Finite Mathematics
Chapter 8
Name ________________________________ Date ______________ Class ____________
Section 8-1 Sample Spaces, Events, & Probability
Goal: To find sample spaces for multiple experiments. To find events and probabilities for
the sample spaces
Empirical Probability:
P( E ) ≈
frequency of occurance of E f ( E )
=
total number of trials
n
Theoretical Probability:
P( E ) =
number of elements in E n( E )
=
number of elements in S n( S )
An experiment consists of drawing a card from a standard 52-card deck. In problems 1–6
what is the probability of drawing
1. a black card?
P(Black) =
2. the six of clubs?
26 1
=
52 2
P(6 of clubs) =
3. a black 6?
P(Black 6) =
1
52
4. a six?
2
1
=
52 26
P(6) =
5. a six of clubs or a six of diamonds?
P(6 of clubs or dia.) =
4
1
=
52 13
6. a card that is not a black six?
2
1
=
52 26
P(Not black 6) =
8-1
50 25
=
52 26
Finite Mathematics
Chapter 8
An experiment consists of dealing 4 cards from a standard 52-card deck. In problems 7–12
what is the probability of being dealt a hand with
Denominator for problems 7–12 is:
7. 4 aces?
Num: C4,4 =
C52,4 =
52!
= 270, 725
4!(52 − 4)!
8. 3 aces and 1 king?
4!
=1
4!(4 − 4)!
Num: C4,3 ⋅ C4,1 =
4!
4!
⋅
3!(4 − 3)! 1!(4 − 1)!
= 4 ⋅ 4 = 16
P(4 aces) =
1
= 0.0000037
270, 725
9. 2 aces and 2 kings?
Num: C4,2 ⋅ C4,2 =
4!
4!
⋅
2!( 4 − 2)! 2!(4 − 2)!
P(3 aces and 1 king) =
16
= 0.000059
270, 725
10. 1 ace and 3 kings?
Num: C4,1 ⋅ C4,3 =
4!
4!
⋅
1!( 4 − 1)! 3!( 4 − 3)!
= 6 ⋅ 6 = 36
= 4 ⋅ 4 = 16
36
16
P(2 aces and 2 kings) =
= 0.00013 P(1 ace and 3 kings) =
= 0.000059
270, 725
270, 725
11. no aces?
12. no face cards?
48!
= 194,580
4!( 48 − 4)!
194,580
P(no aces) =
= 0.7187
270, 725
Num: C40,4 =
Num: C48,4 =
40!
= 91,390
4!( 40 − 4)!
91,390
P(no face cards) =
= 0.3376
270, 725
8-2
Finite Mathematics
Chapter 8
An experiment consists of rolling two fair dice and adding the dots on the two sides facing
up. Assuming each simple event is equally likely, find the probability of the sum of the dots
indicated in problems 13–18.
13. Sum is 12
P(12) =
14. Sum is 7
1
36
P (7) =
15. Sum is not 7.
6 1
=
36 6
16. Sum is divisible by 5.
P(not 7) =
30 5
=
36 6
P(divisible by 5)
P(5 or 10) =
17. The sum is odd.
P(odd) =
18 1
=
36 2
7
36
18. Sum is greater than 1 and less than 13.
P(> 1 or <13) =
8-3
36
=1
36
Finite Mathematics
Chapter 8
19. An English 101 class at a junior college must select 3 students to serve on an advisory
committee to the dean. There are 14 freshmen and 12 sophomores in the class.
a.
i. How many 3-person committees can be chosen from the class?
26!
C26,3 =
= 2600
3!( 26 − 3)!
ii. How many 3-person committees can be chosen from the class if the committee
consists of only freshmen?
14!
C14,3 =
= 364
3!(14 − 3)!
iii. How many 3-person committees can be chosen from the class if the committee
consists of only sophomores?
12!
C12,3 =
= 220
3!(12 − 3)!
iv. How many 3-person committees can be chosen from the class if the committee
consists of 2 freshmen and 1 sophomore?
14!
12!
C14,2 ⋅ C12,1 =
⋅
2!(14 − 2)! 1!(12 − 1)!
= 91⋅12 = 1092
b. What is the probability that the committee will consists of
i. all freshmen?
Using parts i and ii:
P(all fresh.)=
364
= 0.14
2600
P(all soph.)=
220
= 0.085
2600
ii. all sophomores?
Using parts i and iii:
iii. 2 freshmen and 1 sophomore?
Using parts i and iii:
P(2 fresh. and 1 soph.)=
1092
= 0.42
2600
iv. 1 freshman and 2 sophomores?
14!
12!
Num: C14,1 ⋅ C12,2 =
⋅
1!(14 − 1)! 2!(12 − 2)!
= 14 ⋅ 66 = 924
924
P(1 fresh. and 2 soph.)=
= 0.355
2600
v. Find the sum of the probabilities in parts (i)–(iv) and write an explanation for the
sum.
The sum is 1 because the four parts consist of all possible outcomes for the
committee.
8-4
Finite Mathematics
Chapter 8
20. A computer store has 10 copies of a software package. Unknown to the store, 3 of the
packages have corrupted disks. What is the probability that
Denominator for parts c–f is: C10,3 =
10!
= 120
3!(10 − 3)!
a. a customer buys one package, that it will be one of the packages that contains a
corrupted disk?
3
P(corrupt) =
10
b. a customer buys one package, that it will be one of the packages that is not
corrupted?
7
P(not corrupt) =
10
c. a customer buys three packages and all are corrupted.
3!
1
= 1 P(all corrupt) =
= 0.0083
Num: C3,3 =
3!(3 − 3)!
120
d. a customer buys three packages and two of them are corrupted.
Num: C3,2 ⋅ C7,1 =
3!
7!
⋅
2!(3 − 2)! 1!(7 − 1)!
P(2 corrupt) =
21
= 0.175
120
= 3 ⋅ 7 = 21
e. a customer buys three packages and one of them is corrupted.
Num: C3,1 ⋅ C7,2 =
3!
7!
⋅
1!(3 − 1)! 2!(7 − 2)!
P(1 corrupt) =
63
= 0.525
120
= 3 ⋅ 21 = 63
f. a customer buys three packages and none of them are corrupted.
Num: C3,0 ⋅ C7,3 =
3!
7!
⋅
0!(3 − 0)! 3!(7 − 3)!
P(0 corrupt) =
35
= 0.2917
120
= 1⋅ 35 = 35
g. Find the sum of the probabilities in parts c–f and write an explanation for the sum.
The sum is 1 because the four parts consist of all possible outcomes for the number of
corrupt packages.
8-5
Finite Mathematics
Chapter 8
8-6
Finite Mathematics
Chapter 8
Name ________________________________ Date ______________ Class ____________
Section 8-2 Unions, Intersections, and
Complement Events; Odds
Goal: To find probabilities of compound events. To find odds.
Theorem: Probability of a Union of Two Events
For any two events:
P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
For two mutually exclusive events:
P ( A ∪ B ) = P ( A) + P ( B )
Complement:
P ( E ) = 1 − P ( E ′ ) or P ( E ′ ) = 1 − P ( E )
Odds: In favor of event E:
P( E )
P( E )
=
1 − P( E ) P( E ′)
Against event E:
1 − P( E ) P( E ′)
=
P( E )
P( E )
1. A container contains 5 red balls, 3 green balls, and 2 white balls. A single ball is drawn at
random from the container. Find the probability that
a) the ball is red.
P(R) =
b) the ball is green or white.
5
= 0.5
10
c) the ball is red or green.
P( R or G) =
8
= 0.8
10
e) the ball is not green.
P(not G) =
7
= 0.7
10
P(G or W) =
5
= 0.5
10
d) the ball is not white.
P(not W) =
8
= 0.8
10
f) the ball is not red.
P(not R) =
8-7
5
= 0.5
10
Finite Mathematics
Chapter 8
2. An experiment consists of rolling two fair dice and adding the dots on the two sides facing
up. Compute the probability of obtaining the following:
a) The sum is 12.
P (12) =
b) The sum is 8.
1
36
P (8) =
c) The sum is not 12.
P (not 12) =
d) The sum is not 8.
35
36
e) The sum is 8 or 12.
P (not 8) =
31
36
f) The sum is 8 and 12.
P(8 or 12) = P (8) + P (12)
5
1
=
+
36 36
6 1
=
=
36 6
g) The sum is divisible by 2 or 3.
5
36
P (8 and 12) = 0
h) The sum is divisible by 2 and 3.
P (div. 2 and 3) = P (6,12)
P(div. 2 or 3)
P(div. 2) + P (div. 3) − P(div. 2 and 3)
P(even) + P(3, 6,9,12) − P(6,12)
18 12 6
=
+ −
36 36 36
24 2
=
=
36 3
=
8-8
6 1
=
36 6
Finite Mathematics
Chapter 8
3. A survey was taken at a local college campus. The survey asked if the student was a fulltime student or a part-time student and if he or she worked full time or part time. The
following chart illustrates the (empirical) results.
Did not work
Full-time student
Part-time student
Total
0.4
0.05
0.45
Worked part
time
0.15
0.2
0.35
Worked full
time
0.05
0.15
0.20
If a student is picked at random what is the probability that
a) he or she is a full-time student or works part time?
P (FTS) + P (PTW) − P (FTS and PTW)
0.60 + 0.35 − 0.15 = 0.8
b) he or she is a full-time student and works part time?
P (FTS and PTW) =0.15
c) he or she is a part-time student or does not work?
P (PTS) + P (No W) − P (PTS and No W)
0.40 + 0.45 − 0.05 = 0.8
d) he or she is a part-time student and does not work?
P (PTS and No W) =0.05
e) he or she is a full-time student and does not work?
P (FTS and No W) =0.4
f) he or she is a part-time student or works full time?
P (PTS) + P (FTW) − P (PTS and FTW)
0.40 + 0.20 − 0.15 = 0.45
8-9
Total
0.60
0.40
1.00
Finite Mathematics
Chapter 8
4. A class is made up of 4 male and 5 female seniors, 3 male and 6 female juniors, and 2
male and 5 female sophomores. Use the following chart.
Male
Female
Total
Senior
4
5
9
Junior
3
6
9
Sophomore
2
5
7
Total
9
16
25
If one person from the class is picked at random to serve on an advisory committee, what is
the probability of:
a) choosing a male or a junior?
P (M) + P (Jr.) − P (M and Jr.)
9
9
3 15
+ −
=
= 0.6
25 25 25 25
b) choosing a female sophomore?
P (F and Soph) =
5
= 0.2
25
c) not choosing a senior?
P (not Sr) =
16
= 0.64
25
d) not choosing a male?
P (not M) =
16
= 0.64
25
e) choosing a female or a sophomore?
P (F) + P (Soph.) − P(F and Soph.)
16 7
5 18
+ −
=
= 0.72
25 25 25 25
f) choosing a senior or a female?
P (Sr.) + P(F) − P (Sr. and F)
9 16 5 20
+ −
=
= 0.8
25 25 25 25
8-10
Finite Mathematics
Chapter 8
5. A candidate for mayor of a small town predicts that he has a 70% chance of winning the
election.
a) what is the probability that he will lose the election?
P (losing) = 100% − P (winning)
= 100% − 70%
= 30%
b) what are the odds that he will win the election?
Odds Win =
P(win) 70% 7
=
=
P(lose) 30% 3
c) what are the odds that he will lose the election?
Odds Lose =
P(lose) 30% 3
=
=
P(win) 70% 7
8-11
Finite Mathematics
Chapter 8
8-12
Finite Mathematics
Chapter 8
Name ________________________________ Date ______________ Class ____________
Section 8-3 Conditional Probability
Goal: To find conditional probabilities using contingency tables
P( A ∩ B)
P( B ∩ A)
or P( B | A) =
P ( B)
P ( A)
Conditional Probability:
P( A | B) =
Product Rule:
P ( A ∩ B ) = P ( A) P ( B | A) = P ( B ) P ( A | B )
Two events are independent if and only if P ( A ∩ B ) = P ( A) P ( B )
Given the probabilities in the table below for events in a sample space S, find the
probabilities in 1 – 9 relative to the probabilities in the table.
E
F
TOTALS
A
0.05
0.02
0.07
B
0.1
0.08
0.18
C
0.30
0.06
0.36
D
0.35
0.04
0.39
1. P (C)
2. P(F)
3. P ( A ∩ F )
P (C ) = 0.36
P ( F ) = 0.15
P ( A ∩ F ) = 0.02
4. P (C ∩ E )
5. P ( A F )
6. P ( F A)
P (C ∩ E ) = 0.30
P( A | F ) =
P( F | A) =
7. P ( D E )
8. P ( E B )
9. P (C C )
P( D ∩ E )
P( E )
0.35
P( D | E ) =
= 0.412
0.85
P( D | E ) =
P( A ∩ F )
P( F )
0.02
P( A | F ) =
= 0.133
0.15
P( E ∩ B)
P( B)
0.1
P( E | B) =
= 0.556
0.18
P( E | B) =
8-13
TOTALS
0.85
0.15
1.00
P( F ∩ A)
P( A)
0.02
P( F | A) =
= 0.286
0.07
P(C ∩ C )
P(C )
0.36
P(C | C ) =
=1
0.36
P(C | C ) =
Finite Mathematics
Chapter 8
Refer to the table above to answer 10 and 11.
10. Are events C and E independent? Why or why not?
If events are independent, the outcome of the first will not affect the outcome of the
second. Since the outcome of the first could affect the outcome of the second, the events are
not independent.
11. Are events B and F independent? Why or why not?
If events are independent, the outcome of the first will not affect the outcome of the
second. Since the outcome of the first could affect the outcome of the second, the events are
not independent.
12. Two balls are drawn from a container that holds 3 red balls, 2 green balls and 5 white
balls.
A) If the first ball drawn is replaced before the second ball is drawn then find the
following probabilities:
a) that both balls are green.
P (2G ) = P (G ) P (G )
=
2 2
4
⋅ =
= 0.04
10 10 100
b) that neither ball is green.
P (No G ) = P(No G ) P (No G )
=
8 8
64
⋅ =
= 0.64
10 10 100
c) that at least one ball is green.
P (At least 1 G) = 1 − P (No G)
= 1 − 0.64
= 0.36
d) that the two balls are not the same color.
P(Diff. Color) = P ( R) P(No R) + P(G ) P(No G) + P(W ) P(No W)
3 7 2 8 5 5
= ⋅ + ⋅ + ⋅
10 10 10 10 10 10
21 16
25
62
=
+
+
=
= 0.62
100 100 100 100
8-14
Finite Mathematics
Chapter 8
B) If the first ball drawn is not replaced before the second ball is drawn then find the
following probabilities:
a) that both balls are green.
P (2G ) = P (G ) P (G )
=
2 1 2
⋅ =
= 0.022
10 9 90
b) that neither ball is green.
P (No G ) = P(No G ) P (No G )
=
8 7 56
⋅ =
= 0.622
10 9 90
c) that at least one ball is green.
P (At least 1 G) = 1 − P (No G)
= 1 − 0.622
= 0.378
d) that the two balls are not the same color.
P(Diff. Color) = P ( R) P(No R) + P(G ) P(No G) + P(W ) P(No W)
3 7 2 8 5 5
= ⋅ + ⋅ + ⋅
10 9 10 9 10 9
21 16 25 62
=
+ +
=
= 0.689
90 90 90 90
8-15
Finite Mathematics
Chapter 8
13. A certain state allows a person three tries to pass the written portion of the test to obtain
a driver’s license. If a person fails the test, he or she may retake the test a week later. If a
person fails the written portion 3 times, he or she must wait 1 year from the date of the third
try before he or she can try again. From past records 60% pass the test on their first attempt,
20% then pass the test on the second attempt, and only 5% pass the test on their third attempt.
For a person taking the test:
a) What is the probability of passing the written portion of the driving test on the first
or second attempt?
P (Pass 1st or 2nd) = P (Pass) + P (Fail)P (Pass)
= 0.60 + (0.40)(0.20)
= 0.60 + 0.08 = 0.68
b) What is the probability of failing the test on the first and second attempt but
passing on the third attempt?
P (Fail, Fail, Pass) = P (Fail)P (Fail)P (Pass)
=(0.40)(0.80)(0.05)
=0.016
c) What is the probability that the person will have to wait a year (failing all three
attempts) before trying again?
P (Fail, Fail, Fail) = P (Fail)P(Fail)P(Fail)
=(0.40)(0.80)(0.95)
=0.304
8-16
Finite Mathematics
Chapter 8
14. In a study to determine the frequency and dependency of left-handedness relative to
females and males, 1,000 people were chosen at random and the following results recorded:
LEFT-HANDED L
RIGHT-HANDED L′
TOTALS
FEMALE
F
52
388
440
MALE
F′
78
482
560
TOTALS
130
870
1000
a) Convert this table to a probability table by dividing each entry by 1000.
LEFT-HANDED L
RIGHT-HANDED L′
TOTALS
FEMALE
F
0.052
0.388
0.440
MALE
F′
0.078
0.482
0.560
TOTALS
0.130
0.870
1.000
b) What is the probability that a person chosen at random is left-handed?
P ( LH ) = 0.130
c) What is the probability that a person chosen at random is a left-handed woman?
P (LH and F) = 0.052
d) What is the probability that a person is a woman given that the person is lefthanded?
P( F ∩ LH )
P( LH )
0.052
P( F | LH ) =
= 0.4
0.130
P( F | LH ) =
e) What is the probability that a person is left-handed given that the person is a male?
P( LH ∩ M )
P(M )
0.078
P( LH | M ) =
= 0.139
0.560
P( LH | M ) =
f) Are the events left-handedness and male independent? Why or why not?
Since the value of the conditional probability in part e is not the same as the
value of the probability of left-handed, the events are not independent.
8-17
Finite Mathematics
Chapter 8
8-18
Finite Mathematics
Chapter 8
Name ________________________________ Date ______________ Class ____________
Section 8-4 Bayes’ Theorem
Goal: To find probabilities using Bayes’ Theorem
Bayes’ Formula:
P(U1 | E ) =
P(U1 ∩ E )
P(U1 ∩ E )
=
P( E )
P (U1 ∩ E ) + P(U 2 ∩ E ) + " P(U n ∩ E )
Similar results will hold for U 2 ,U 3 ,",U n
Find the probabilities in problems 1–9 by referring to the following tree diagram and using
Bayes’ formula.
0.2
D
0.6
A
E
0.4
0.2
F
0.1
D
0.5
Start
B
0.1
C
0.5
0.4
E
F
0.3
D
0.6
E
0.1
8-19
F
Finite Mathematics
Chapter 8
1. P ( B F )
2. P ( A E )
P( B ∩ F )
P( A ∩ E )
P( A | E ) =
P( A ∩ F ) + P( B ∩ F ) + P(C ∩ F )
P( A ∩ E ) + P( B ∩ E ) + P (C ∩ E )
(0.5)(0.4)
(0.4)(0.6)
=
=
(0.4)(0.2) + (0.5)(0.4) + (0.1)(0.1)
(0.4)(0.6) + (0.5)(0.5) + (0.1)(0.6)
0.20
0.24
=
=
0.08 + 0.20 + 0.01
0.24 + 0.25 + 0.06
0.20
0.24
=
= 0.69
=
= 0.44
0.29
0.55
P( B | F ) =
4. P( B E )
3. P(C D)
P(C ∩ D)
P( B ∩ E )
P( B | E ) =
P( A ∩ D ) + P( B ∩ D) + P(C ∩ D)
P( A ∩ E ) + P( B ∩ E ) + P (C ∩ E )
(0.1)(0.3)
(0.5)(0.5)
=
=
(0.4)(0.2) + (0.5)(0.1) + (0.1)(0.3)
(0.4)(0.6) + (0.5)(0.5) + (0.1)(0.6)
0.03
0.25
=
=
0.08 + 0.05 + 0.03
0.24 + 0.25 + 0.06
0.03
0.25
=
= 0.1875
=
= 0.45
0.16
0.55
P(C | D) =
6. P(C E )
5. P( A D )
P( A ∩ D)
P(C ∩ E )
P(C | E ) =
P( A ∩ D) + P( B ∩ D) + P(C ∩ D)
P( A ∩ E ) + P( B ∩ E ) + P(C ∩ E )
(0.4)(0.2)
(0.1)(0.6)
=
=
(0.4)(0.2) + (0.5)(0.1) + (0.1)(0.3)
(0.4)(0.6) + (0.5)(0.5) + (0.1)(0.6)
0.08
0.06
=
=
0.08 + 0.05 + 0.03
0.24 + 0.25 + 0.06
0.08
0.06
=
= 0.5
=
= 0.11
0.16
0.55
P( A | D) =
8-20
Finite Mathematics
Chapter 8
7. P( B D )
8. P( A F )
P( B ∩ D)
P( A ∩ F )
P( A | F ) =
P( A ∩ D) + P( B ∩ D) + P(C ∩ D)
P( A ∩ F ) + P( B ∩ F ) + P(C ∩ F )
(0.5)(0.1)
(0.4)(0.2)
=
=
(0.4)(0.2) + (0.5)(0.1) + (0.1)(0.3)
(0.4)(0.2) + (0.5)(0.4) + (0.1)(0.1)
0.05
0.08
=
=
0.08 + 0.05 + 0.03
0.08 + 0.20 + 0.01
0.05
0.08
=
= 0.3125
=
= 0.28
0.16
0.29
P( B | D) =
9. P(C F )
P(C ∩ F )
P( A ∩ F ) + P( B ∩ F ) + P(C ∩ F )
(0.1)(0.1)
=
(0.4)(0.2) + (0.5)(0.4) + (0.1)(0.1)
0.01
=
0.08 + 0.20 + 0.01
0.01
=
= 0.034
0.29
P(C | F ) =
For problems 10–13, a container has 3 red balls, 2 black balls, and 5 white balls. Two balls
are drawn in succession with replacement.
10. If the second ball is white, what is the probability that the first ball was black?
P( B ) =
2
= 0.2
10
11. If the second ball is black, what is the probability that the first ball was white?
P(W ) =
5
= 0.5
10
12. If the second ball is black, what is the probability that the first ball was red?
P( R ) =
3
= 0.3
10
8-21
Finite Mathematics
Chapter 8
13. If the second ball is red, what is the probability that the first ball was red?
P( R ) =
3
= 0.3
10
For problems 14–17, a container has 3 red balls, 2 black balls, and 5 white balls. Two balls
are drawn in succession without replacement.
14. If the second ball is white, what is the probability that the first ball was black?
Since the second ball is white, the number of black balls was not affected, P( B) =
2
= 0.22.
9
15. If the second ball is black, what is the probability that the first ball was white?
Since the second ball is black, the number of white balls was not affected, P(W ) =
5
= 0.56.
9
16. If the second ball is black, what is the probability that the first ball was red?
Since the second ball is black, the number of red balls was not affected, P( R) =
3
= 0.33.
9
17. If the second ball is red, what is the probability that the first ball was red?
Since the second ball is red, the number of red balls was affected, P( R) =
2
= 0.22.
9
18. A local university offers free flu shots to its employees. Of the eligible employees, 65%
have the flu shot and 35% do not. Of the employees that take the flu shot, 5% actually get
the flu. Of the employees that do not take the flu shot, 55% actually get the flu. What is the
probability that a person who gets the flu took the flu shot? What is the probability that a
person who does not get the flu did not have a flu shot?
P( FS ∩ F )
P( FS ∩ F ) + P( NFS ∩ F )
(0.65)(0.05)
=
(0.65)(0.05) + (0.35)(0.55)
0.0325
=
0.0325 + 0.1925
0.0325
=
= 0.14
0.225
P( FS | F ) =
P( NFS ∩ NF )
P( FS ∩ NF ) + P( NFS ∩ NF )
(0.35)(0.45)
=
(0.65)(0.95) + (0.35)(0.45)
0.1575
=
0.6175 + 0.1575
0.1575
=
= 0.20
0.775
P( NFS | NF ) =
8-22
Finite Mathematics
Chapter 8
19. An office supply company sells 3 different computer systems, A, B, and C. 55% of the
customers who buy a computer buy brand A, 20% buy brand B, and 25% buy brand C. At
the time of purchase the customer is offered an extended warranty for an added fee. 30% of
the customers who bought brand A, 25% of the customers who bought brand B, and 40% of
the customers who bought brand C bought the extended warranty. If a customer brings a
computer system in for repair under the extended warranty, what is the probability that the
computer was a brand A computer? A brand B computer? A brand C computer?
P( A ∩ W )
P( A ∩ W ) + P( B ∩ W ) + P(C ∩ W )
(0.55)(0.3)
=
(0.55)(0.3) + (0.2)(0.25) + (0.25)(0.4)
0.165
=
0.165 + 0.05 + 0.1
0.165
=
= 0.52
0.315
P( A | W ) =
P( B ∩ W )
P( A ∩ W ) + P( B ∩ W ) + P(C ∩ W )
(0.2)(0.25)
=
(0.55)(0.3) + (0.2)(0.25) + (0.25)(0.4)
0.050
=
0.165 + 0.05 + 0.1
0.050
=
= 0.16
0.315
P( B | W ) =
P(C ∩ W )
P( A ∩ W ) + P( B ∩ W ) + P(C ∩ W )
(0.25)(0.4)
=
(0.55)(0.3) + (0.2)(0.25) + (0.25)(0.4)
0.1
=
0.165 + 0.05 + 0.1
0.1
=
= 0.32
0.315
P(C | W ) =
8-23
Finite Mathematics
Chapter 8
20. A survey of registered voters asked the question “Do you support legislation to ban
smoking in all public facilities?” 62% agreed, 30% did not agree, and 8% had no opinion.
Then each person was asked “Will you vote for Joe Vader (a candidate who supports the ban)
in the upcoming election?” Of those who agreed with the smoking ban, 80% said that they
would vote for Joe Vader, 7% said they would not, and 13% said they did not know who they
would vote for. Of those who disagreed with the ban, 25% said they would vote for Joe
Vader, 52% said they would not vote for him, and 23% said they did not know who they
would vote for. Of those who had no opinion on the ban, 18% said they would vote for Joe,
28% said they would not vote for him, and 54% said they did not know who they would vote
for. If a person votes for Joe Vader, what is the probability that they support the smoking
ban? If a person votes for Joe Vader, what is the probability that they have no opinion on the
smoking ban? If a person does not vote for Joe, what is the probability that they do not
support the smoking ban?
P( A ∩ V )
P( A ∩ V ) + P( D ∩ V ) + P( N ∩ V )
(0.62)(0.8)
=
(0.62)(0.8) + (0.3)(0.25) + (0.08)(0.18)
0.496
=
0.496 + 0.075 + 0.0144
0.496
=
= 0.85
0.5854
P( A | V ) =
P( N ∩ V )
P( A ∩ V ) + P( D ∩ V ) + P( N ∩ V )
(0.08)(0.18)
=
(0.62)(0.8) + (0.3)(0.25) + (0.08)(0.18)
0.0144
=
0.496 + 0.075 + 0.0144
0.0144
=
= 0.02
0.5854
P( N | V ) =
P( D ∩ NV )
P( A ∩ NV ) + P( D ∩ NV ) + P( N ∩ NV )
(0.3)(0.52)
=
(0.62)(0.07) + (0.3)(0.52) + (0.08)(0.28)
0.156
=
0.0434 + 0.156 + 0.0224
0.156
=
= 0.70
0.2218
P( D | NV ) =
8-24
Finite Mathematics
Chapter 8
Name ________________________________ Date ______________ Class ____________
Section 8-5 Random Variables, Probability
Distribution, and Expected Value
Goal: To find probabilities and expectation given a probability distribution
Expected Value of a Random Variable:
E ( X ) = x1 p1 + x2 p2 + x3 p3 + " + xn pn
In problems 1–3 if the probability distribution for the random variable X is given in the table,
what is the expected value of X?
1.
xi
–2
0
2
4
pi
0.6
0.1
0.2
0.1
n
E ( X ) = ∑ X i P( X i )
i =1
= (−2)(0.6) + (0)(0.1) + (2)(0.2) + (4)(0.1)
= −1.2 + 0 + 0.4 + 0.4
= −0.4
2.
xi
–2
0
2
4
pi
0.2
0.1
0.1
0.6
n
E ( X ) = ∑ X i P( X i )
i =1
= (−2)(0.2) + (0)(0.1) + (2)(0.1) + (4)(0.6)
= −0.4 + 0 + 0.2 + 2.4
= 2.2
8-25
Finite Mathematics
3.
Chapter 8
xi
–5
–3
–1
0
1
2
3
pi
0.1
0.05
0.4
0.15
0.15
0.1
0.05
n
E ( X ) = ∑ X i P( X i )
i =1
= (−5)(0.1) + (−3)(0.05) + (−1)(0.4) + (0)(0.15) + (1)(0.15) + (2)(0.1) + (3)(0.05)
= −0.5 − 0.15 − 0.4 + 0 + 0.15 + 0.2 + 0.15
= −0.55
4. The following table gives the probability distribution of the number of packages (in
thousands) handled by a shipping company. Find the expected number of packages handled
on any day.
Number of Packages
4
8
12
16
20
Probability
0.1
0.2
0.2
0.4
0.1
n
E ( X ) = ∑ X i P( X i )
i =1
= (4)(0.1) + (8)(0.2) + (12)(0.2) + (16)(0.4) + (20)(0.1)
= 0.4 + 1.6 + 2.4 + 6.4 + 2
= 12.8
You can expect to handle an average of 12.8 packages per day.
5. The following table gives the probability distribution of the number of days a student is
absent from a math class each semester. Find the expected number of days a student will be
absent from a math class per semester.
Number of Days Absent
0
1
2
3
4
Probability
0.2
0.4
0.2
0.1
0.1
n
E ( X ) = ∑ X i P( X i )
i =1
= (0)(0.2) + (1)(0.4) + (2)(0.2) + (3)(0.1) + (4)(0.1)
= 0 + 0.4 + 0.4 + 0.3 + 0.4
= 1.5
You can expect a student to be absent an average of 1.5 days for the semester.
8-26
Finite Mathematics
Chapter 8
6. The following table gives the probability distribution of the number of flights that are
overbooked per day for Flying High Airlines. Find the expected number of flights that are
overbooked per day.
Number of Flights Overbooked
0
1
2
3
Probability
0.4
0.3
0.2
0.1
n
E ( X ) = ∑ X i P( X i )
i =1
= (0)(0.4) + (1)(0.3) + (2)(0.2) + (3)(0.1)
= 0 + 0.3 + 0.4 + 0.3
= 1.0
You can expect an average of 1 flight per day to be overbooked.
7. The following table gives the probability distribution of the number of people who ride
on the public transportation system on the main route on Saturdays in a large city. Find the
expected number of people on the transportation system on the main route on a Saturday.
Number of Riders
25
50
75
100
Probability
0.05
0.10
0.2
0.25
125
150
175
200
0.2
0.1
0.05
0.05
n
E ( X ) = ∑ X i P( X i )
i =1
= (25)(0.05) + (50)(0.1) + (75)(0.2)
+ (100)(0.25) + (125)(0.2) + (150)(0.1) + (175)(0.05) + (200)(0.05)
= 1.25 + 5 + 15 + 25 + 25 + 15 + 8.75 + 10
= 105
You can expect 105 riders on average on the main route on any given Saturday.
8-27
Finite Mathematics
Chapter 8
8-28