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Mat 142 College Mathematics
Department of Mathematics and Statistics
Chapter 2 Probability
This lecture note is based on the text book Mathematical Literacy in a numerate
Society – by Matthew A. Isom and Jay Abramson
The probability an event E will occur is the number of desired outcomes divided by the
Number of desired outcomes
total possible number of outcomes. We write P( E ) 
.
Number of total outcomes
Example 1. Roll a die once. The total outcomes are namely 1, 2, 3, 4, 5, and 6. Now the
1
probability of the event for rolling a 3 is P( E )  P(rolling a 3)  , because there are 6
6
total outcomes and 1 desired outcomes.
Example 2. Roll a die once. Write the sample space. Find the following probabilities:
a) P( E )  P(rolling a 3 or a 5)
b) P( E )  P(rolling a 3 or more)
c) P( E )  P(rolling a number greater th an 10)
d) P( E )  P(rolling an even number)
Answer: a) 1/3
b) 2/3
c) 0
d) ½
Example 3. Roll a die twice. Write the sample space. Find the following probabilities:
a) P( E )  P(rolling a sum 3 or a sum 5)
b) P( E )  P(rolling a sum 3 or more)
c) P( E )  P(rolling a sum greater th an 10)
d) P( E )  P(rolling a sum greater th an 12)
The sample space of the event of rolling a die twice:
11
21
31
41
51
61
Answer: a) 6/36
12
22
32
42
52
62
13
23
33
43
53
63
b) 35/36
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
c) 3/36
d) 0
Example 4. Roll a die twice. Write the sample space. Find the following probabilities:
a) P( E )  P(all two rolls are either a 1 or a 3)
b) P( E )  P(all two rolls are not 2)
c) P( E )  P(all two rolls are above 3)
d) P( E )  P(all are a 1 or all are a 5)
The sample space of the event of rolling a die twice:
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Mat 142 College Mathematics
11
21
31
41
51
61
12
22
32
42
52
62
13
23
33
43
53
63
Department of Mathematics and Statistics
14
24
34
44
54
64
15
25
35
45
55
65
16
26
36
46
56
66
Answer: a) Look at the sample space, we have all three rolls either a 1 or a 3 are 11, 13,
31, and 33. Thus P( E )  P(all two rolls are either a 1 or a 3) = 4/36.
We could find this probability using independent event and multiplication principle as
the probability of getting first outcome either 1 or 3 is 2/6, then the probability of getting
second outcome either 1 or 3 is again 2/6. By multiplication principle the probability of
getting all two outcomes either a 1 or a 3 is 2 / 6  2 / 6  4 / 36
b) P( E )  P(all two rolls are not 2) = 25/36, you may verify looking at the sample space,
that we have 12, 21, 22, 23, 24, 25, 26, 32, 42, 52, 62 with outcome 2’s. There are 25
outcomes with no 2’s.
Using independent event and multiplication principle we have
P( E )  P(all two rolls are not 2) = 5 / 6  5 / 6  25 / 36
c) P( E )  P(all two rolls are above 3) = 9/36
Using independent event and multiplication principle we have
P( E )  P(all two rolls are above 3) = 3 / 6  3 / 6  9 / 36
d) P( E )  P(all are a 1 or all are a 5) = 1/ 6 1/ 6  1/ 6 1/ 6  2 / 36 . You look at the
sample space we have only two outcomes 11 and 55.
Example 5. Roll a die three times. How many outcomes do we have in the sample space?
Find the following probabilities:
a) P( E )  P(all three rolls are either a 1 or a 3)
b) P( E )  P(all three rolls are not 2)
c) P( E )  P(all three rolls are above 3)
d) P( E )  P(all are a 1 or all are a 5)
Answer: a) 2 / 6  2 / 6  2 / 6  8 / 216
b) 5 / 6  5 / 6  5 / 6  125 / 216
c) 3 / 6  3 / 6  3 / 6  27 / 216
d) 1/ 6 1/ 6 1/ 6  1/ 6 1/ 6 1/ 6  2 / 216
Example 6. Roll a die five times. How many outcomes do we have in the sample space.
Find the following probabilities:
a) P( E )  P(all five rolls are either a 1 or a 3)
b) P( E )  P(all five rolls are not 2)
c) P( E )  P(all five rolls are above 3)
d) P( E )  P(all are a 1 or all are a 5)
Answer: a) 2 5 / 6 5
b) 5 5 / 6 5
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c) 3 5 / 6 5
2
d) 2 / 6 5
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Mat 142 College Mathematics
Department of Mathematics and Statistics
Example 7. A card is chosen at random from a deck of 52 cards. It is then replaced, the
deck reshuffled and a second card is chosen. What is the probability of getting a jack and
an eight?
Solution. The event is independent. The probability of drawing first card a jack is 4/52
and second card an eight is 4/52. Also drawing a first card an eight is 4/52 and second
card a jack is 4/52. The probability of drawing a jack and an eight is
4 / 52  4 / 52  4 / 52  4 / 52  2 / 169
Exercise: A card is chosen at random from a deck of 52 cards. It is then replaced, the
deck reshuffled and a second card is chosen.
a) What is the probability of getting a jack and then an eight? Ans: 1/169
b) What is the probability of getting a diamond and then a heart? Ans: 1/16
Example 8. A family has two children. Using b to stand for boy and g for girl in ordered
pairs, give each of the following.
a) the sample space
b) the event E that the family has exactly one daughter.
c) the event F the family has at least one daughter
d) the event G that the family has two daughters
e) p(E)
f) p(F)
g) p(G)
Example 9. A group of three people is selected at random. 1)What is the probability that
all three people have different birthdays. 2) What is the probability that at least two of
them have the same birthday?
1) The probability that all three people have different birthdays is
365 P3 365 364 363



 0.992
365 365 365
365 3
2) The probability that all three people have same birthday is
365 364 363
1


 1  0.992  0.008
365 365 365
Example 10. Given the binomial probability formula b(n, k , p)  C (n, k ) p k (1  p) nk .
Evaluate the following. 1) b(10, 7, 0.7)
2) b(5, 3, 1/36)
c) b(8, 8, 0.5)
Solution: 1) b(10, 7, 0.7)  C (10, 7)0.7 7 (1  0.7)107  ? Use calculator for simplification.
Try yourself for 2) and 3).
Example 11. A fair coin is tossed 7 times. 1) What is the probability that of getting 7
tails? 2) What is the probability of getting at most 6 tails?
Solution: 1) The probability of getting 7 tails is
b(7, 7, 0.5)  C (7, 7)0.57 (1  0.5) 77  0.57
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Mat 142 College Mathematics
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2) the probability of getting at most 6 tails =
 1  C (7, 7)0.57 (1  0.5) 77  1  0.57  0.992
(One can solve this problem without considering binomial probability, try yourself)
Conditional probability. A conditional probability is a probability whose sample space
has been limited to only those outcomes that fulfill a certain condition. A conditional
n( A  B )
probability of an event A, given event B is p( A | B) 
n( B )
Example 12. In a newspaper poll concerning violence on television, 600 people were
asked, “what is your opinion of the amount of violence on prime time television – is there
too much violence on television?” Their responses are indicated in the table below.
Yes (Y)
No (N)
Don’t know
Total
Men (M)
162
95
23
280
Women (W)
256
45
19
320
Total
418
140
42
600
Suppose we label the events in the following manner: W is the event that a response is
from a woman, M is the event that a response is from a man, Y is the event that a
response is yes, and N is the event that a response is no, then the event that a woman
responded yes would be written as Y | W and p(Y | W) = 256/320 = 0.8.
Use the given table to answer following questions.
a) p(N)
b) p(W)
c) p(N | W)
f) p(W  N ) g) p(Y)
h) p(M)
k) p (Y  M )
l) p ( M  Y )
d) p(W | N) e) p( N  W )
i) p(Y | M)
j) p(M | Y)
m) p(W  Y )
n) p(W | Y )
Answer: a) 0.23
b) 0.53
c) 0.14
d) 0.32
e) 0.08
f) 0.08
g) 0.70
h) 0.47
i) 0.58
j) 0.39
k) 0.27
l) 0.27
m) 0.43
n) 0.61
Expected Value: The standard way of finding expected value of an experiment is to
multiply the value of each outcome of the experiment by the probability of that outcome
and add the results.
We write expected value E =  xi pi x1 p1  x2 p2  x3 p3  
Example 13. Your midterm exam grades are 88, 90. Average midterm exam is 50
percent of your grade. Your homework average is 80, it is 15 percent of your grade, your
quiz average is 85, which is 10 percent of your grade and you earned a 78 on the final,
which is 25 percent of your grade. What is your expected grade? What should be your
average midterm grade to have your expected grade 90?
Solution: The expected grade is E = (88+90)/2(0.5)+80(0.15)+85(0.10)+78(0.25) = 84.5
Try yourself to find the next answer.
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Mat 142 College Mathematics
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Independent Events: Two events A and B are called independent if and only if
P( A  B)  P( A) P( B) , otherwise A and B are dependent.
Example 14. In two tosses of a single fair coin show that the events “A head on the first
toss” and “A head on the second toss” are independent.
Solution: The sample space S S  {HH , HT , TH , TT } , the event with a head on the first
toss A  {HH , HT } and an event with a head on the second toss B  {HH , TH } .
Now show that P( A  B)  P( A) P( B) .
Bayes’ Formula: Let A, B, C are mutually exclusive events whose union is the sample
space S. Let E be the arbitrary event in S such that P ( E )  0 , then
P( A  E )
P( B  E )
P(C  E )
,
,
P( A | E ) 
P( B | E ) 
P(C | E ) 
P( E )
P( E )
P( E )
where P( A  E )  P( E | A) P( A) and so on.
Example 15. A company produces 1,000 refrigerators a week at three plants. Plant A
produces 350 refrigerators a week, plant B produces 250 refrigerators a week, and plant C
produces 400 refrigerators a week. Production records indicate thatb5% of the
refrigerators at plant A will be defective, 3% of those produced at plant B will be
defective, and 7% of those produced at plant C will be defective. All refrigerators are
shipped to a central warehouse. If a refrigerator at the warehouse is found to be defective,
what is the probability that it was produced a) at plant A? b) at plant B? c) at plant C?
We consider D as defective and D’ as non defective.
0.05
D
A
0.95
D’
0.03
B
0.97
D
0.07
D
0.35
Start
0.25
D’
0.40
C
a)
0.93 D’
We now answer all questions from the tree diagram.
P ( A  D)
0.35(0.05)
1
P ( A | D) 


P ( D)
0.35(0.05)  0.25(0.03)  0.40(0.07) 3
You now try to find b) P( B | D) 
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P ( B  D)
0.25(0.03)

?
P ( D)
0.35(0.05)  0.25(0.03)  0.40(0.07)
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Mat 142 College Mathematics
c) P(C | D) 
Department of Mathematics and Statistics
P(C  D)
0.40(0.07)

?
P ( D)
0.35(0.05)  0.25(0.03)  0.40(0.07)
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