Section 4.5 Solutions
Important Note: These
solutions do not include tree diagrams, but trees can be a very important part
of the solution process.
4.86
A = {household is prosperous}
B = {household is educated}
We know P(A)
= 0.138, P(B) = 0.261, and P(A
and B) = 0.082. To determine the
union probability, simply apply the general addition rule:
P(A or B) = P(A) + P(B) – P(A and B) = 0.138 + 0.261 – 0.082 = 0.317
4.87
From exercise 4.86,
A = {household is prosperous}
B = {household is educated}.
Furthermore, P(A)
= 0.138, P(B) = 0.261, and P(A and B) = 0.082.
Then, 
.
Since this
conditional probability does not equal the unconditional probability of A, 0.138, the events A and B are not independent (i.e.,
they are dependent).
4.92
I’m not including
the Venn diagram (only because I can’t create it in Word and post it to the
Web). Please see me with questions about the diagram itself.
The probabilities
are
P(A only) = 0.45
P(A and B only) = 0.1
P(A and C only) = 0.05
P(B only) = 0.25
P(B and C only) = 0.05
P(C only) = 0.1
P(A and B and C) = 0
4.93
To answer this, add
up all the disjoint probabilities listed above:
P(A or B or C) = 0.45 + 0.1 +
0.05 + 0.25 + 0.05 + 0.1 = 1
(Note: You must answer this via the Venn
diagram, as we don’t have a general addition rule for more than two events –
such a rule exists, but it is more complicated than the two-event rule.)
4.94
From above,
P(A and B only) = 0.1
4.96
a. To determine the probability of choosing a
woman, simply divide the total number of women by the total number of people: P(woman) = (1119)/(1944) 
0.576.
b. From the table, P(woman and professional) = 39/1944, and P(professional) = 83/1944. Then P(woman
| professional) = 
. (Note this probability could have been found directly from
the table, by looking only at the professional column—which is what was given—and
then determining the probability of being a woman: 39/83.)
c. Note that P(woman)
= 0.576 
0.470 = P(woman |
professional). Therefore, by definition, the events woman and professional are
not independent—they are dependent (knowing a person is professional changes—decreases—the
probability the person is a woman).
4.102
Let E represent a student
studying education and W represent a
student who is a woman. Then P(E) = 0.15, P(W) = 0.6, and P(W|E) = 0.8. Hence, by the definition of
conditional probability and the general multiplication rule,
4.103
I’m not including
the tree diagram (only because I can’t create it in Word and post it to the
Web). Please see me with questions about the tree diagram itself.
Let NW represent a non-word
error and W represent a word error.
Furthermore, let C represent a caught
error and NC represent a non-caught
error.
Then, P(NW) = 0.25, P(W) = 0.75, P(C|NW) = 0.9, and P(C|W)
= 0.7.
Now, by the general
multiplication rule,
P(C) = P(C and NW) + P(C and W) = P(NW)P(C|NW)
+ P(W)P(C|W) = (0.25)(0.9) +
(0.75)(0.7) = 0.75.
4.105
I’m not including
the tree diagram (only because I can’t create it in Word and post it to the
Web). Please see me with questions about the tree diagram itself.
Let R represent regular gas, MG represent midgrade gas, and PR represent premium gas. Furthermore,
let $20 represent a person spending
at least $20.
Then, P(R)
= 0.4, P(MG) = 0.35, P(PR) = 0.25, P($20|R) = 0.3, P($20|MG)
= 0.5, and P($20|P) = 0.6.
Now, by the general
multiplication rule,
P($20) = P(R and $20) + P(MG and $20) + P(PR and $20) = P(R)P($20|R)
+ P(MG)P($20|MG) + P(PR)P($20|PR) = (0.4)(0.3) +
(0.35)(0.5) + (0.25)(0.6) = 0.445.
4.108
I’m not including
the tree diagram (only because I can’t create it in Word and post it to the
Web). Please see me with questions about the tree diagram itself.
Let B represent a bachelor’s
degree, M represent a master’s
degree, and D represent a doctoral
degree. Furthermore, let W represent
a woman and M represent a man.
Then, P(B) = 0.73, P(M) = 0.21, P(D) = 0.06, P(W|B) =
0.48, P(W|M) = 0.42, and P(W|D) = 0.29.
Now, by the general multiplication rule,
P(W and B) = P(B)P(W|B) = (0.73)(0.48) =
0.3504
and
P(W) = P(W and B) + P(W and M) + P(W and D) = P(B)P(W|B) +
P(M)P(W|M) + P(D)P(W|D) = (0.73)(0.48) + (0.21)(0.42) + (0.06)(0.29) = 0.456.
Hence, 
.
I’m not including
the tree diagram (only because I can’t create it in Word and post it to the
Web). Please see me with questions about the tree diagram itself.