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45733 Probability and Statistics I
(3rd Mini AY 19992000 FlexMode and FlexTime)
Assignment#3 Answers
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 X_{1 } = # of contracts assigned to Firm 1;
X_{2 } = # 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
X_{1} and X_{2}, we have,
{ 4/9 x_{1} = 0 { 4/9 x_{2} = 0
{ {
{ 4/9 x_{1} = 1 { 4/9 x_{2} = 1
f(x_{1}) = { f(x_{2}) = {
{ 1/9 x_{1} = 2 { 1/9 x_{2} = 2
{ {
{ 0 otherwise { 0 otherwise
Hence, E(X_{1}) = E(X_{2}) =
0(4/9) + 1(4/9) + 2(1/9) = 6/9 = 2/3
Let Y = profit for Firm 1 = 90,000X_{1}. Therefore,
the expected profit is:
E[profit for Firm 1] = E(Y) = E[90,000X_{1}] =
90,000E(X_{1}) = 90,000(2/3) = 60,000
Now let W = profit for both Firm 1 and 2 =
90,000(X_{1} + X_{2}). Thus the expected profit is:
E(W) = E[90,000(X_{1} + X_{2})] =
90,000E(X_{1} + X_{2}) =
90,000[E(X_{1}) + E(X_{2})] =
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(Y^{2}) = 0*0.1 + 1*0.5 + 4*0.4 = 2.1
VAR(Y) = 2.1  1.3^{2}
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) =
F(y) = P(Y<y) =
Therefore:
F(y) =


P[1 £
Y £ 2] = F(2)  F(1) =
1 
 Area of triangle = * base * height =
1 
4.6
 Graph f(y)
 For y < 0, F(y) = 0
For y > 2, F(y) = 2
For 0 < y < 1 F(y) =
ò_{0}^{y}tdt = y^{2}/2
For 1 £ y £ 2
F(y) =
ò_{0}^{1}tdt +
ò_{1}^{y}(2  t)dt =
1/2 + [2t  t^{2}/2 _{1}^{y} = 2y  y^{2}/2  1
For y > 1 F(y) = 1
 P[0.8£
Y£
1.2] = F(1.2)  F(0.8) = (2.4  0.72  1) = 0.36
 P(Y > 1.5  Y > 1) = P(Y > 1.5)/P(Y > 1) =
[1  (3  1.125 1)]/(1/2) = .25
4.8
 Find c.
Þ
c =
 F(y) = P(Y £ y) =
Hence:
F(y) =
 Functions"
 P(1 £ Y
£ 2) = F(2)  F(1) =
1 
 Area of triangle = * base * height =
4.12 f(y) =
E[Y] =
=
E[Y^{2}] =
=
VAR[Y] =