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

Assignment #4 Answers
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