**(b)** **P(Y**_{1} = 1 | Y_{2} = 2) =
P(Y_{1} = 1 Ç Y_{2} = 2)/
P(Y_{2} = 2) = (12/84)/(18/84)

Using the table shown in 5.3.

**(c)** **P(Y**_{3} = 1 | Y_{2} = 1) =
P(Y_{3} = 1 Ç Y_{2} = 1)/
P(Y_{2} = 1) = (24/84)/(45/84)

Where

**
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P(Y**_{3} = 1 Ç Y_{2} = 1) = --------
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**(d)** The probabilities are identical.

**5.31** **Y**_{1} and **Y**_{2} are not
independent. From the table shown in 5.3 above, it is obvious that:

**f(0, 0) ¹ f**_{1}(0)f_{2}(0)

**5.44 (a)** Use **E[Y**_{1}] = np = 2*

**(b)** Use **VAR[Y**_{1}] = np(1-p) =

**(c)** **E[Y**_{1} - Y_{2}] =
E[Y_{1}] - E[Y_{2}] = 0

Because both distributions are identical.

**5.45** Using **f(y**_{1}) from 5.15a:

**E(Y**_{1}) = 0*(10/84) + 1*(40/84) + 2*(30/84) + 3*(4/84) =
112/84 = 4/3