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

5.1 (a)
```
Y2

1   2   3
------------
1 | 1   2   1 | 4
|           |
Y1  2 | 2   2   0 | 4
|           |
3 | 1   0   0 | 1
|           |
---------------
4   4   1 | 9
```

(b) F(1,0) = P[Y1£ 1, Y2£ 0] = 5.2 (a)
```
Y2 = Side Bet

-1  1  2  3
--------------
0 | 1  0  0  0 | 1
|            |
1 | 0  1  1  1 | 3
Y1   |            |
2 | 0  2  1  0 | 3
|            |
3 | 0  1  0  0 | 1
|            |
----------------
1  4  2  1 | 8
```

(b) F(2,1) = P[Y1 £ 2, Y2 = 1] = 5.3 The Sample Space has elements. Ranges:
Y1 = 0, 1, 2, 3 Number of married executives in the sample
Y2 = 0, 1, 2, 3 Number of never married executives in the sample

To get the joint distribution, f(y1, y2), you have to do a series of "committee selection" problems like those we did in class. For example:

P(Y1=2, Y2=0) = f(y1=2, y2=0) = and so on.

This produces:
```
Y2

0  1  2  3
--------------
0 | 0  3  6  1 | 10
|            |
1 | 4 24 12  0 | 40
Y1   |            |
2 |12 18  0  0 | 30
|            |
3 | 4  0  0  0 |  4
|            |
----------------
20 45 18  1 | 84
```

5.4 (a) = [k(y22)/4]|01 = k/4
Hence, k = 4

(b) F(y1,y2) = = Hence: F(y1,y2) = (c) = 5.13 (a) Using results in 5.1

f1(y1) = (b) No. Note f(y1) = Which yields: f(0) = , f(1) = , f(2) = 5.14 (a) Using the results in 5.2

f2(y2) = (b) P[Y1=3|Y2=1] = 5.15 (a) Using results from 5.3
```
{ 10/84  y1 = 0
{
{ 40/84  y1 = 1
{
f(y1) = { 30/84  y1 = 2
{
{  4/84  y1 = 3
{
{ 0 otherwise

```

(b) P(Y1 = 1 | Y2 = 2) = P(Y1 = 1 Ç Y2 = 2)/ P(Y2 = 2) = (12/84)/(18/84)
Using the table shown in 5.3.

(c) P(Y3 = 1 | Y2 = 1) = P(Y3 = 1 Ç Y2 = 1)/ P(Y2 = 1) = (24/84)/(45/84)
Where
```
æ4öæ2öæ3ö
ç ÷ç ÷ç ÷
è1øè1øè1ø
P(Y3 = 1 Ç Y2 = 1) =  --------
æ9ö
ç ÷
è3ø
```
(d) The probabilities are identical.

5.31 Y1 and Y2 are not independent. From the table shown in 5.3 above, it is obvious that:

f(0, 0) ¹ f1(0)f2(0)

5.44 (a) Use E[Y1] = np = 2* (b) Use VAR[Y1] = np(1-p) = (c) E[Y1 - Y2] = E[Y1] - E[Y2] = 0

Because both distributions are identical.

5.45 Using f(y1) from 5.15a:

E(Y1) = 0*(10/84) + 1*(40/84) + 2*(30/84) + 3*(4/84) = 112/84 = 4/3