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

3.2 The sample space is:

S = {HH, HT, TH, TT}

and the corresponding values of Y are 2, -1, -1, and 1, respectively.

Now, since each outcome in S is equally likely

```                            { 1/2  y = -1
{
{ 1/4  y =  1
f(y) = {
{ 1/4  y =  2
{
{  0 otherwise
```

3.3 Since there are only 4 components, clearly Y = 2, 3, 4. Y = 2 occurs only if the first and second components tested are both defective. Let D indicate "Defective" and G indicate "Good". Hence:

P(Y = 2) = P(D Ç D) = (2/4)(1/3) = 1/6
P(Y = 3) = P(D Ç G Ç D) + P(G Ç D Ç D) = 2(2/4)(2/3)(1/2) = 2/6
P(Y = 4) = P(G Ç G Ç D Ç D) + P(D Ç G Ç G Ç D) + P(G Ç D Ç G Ç D) =
3(2/4)(1/3)(2/2) = 1/2

Hence, the distribution of Y is
```                            { 1/6  y = 2
{
{ 2/6  y = 3
f(y) = {
{ 3/6  y = 4
{
{  0 otherwise
```

3.5 Assume that the correct ordering of the animal words is ABC. If the child is guessing, there are 6 equally likely permutations she could choose, with Y = number of matches associated with each permutation". Therefore
```
A    B    C    Y
---------------------
A    B    C    3
A    C    B    1
B    A    C    1
B    C    A    0
C    A    B    0
C    B    A    1
```
Hence, P(0) = 2/6 = 1/3, P(1) = 3/6 = 1/2,
P(2) = 0, and P(3) = 1/6. The probability function is:
```                            { 1/3  y = 0
{
{ 1/2  y = 1
{
f(y) = {  0   y = 2
{
{ 1/6  y = 3
{
{  0 otherwise
```

3.17 Let X1 = # of contracts assigned to Firm 1; X2 = # of contracts assigned to Firm 2. There are 9 possible ways the contracts can be assigned to the 3 firms, each with a probability of 1/9. Considering the probability distributions for X1 and X2, we have,
```              { 4/9  x1 = 0              { 4/9  x2 = 0
{                          {
{ 4/9  x1 = 1              { 4/9  x2 = 1
f(x1) = {                  f(x2) = {
{ 1/9  x1 = 2              { 1/9  x2 = 2
{                          {
{  0 otherwise             {  0 otherwise
```

Hence, E(X1) = E(X2) = 0(4/9) + 1(4/9) + 2(1/9) = 6/9 = 2/3

Let Y = profit for Firm 1 = 90,000X1. Therefore, the expected profit is:

E[profit for Firm 1] = E(Y) = E[90,000X1] = 90,000E(X1) = 90,000(2/3) = 60,000

Now let W = profit for both Firm 1 and 2 = 90,000(X1 + X2). Thus the expected profit is:

E(W) = E[90,000(X1 + X2)] = 90,000E(X1 + X2) = 90,000[E(X1) + E(X2)] =
90,000[2/3 + 2/3] = 120,000

3.19 Let Y = "insurance company's loss in dollars". The probability distribution given is:
```                            { .001  y = 85,000
{
{ .01   y = 42,500
f(y) = {
{ .989  y = 0
{
{  0 otherwise
```

E(Loss) = E(Y) = .001*85,000 + .01*42,500 + .989*0 = 510

Hence, the insurance company should charge a premimum of \$510.00 for it to break exactly even.

3.22 Given

f(y) = Let X = Daily Cost = 10Y, then

E(Y) = 0*0.1 + 1*0.5 + 2*0.4 = 1.3
E(Y2) = 0*0.1 + 1*0.5 + 4*0.4 = 2.1
VAR(Y) = 2.1 - 1.32
Therefore: E(X)= E(10Y) = 10E(Y) = 10*(0*0.1 + 1*0.5 + 2*0.4) = 10*1.3 = \$13

and VAR(X) = VAR(10Y) = 100VAR(Y) = 100*.41 = 41

4.3 Given

f(y) = 1.  2. F(y) = P(Y<y) = Therefore:

F(y) = 3. 4.        P[1 £ Y £ 2] = F(2) - F(1) = 1 - 5. Area of triangle = * base * height = 1 - 4.6
1. Graph f(y) 2. For y < 0, F(y) = 0
For y > 2, F(y) = 2
For 0 < y < 1       F(y) = ò0ytdt = y2/2
For 1 £ y £ 2        F(y) = ò01tdt + ò1y(2 - t)dt = 1/2 + [2t - t2/2 |1y = 2y - y2/2 - 1
For y > 1 F(y) = 1

3. P[0.8£ Y£ 1.2] = F(1.2) - F(0.8) = (2.4 - 0.72 - 1) = 0.36

4. P(Y > 1.5 | Y > 1) = P(Y > 1.5)/P(Y > 1) = [1 - (3 - 1.125 -1)]/(1/2) = .25
4.8
1. Find c. Þ c = 2. F(y) = P(Y £ y) = Hence:

F(y) = 3. Functions" 4. P(1 £ Y £ 2) = F(2) - F(1) = 1 - 5. Area of triangle = * base * height = 4.12 f(y) = E[Y] = = E[Y2] = = VAR[Y] = 