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