1.
PROBABILITY THEORY
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UNIT 1
PROBABILITY THEORY
OBJECTIVES
General Objective
To understand the concept of probability.
Specific Objectives
At the end of the unit you should be able to:
Define Classical Probability and state the concept of Experiments
and Events
Define the following events:
• Conditional Events
• Independent Events
• Mutually Exclusive Events
List the elements in the sample spaces
To find the probability of an event based on Classical probability
Use the set theory to explain:
• Venn diagrams
• complementary of sets
• Union of sets
• Intersection of sets
• Null sets
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PROBABILITY THEORY
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INPUT
1.0 INTRODUCTION
So much in people’s lives is affected by chance. From the time
a person awakes until he or she goes to bed, that person makes
decision regarding the possible events that are governed at least in
part by chance. For example, should I carry an umbrella to class
today?
Will my motorcycle battery last until the end of the semester?
Should I accept a new job?
Probability as a general concept can be
defined as the chance of an event occurring. Probability
is the basis of inferential statistics. For example,
predictions are based on probability, and hypotheses are
tested by using probability.
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1.1 PROBABILITY
A probability experiment is a chance process that leads to well-defined
results called outcomes.
An outcome is the result of a single trial of a probability experiment.
A sample space is the set of all possible outcomes of a probability
experiment.
An event consists of the outcomes of a probability experiment.
Some sample spaces for various probability experiments are shown here.
Experiment
Toss one coin
Roll a die
Answer a true-false question
Toss two coins
Sample space
Head, Tail
1, 2, 3, 4, 5, 6
True, False
Head-head, tail-tail, head-tail, tail-head
Example 1.1
1.
a) Find the sample space for rolling two dice.
b) Find the sample space for the gender of the children if a family has three
children. Use B for boy and G for girl.
2.
Use a tree diagram to find the sample space for the gender of three children
in a family, as in ACTIVITY 1B
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PROBABILITY THEORY
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Solution to Example 1.1
1. a
Since each die can land in six different ways, and two dice are rolled, the
sample space can be presented by a rectangular array. The sample space
is the list of pairs of numbers in the chart.
Die 1
1
2
3
4
5
6
Die 2
1
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
(6, 1)
2
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
3
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
4
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
(6, 4)
5
(1, 5)
(2, 5)
(3, 5)
(4, 5)
(5, 5)
(6, 5)
6
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 6)
1. b
There are two genders, male and female, and each child could be either
gender. Hence there are eight possibilities, as shown here.
BBB
BBG
BGB
GBB
GGG
GGB
GBG
BGG
2.
children
B
G
B
B
G
G
B
B
G
B
G
G
B
G
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PROBABILITY THEORY
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ACTIVITY 1A
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INTPUT…!
1. Find the sample space for tossing two coins.
2. A die is rolled and a coin is tossed. Show the sample space.
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PROBABILITY THEORY
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FEEDBACK TO ACTIVITY 1A
1. S = {(H, H), (H, T), (T, H),(T,T)}
2.
Die
Coin
Head
(H))
Tail (T)
1
(H, 1)
(T, 1)
2
3
4
5
6
(H, 2)
(H, 3)
(H, 4)
(H, 5)
(H, 6)
(T, 2)
(T, 3)
(T, 4)
(T, 5)
(T, 6)
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PROBABILITY THEORY
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INPUT
1.2
CERTAIN AND COMPLEMENTARY EVENTS
When a die is rolled, the sample space is S = {1, 2, 3, 4, 5, 6}. Now let us define
event A as ‘number 1 appears on the die’s surface, therefore complement of A
(written as A’) consists of all the number on the die’s surface excluding 1,
therefore A’ = {2, 3, 4, 5, 6}.
Events can be presented pictorially by Venn diagrams. Figure (a) shows a simple
event E. The area inside the rectangle represents all the events in the sample
space(S). Figure (b)
−
Shows the complement of an event ( E ), which is the area inside the rectangle
but outside the circle representing E.
S
E
S
E
E
Eee
Fig a
Fig b
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PROBABILITY THEORY
1.2.1
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SET DESIGNATIONS
1. Sample space, S is represented by elements in a rectangle. Any event, E
is represented by its elements in a circle.
S
E
−
2. E’ or E is the complement of E. E’ means event E never occurred.
E’
E
3. E1 ∪ E 2 means either E1 or E2 or both have occurred.
S
E1
E2
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PROBABILITY THEORY
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4. E1 ∩ E 2 means both occurred.
S
E1 ∩ E 2
5. E1 and E2 are two mutually exclusive events in which E1 ∩ E 2 = φ . They
have no shared outcomes.
S
E
1
E2
6. E1 , E 2 , E3, .....E n are mutually exclusive and finite if and only if
i) E i ∩ E j = φ for every i and j,
ii) E1 ∪ E 2 ∪ E 3 ∪ ...... ∪ E n = S
E1
E2
E3
E4
E5
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1.2.2 SET IDENTITIES
The following identities can be used if there is a need.
1.
2.
3.
4.
5.
6.
7.
A∪ A = A
A ∪φ = φ
A∪S = S
A∩ A = A
A∪ B = B ∪ A
A∩S = A
A ∩φ = φ
8. A ∩ B = B ∩ A
9. A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
10. A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
11. ( A ∪ B )' = A∩' B '
12. ( A ∩ B )' = A'∪ B '
13. A ∪ B = A ∪ ( A'∩ B )
14. B = ( A ∩ B ) ∪ ( A'∩ B )
Example 1.2
1. A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 6, 7}, find:
i) A ∩ B
ii) A ∪ B
iii) n( A ∪ B )
2. A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6} and C = {3, 5, 7, 9}, find:
i) A ∩ B
ii) A ∩ C
iii) B ∩ C
iv) A ∩ B ∩ C
v) n ( A ∪ B ∪ C )
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PROBABILITY THEORY
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Solution to example 1.2
1.
i) A ∩ B = {1, 3, 6}
ii) A ∪ B = {1, 2, 3, 4, 5, 6, 7}
iii) n ( A ∪ B ) = n(A) + n(B) – n ( A ∩ B ) = 6 + 4 – 3 = 7
*If A, B and C are finite sets, therefore:
n( A ∪ B ∪ C ) = n( A) + n( B ) + n(C ) − n( A ∩ B ) − n( A ∩ C ) − n( B ∩ C ) + n( A ∩ B ∩ C )
2.
i)
ii)
iii)
iv)
v)
A ∩ B = {2, 4, 6}
A ∩ C = {3, 5}
B ∩ C = { φ}
A ∩ B ∩ C = { φ}
n ( A ∪ B ∪ C)
n( A) + n( B ) + n(C ) − n( A ∩ B ) − n( A ∩ C ) − n( B ∩ C ) + n( A ∩ B ∩ C )
=6+3+4–3–2–0+0
=8
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PROBABILITY THEORY
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ACTIVITY 1B
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1. Find the sample space for tossing two coins.
2. Nasir normally has one type of drink after lunch everyday. He randomly
drinks tea, coffee or simply plain water. If event A represents ‘Mamat has
one type of drink after lunch’, list down the elements in the sample space
S and event A. Find the relationship between the sample space and the A
set.
3. If n (A ∪ B) = 45, n( A ∩ B ) = 5 , and n(B) = 22, find n(A).
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PROBABILITY THEORY
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FEEDBACK TO ACTIVITY 1B
1.
S = {HH, HT,TH,TT)
2.
S={tea, coffee, water}, A = {tea, coffee, water}, therefore S = A
3.
28
14.
PROBABILITY THEORY
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INPUT
1.3 CLASSICAL PROBABILITY
Classical probability uses sample spaces to determine the numerical probability
that an event will happen. One does not actually have to perform the experiment
to determine that probability. Classical probability assumes that all outcomes in
the sample space are equally likely to occur. For example, when a single die is
rolled, each outcome has the same probability of occurring. Since there are six
outcomes, each outcome has a probability of 1 .
6
Equally likely events are events that have the same probability of
occurring
Formula for Classical Probability
The probability of any event E is
_____Number of outcomes in E____________
Total number of outcomes in the sample space
This probability is denoted by
n( E )
n( S )
This probability is called classical probability, and it uses
the sample space S.
P( E ) =
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PROBABILITY THEORY
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Probability Rule 1
The probability of any event E is a number (either a fraction or decimal) between
and including 0 and 1. This is denoted by 0 ≤ P ( E ) ≤ 1 .
Probability Rule 2
If an event E cannot occur (i.e., the even contains no members in the sample
space), the probability is zero.
Probability Rule 3
If an event is certain, the probability of E = 1.
Probability Rule 4
The sum of the probabilities of the outcomes in the sample space is 1
The next example illustrates
the probability rules
Example 1.3
1. If a family has three children, find the probability that all the children are girls.
2. When a single die is rolled, find the probability of getting a 9.
3. When a single die is rolled, what is the probability of getting a number less
than 7?
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PROBABILITY THEORY
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Solution to Example 1.3
1. The sample space for the gender of children for a family that has three
children is BBB, BBG, BGB, GBB, GGG, GGB, GBG, and BGG. (see t6he
tree diagram in the basic concept’s section). Since there is one way in eight
possibilities for all three children to be girls,
P(GGG) =
1
8
2. Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9.
Hence, the probability is P(9) = 0 = 0
6
3.
Since all outcomes, 1, 2, 3, 4, 5, and 6, are less than 7, the probability is
P(number less than 7) =
6
6
=1
** (Rule 4) For example, in a roll of a fair die, each outcome in the sample
space has a probability of 1 . Hence, the sum of the probabilities of the
6
outcomes is as shown.
Outcome
Probability
Sum
1
2
3
4
5
6
1
6
1
6
1
6
1
6
1
6
1
6
1
6
+ 1 + 1 + 1 + 1 + 1 =1
6
6
6
6
6
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PROBABILITY THEORY
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ACTIVITY 1C
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1. What is the probability of throwing a number greater than 4 with a die whose
faces are numbered from 1 to 6?
2. In a competition a prize is given for correctly forecasting the results of six
football matches. If a competitor sends in ten different forecasts, what is the
probability, that he receives the prize?
3. a)
b)
One red and one black marble are concealed in a bag. Find the
probability of drawing a red marble.
When three red and one black marbles are placed in the bag, find the
probability of drawing one red marble. What is the probability of drawing
one black marble?
4. A box contains 132 rivets of which 32 are undersized, 47 are oversized and
62 are satisfactory. Determine the probability of drawing at random:
(a) one undersized; (b) one oversized; and
(c) one satisfactory rivet from the box.
5. Four hundred resistors are examined and 6% are found to be defective.
Determine the probability that one selected at random will be defective and
also the probability that it will not be defective.
6. A purse contains 7 copper and 13 silver coins. Determine the probability of
selecting a copper coin when one is taken at random.
7. Determine the probability of winning a prize in a raffle by buying 3 tickets,
when there are 7 prizes and a total of 450 tickets sold.
8. Determine the probability of an event not happening when the probability of it
happening is 7/93.
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PROBABILITY THEORY
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FEEDBACK TO ACTIVITY 1C
1. 1/3 or 0.33 or 33 1 %
3
2. 0.0137 or 1.37%
3. (a) ½,
4. (a)
23
47
31
(b)
(c)
132
132
66
3 47
5. ,
30 50
6.
(b) ¾ and ¼
7
20
7. 7/150
8. 86/93
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PROBABILITY THEORY
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INPUT
1.4 COMPLEMENTARY EVENTS
The complement of an event E is the set of outcomes in the sample space
that are not included in the outcomes of event E. The complement of E is
−
denoted by E’ or E (read “E bar”).
Rule for complementary events
−
P( E ) = 1 − P( E )
Formula for empirical probability
Given a frequency distribution, the probability of an event being in a given
class is
f
, of which f is frequency for the class and n is the total
n
frequencies in the distribution
P(E)= =
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PROBABILITY THEORY
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Example 1.4
1. 25 students were asked if they like this module. The responses were
classified as “yes”, “no”, or “undecided”. The results were categorized in a
frequency distribution, as shown. Find the probability that a person
responded “no”.
Response
Yes
No
Undecided
Total
Frequency
15
8
2
25
2. In a sample of 50 students, 21 had type O blood, 22 had type A blood, 5
had type B blood, and 2 had type AB blood. Set up a frequency
distribution and find the following probabilities:
a. A student has type O blood
b. A person has type A or type B blood
c. A person has neither type A nor type O blood
d. A person does not have type AB blood
3. Hostel records indicated that students stayed in the hostel for the number
of days during school break shown in the distribution.
Number of
days stayed
3
4
5
6
7
Total
Frequency
15
32
56
19
5
127
Find the probabilities.
a. A student stayed exactly 5 days.
b. A student stayed less than 6 days
c. A student stayed at most 4 days.
d. A student stayed at least 5 days.
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PROBABILITY THEORY
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4. Based on the Venn diagram below, A and B are two events in the sample
space S. Find:
S
40
A
aAr 35
a.
b.
c.
d.
e.
f.
P(A)
P(A’)
P(B)
P(B’)
P( A ∩ B)
P( A ∩ B' )
(g) P ( A'∩ B )
(h) P ( A'∩ B ' )
(i) P ( A ∪ B )
(j) P ( A ∪ B ' )
(k) P ( A'∪ B )
(l) P ( A'∪ B ' )
5
B
20
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PROBABILITY THEORY
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Solution to Example 1.4
f
8
1. P(E)= n = 25
2.
Type
A
B
AB
O
Total
Frequency
22
5
2
21
50
f 21
=
n 50
5
22
ii. P(A or B) = 50 + 50 = 27 (Add the frequencies of the two
50
classes)
5
7
2
iii. P(neither A nor O) = 50 + 50 = 50 (Neither A nor O means tat a
student has either type B or type AB blood).
48
2
iv. P(not AB) = 1 – P(AB) = 1 − 50 = 50 = 24
25
i. P(O) =
3.
a) P(5) = 56
127
b) P(less than 6 days) = 15
+ 32
+ 56
= 103
(Less than
127
127
127
127
6 days means either 3, or 4 or 5 days.)
c) P(at most 4 days) =
days.
15
127
32
47
+ 127 = 127 (At most 4 days means 3 or 4
56
19
5
80
d) P(at least 5 days) = 127 + 127 + 127 = 127 (At least 5 days means
either 5, or 6, or 7 days.)
23.
PROBABILITY THEORY
4.
n(S) = 35 + 5 + 20 + 40
n( A) 40 2
=
=
n( S ) 100 5
n( A' ) 60 3
b. n(A’) = 60 and P(A’) =
=
=
n( S ) 100 5
n( B ) 25 1
c. n(B) = 25 and P(B) =
=
=
n( S ) 100 4
n( B ' ) 75 3
d. n(B’) = 75 and P(B’) =
=
=
n( S ) 100 4
n( A ∩ B )
5
1
e. n(A ∩ B) = 5 and P(A ∩ B) =
=
=
n( S )
100 20
n( A ∩ B ' ) 35
7
f. n(A ∩ B’) = 35 and P(A ∩ B ’) =
=
=
n( S )
100 20
P ( A'∩ B ) 20 1
g. n(A’ ∩ B ) = 20 and P(A’ ∩ B) =
=
=
n( S )
100 5
P ( A'∩ B ' ) 40 2
h. n(A’ ∩ B’) = 40 and P(A’ ∩ B’) =
=
=
n( S )
100 5
P ( A ∪ B ) 60 3
i. n(A ∪ B) = 60 and P(A ∪ B) =
=
=
n( S )
100 5
P ( A ∪ B ' ) 80 4
j. n(A ∪ B’) = 80 and P(A ∪ B’) =
=
=
n( S )
100 5
P ( A'∪ B ) 65 13
k. n(A’ ∪ B ) = 65 and P(A’ ∪ B ) =
=
=
n( S )
100 20
n( A'∪ B ' ) 95 19
l. n(A’ ∪ B ' ) = 95 and P(A’ ∪ B’) =
=
=
n( S )
100 20
a. n(A) = 40 and P(A) =
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24.
PROBABILITY THEORY
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ACTIVITY 1D
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1. If there are 50 tickets sold at a raffle and on person buys 7 tickets, what is
the probability of that person winning a price?
2. A survey found that 53% of Polytechnic students think this module is the best
of all the modules ever published. If a student is selected at random, find the
probability that he or she will disagree or have no opinion at all.
3. A couple has three children. Find each probability.
a) Of all boys
b) Of all girls
c) Of exactly two boys or two girls
d) Of at least one child of each gender
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FEEDBACK TO ACTIVITY 1D
1.
2.
3
7
50
47%
a. 1 b.
8
1
4
c.
3
4
d.
3
4
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INPUT
1.5 THE ADDITION RULES FOR PROBABILITY
Many problems involve finding the probability of two or more events. For
example, in your class gathering, one might wish to know, for a person selected
at random, that a person is a female or is wearing glasses. In this case there are
three possibilities to consider.
1. The person is a female.
2. The person is wearing glasses.
3. The person is a female and she is wearing glasses.
Consider another example. In the same gathering there are male and female
students. If a person is selected at random, what is the probability that the person
is a male or a female student? In this case, there are only two possibilities:
1. The person is a female.
2. The person is a male.
The difference between the two examples is that in the first case, the person
selected can be a female and is wearing glasses at the same time. In the second
case, the person selected cannot be both a female and a male at the same time.
In the second case, the two events are said to be mutually exclusive; in the first
case, they are not mutually exclusive.
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Two events are mutually exclusive if they cannot occur at the same time (i.e.,
they have no outcomes in common)
Addition Rule 1
When two events A and B are mutually exclusive, the probability that A and B
will occur is
P(A or B) =P(A) + P(B)
Addition Rule 2
If A and B are not mutually exclusive, then
P(A or B) = P(A) + P(B) – P(A ∩ B)
Example 1.5
1. A restaurant has 3 pieces of chicken karipap, 5 pieces of potato karipap and
4 pieces of fish karipap. If a customer selects a piece of karipap for dessert,
find the probability that it will be either potato or fish.
2. There are 20 buffaloes, 13 cows and 6 goats in a lorry. If an animal is selected
at random, find the probability that that animal is either a cow or a goat.
3. A day of the week is selected at random; find the probability that it is a
weekend day (Saturday or Sunday).
4. In a hospital unit there are eight nurses and five doctors. Seven nurses and
three doctors are females. If a staff person is selected the probability that the
subject is a nurse or a female.
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Solution to Example 1.5
1. Since there are 12 pieces of karipap,
P(potato or fish) = P(potato) + P(fish) =
The events are mutually exclusive.
2. P(cow or goat) = P(cow) + P(goat) =
13
39
6
+ 39 =
5
12
9
4
+ 12 = 12 =
3
4
19
39
3. P(Saturday or Sunday) = P(Saturday) + P(Sunday) =
1
7
+1=
7
2
7
4. The sample space is shown below:
Staff
Nurses
Doctors
Total
Females Males
7
1
3
2
10
3
Total
8
5
13
The probability is P (nurse or male) = P(nurse) + P(male) – P(male nurse)
8
3
1
= 13 + 13 − 13 = 10
13
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ACTIVITY 1E
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1. At a convention there are seven mathematics instructors, five computer
science instructors, three statistics instructors, and four science instructors. If
an instructor is selected, find the probability of getting a science instructor or a
math instructor.
2. In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are
females, and 12 of the juniors are males. If a student is selected at random,
find the probabilities of selecting the following:
a. a junior or a female
b. A senior or a female
c. A junior or a senior
3. A woman’s clothing store owner buys from three companies: A, B, and C. The
most recent purchases are shown here.
Product
Dresses
Blouses
A
24
13
B
18
36
C
12
16
If one item is selected at random, find the following probabilities.
a. It was purchased from company A or is a dress.
b. It was purchased from company B or company C.
c. It is a blouse or it was purchased from company A.
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4. A grocery store employs cashiers, stock clerks, and deli personnel. The
distribution of employees according to marital status is shown next.
Marital
status
Married
cashiers Stock
clerks
8
12
Deli
personnel
3
Not
married
5
2
15
If an employee is selected at random, find these probabilities:
a. The employee is a stock clerk or married.
b. The employee is not married.
c. The employee is a cashier or is not married.
5. RTM, TV3, and NTV7 have quiz shows, comedies, and dramas. The number
of each is shown here.
Type of
show
Quiz
show
Comedy
Drama
RTM Tv3
Ntv7
5
2
1
3
4
2
4
8
2
If a show is selected at random, find these probabilities.
a. The show is a quiz show or it is shown on TV3.
b. The show is a drama or a comedy.
c. The show is shown on NTV7 or it is a drama.
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FEEDBACK TO ACTIVITY 1E
1.
11
19
6
7
67
3. a.
118
38
4. a.
45
14
5. a.
31
2. a.
4
7
81
b.
118
22
b.
45
23
b.
31
b.
c. 1
44
59
2
c.
3
19
c.
31
c.
32.
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INPUT
1.6 THE MULTIPLICATION RULES AND CONDITIONAL
PROBABILITY
The previous section showed that the addition rules are used to compute
probabilities for mutually exclusive and not mutually exclusive events. This
section introduces two more rules, the multiplication rules. These rules can be
used to find the probability of two or more events that occur in sequence. For
example, if a coin is tossed and then a die is rolled, one can find the probability of
getting a head on the coin and a 4 on the die. These two events are said to be
independent since the outcome of the first event (tossing a coin) does not affect
the probability outcome of the second event (rolling a die).
Two events A and B are independent if the fact that A occurs does not affect the
probability of B occurring.
In order to find the probability of two independent events that occur in sequence,
one must find the probability of each event occurring separately and then multiply
the answers. For example, if a coin is tossed twice, the probability of getting two
heads is 1 . 1 = 1 . The result can be verified by looking at the sample space, HH,
2 2
4
HT, TH, and TT. Then P (HH) = 1 .
4
33.
PROBABILITY THEORY
CN303/1/ 33
Multiplication Rule 1
When two events are independent, the probability of both
occurring is
P(A and B) = P(A).P(B)
Example 1.6
1. A coin is flipped and a die is rolled. Find the probability of getting a head on
the coin and a 4 on the die.
2. An urn contains three red balls, two blue balls, and five white balls. A ball is
selected and its color noted. Then it is replaced. A second ball is selected and
its color noted. Find the probability of each of the following.
a. selecting two blue balls.
b. Selecting a blue ball and then a red ball.
c. Selecting a red ball and then a blue ball.
3. A pool found that 46% of students say they have suffered great stress at least
once in the exam week. If three students are selected at random, find the
probability that all three will say that they suffer great stress al least once in
the exam week.
Solution to Example 1.6
1
1. P(head and 4) = P(head).P(4) = 1 . 1 = 12 , note that the sample space for the
2 6
coin is H, T; and for the die is 1, 2, 3, 4, 5, 6.
2 2
4
1
2. a. P(blue and blue) = P(blue).P(blue) = 10 . 10 = 100 = 25
10
2 5
1
b. P(blue and white) = P(blue).P(white) = 10 . 10 = 100 = 10
c. P (red and blue) = P(red).P(blue) =
3 2
10 10
.
6
= 100 =
3. Let S denote stress, then P(S and S and S) =
P(S).P(S).P(S) = (0.46)(0.46)(0.46) =0.097
3
50
34.
PROBABILITY THEORY
CN303/1/ 34
ACTIVITY 1F
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT
INPUT…!
1. Two balls are drawn in turn with replacement from a bag containing 8 red
balls, 15 white balls, 24 black balls and 17 orange balls. Determine the
probabilities of having:
(a) two red balls;
(b) a red and a white ball;
(c) no orange balls;
(d) a black and red or black and orange ball;
(e) at least one black ball;
(f) a white ball on the first draw but the second ball not white.
2. The probability of three events happening are 1/8 for event A, 1/5 for
event B, and 2/7 for event C. Determine:
(a) the probability of all three events happening;
(b) the probability of event A and B but not C happening;
(c) the probability of only event B happening; and
(d) the probability of event A or event B happening but not
event C.
3. One bag contains 3 red and 5 black marbles and a second bag contains 4
green and 7 white marbles. One marble is drawn from the first bag and
two marbles from the second bag, without replacement. Determine the
probability of having;
(a) one red and two white marbles;
(b) no green marbles; and
(c) either one black and two green or one black and two white
marbles.
36.
PROBABILITY THEORY
CN303/1/ 36
INPUT
1.7
CONDITIONAL PROBABILITY
In the previous section, the events were independent of each other, since the
occurrence of the first event in no way affected the outcome of the second event.
On the other hand, when the occurrence of the first event changes the probability
of the second event, the two events are said to be dependent. Refer to example
2
2 (b). The probability of selecting the blue ball is 10 . If the ball is not replaced,
5
then the probability of selecting the second (white) ball is 9 since there are only
nine balls remaining.
When the outcome or occurrence of the first events affects the outcome or
occurrence of the second event in such a way that the probability is changed, the
events are said to be dependent.
The conditional probability of an event B in relationship to event A is the
probability that event B occurs after event A has already occurred. The notation
for conditional probability is P(B/A). this notation does not mean that B is
divided by A; rather, it means the probability that event B occurs given that event
2
A has already occurred. In the ball example (ex 2b), P(B/A) is 9 since the first
ball was not replaced.
37.
PROBABILITY THEORY
CN303/1/ 37
Multiplication Rule 2
When two events are dependent, the probability of both occurring is
P(A and B) =P(A).P(B/A)
The conditional probability of an event B in relationship to an event A
was defined as the probability that event B occurs after event A has already
occurred. It can be found by dividing both sides of the equation for multiplication
by P(A), as shown:
P(A and B) = P(A).P(B/A) therefore:
P ( AandB )
= P ( B / A)
P ( A)
The Venn diagram for conditional probability is shown in the figure
below. In this case,
P( A & B)
P(B/A) =
which is represented by the area in the
P ( A)
intersection or overlapping part of the circles A and B divided by area of circle A.
The reasoning here is that if one assumes A has occurred, the A become the
sample space for the next calculation and is the denominator of the probability
P( A & B)
fraction
. The numerator P(A & B) represents the total probability of the
P ( A)
part B contained in A. Hence, P(A & B) becomes the numerator of the probability
P( A & B)
. Imposing a condition reduces the sample space.
fraction
P ( A)
38.
PROBABILITY THEORY
CN303/1/ 38
Venn diagram for Conditional probability
P(A)
P(A and B)
P(B/A) =
P(B)
P( A & B)
P ( A)
1.7.1 PROBABILITIES FOR “AT LEAST”
The multiplication rules can be used with the complementary and noncomplementary events to simplify solving probability problems involving “at
least”. The next examples illustrate how this is done.
39.
PROBABILITY THEORY
CN303/1/ 39
Example 1.7
1. In a shipment of 25 radios, 2 are defective. If two radios are randomly selected
and tested, fond the probability that both are defective if the first one is not
replaced after it has been tested.
2. Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls
and on red ball. A coin is tossed. If it falls heads up, box 1 is selected and a
ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the
probability of selecting a red ball.
3. A box contains white chips and black chips. A person selects two chips without
replacement. If the probability of selecting a black chip and a white chip is 15 ,
56
3
and the probability of selecting a black chip on the first draw is 8 , find the
probability of selecting the white chip on the second draw, given that the first
chip selected was a black chip.
4. The probability Ali parks in a no-parking zone and gets a parking ticket is 0.06,
and the probability that Ali cannot find a legal parking space and has to park in
the no-parking zone is 0.20. On Monday, Ali arrives at school and has to park
in a no-parking zone. Find the probability that he will get a parking ticket.
5. A recent survey asked 100 people if they thought Sardin Cap Ayam is the best
sardin. The results of the survey are shown in the table.
Gender
Male
Female
Total
Yes
32
8
40
No
18
42
60
Total
50
50
100
Find these probabilities.
a. The respondent answered “yes,” given that the respondent was a
female.
b. The respondent was a male, given that the respondent answered
“no.”
40.
PROBABILITY THEORY
CN303/1/ 40
6.
A coin is tossed five times. Find the probability of getting at least one tail.
7.
A survey reported that 3% of pens sold in the Politeknik are Pilot pens. If 4
students who purchased a pen are randomly selected, find the probability
that at least one purchased a Pilot pen.
8.
A coin is tossed three times. Find the probability of getting
i) Exactly 2 tails
ii) At least 2 tails
Solution to Example 1.7
1. Since the event are dependent,
P(D1 and D2) =P(D1).P(D2/D1) = (2/25).(1/24) = 2/600 = 1/300
2. With the use of a tree diagram, the sample space can be determined as
shown in the figure. First assign the probabilities to each branch. Next, using
the multiplication rule, multiply the probabilities for each branch.
Ball
Box 2
Red 1 . 2 =
2 3
Blue 1 . 1 =
2 3
P(R/B1) ¼
Red 1 . 1 =
2 4
P(B/B2) ¾
P(B2) ½
Box 1
P(R/B1) 2/3
P(B/B1) 1/3
P(B1) ½
Blue 1 . 3 =
2 4
2
6
1
6
1
8
3
8
Finally, use the addition rule, since a red ball can be obtained from box 1 or
box2;
8
3
P(red) = 2 + 1 = 24 + 24 = 11
6
8
24
Tree diagrams can be used when the events are independent or dependent, and
they can also be used for sequences of three or more events.
41.
PROBABILITY THEORY
CN303/1/ 41
3. Let B = selecting a black chip W = selecting a white chip
P( B & W ) 15 5
Then P(W/B) =
= 56 =
Hence the probability of selecting a white
3
7
P( B)
8
5
chip on the second draw given that the first chip[ selected was black is 7 .
4. Let N= parking in a no-parking zone and T = getting a ticket, then
P ( N & T ) 0.06
=
= 0.30
Hence, Ali has a 0.30 probability of
P( N )
0.20
getting a parking ticket, given that he parked in a no-parking zone.
P(T/N) =
5.
Let M = respondent was a male
Y = respondent answered “yes”
F = respondent was a female N = respondent answered “no”
a)
The problem is to find P(Y/F). The rules states P(Y/F) = P(Y
and F) /P(F). The probability P(F and Y) is the number of
females who responded “yes” divided by the total number of
respondents: P(F and Y) =8/100
The probability P(F) is the probability of selecting a female:
P(F) = 50/100
Then,
P(Y/F) =
b.
4
P ( F & Y ) 8 / 100
=
=
50 / 100 25
P( F )
The problem is to find P(M/N)
P(M/N) =
P ( N & M ) 18 / 100 3
=
=
P( N )
60 / 100 10
42.
PROBABILITY THEORY
6.
CN303/1/ 42
It is easier to find the probability of the complement of the event, which is
“all heads,” and then subtract the probability from 1 to get the probability of
at least one tail.
−
P(E) = 1 – P( E )
P(at least 1 tail) = 1 – P(all heads)
1
P(all heads) = (1/2)5 = 32
Hence,
31
1
P(at least one tail) = 1 - 32 = 32
−
7.
Let E = at least one Pilot pen purchased and E = no Pilot pen purchased.
Then,
−
P(E) = 0.03 and P( E ) = 1 – 0.03 = 0.97
P(no Pilot pen purchased) = (0.97)(0.97)(0.97) = 0.885; hence, P(at
least one Pilot pen purchased) = 1 – 0.885 = 0.115
8.
3
i) P(exactly 2 tails) = 1 + 1 + 1 = 8
8
8
8
ii) P(at least 2 tails) = P(2 tails and 1 head) + P(3 tails)
3
=8 + 1 = 1
8
2
(##Draw a tree diagram to verify your answer)
43.
PROBABILITY THEORY
CN303/1/ 43
ACTIVITY 1G
TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITHY THE NEXT
INPUT…!
1. The probabilities of an engine failing are given by: p1, failure due to
overheating; p2, failure due to ignition problems; p3, failure due to fuel
blockage. When p1 = 1/7, p2 = 2/9 and p3 = 3/11, determine the
probabilities of:
(a) both p1 and p2 happening;
(b) either p2 or p3 happening
(c) both p1 and either p2 or p3 happening.
2. Actuarial tables show that the life expectancy of three men, A, B and C,
over a twenty-year period depends on their age and is given by PA = 4/15,
PB = 11/15 and PC = 14/15. Determine the probabilities that in twenty
years:
(a) all three men will be alive;
(b) A will be alive but B and C will be dead;
(c) At least one man will be alive.
3. When exploration for oil occurs a test hole is drilled. If as a result of this
test drilling it seems likely that really large quantity of oil exist, (a bonanza)
then the well is said to have structure. Examination of past records reveals
the following information;
Bonanza
No Bonanza
Total
Structure
0.20
0.15
0.35
No structure
0.05
0.60
0.65
Total
0.25
0.75
1.00
a) Find P(Bonanza/structure)
b) (No Bonanza/structure)
c) P(Bonanza/no structure)
d) P(No bonanza/no structure)
44.
PROBABILITY THEORY
CN303/1/ 44
4. Suppose we have 100 urns. Type 1 urn (of which there are 70) each contains
5 black and 5 white balls. Type 2 urn (which there are 30) each contains 8
black and 2 white balls. An urn is randomly selected and a ball is drawn from
that urn. If the ball chosen was black, what is the probability the ball came
from a type 1 urn?
46.
PROBABILITY THEORY
CN303/1/ 46
SELF-ASSESMENT 1
You are approaching success. Try all the questions in this self assessment
section and check your answers with those given in the Feedback to the Self Assessment 1 on the next page. If you have problems, consult your instructor.
Good luck.
1. State which events are independent and which are dependent.
i) Tossing a coin and throwing a die
ii) Drawing a ball from an urn, not replacing it, and then drawing a
second ball.
iii) Getting a raise in salary and purchasing a new car.
iv) Driving on ice and having accident.
v) Having a large show size and having a high IQ.
vi) A father being left-handed and a daughter being left-handed.
vii) Smoking excessively and having lung cancer.
viii) Eating an excessive amount of ice cream and smoking an
excessive amount of cigarettes.
2. A survey found that 68% of book buyers are 40 or older. If two book buyers
are selected at random, find the probability that both are 40 or older.
3. A salesman finds that the probability of making a sale is 0.23. If he talks to four
customers today, find the probability that he will make four sales.
4. Find the probability of selecting two people at random who were born in the
same month.
5. If three people are selected, find the probability that all three were born in
January.
6. What is the probability that a husband, wife and daughter have the same
birthday?
47.
PROBABILITY THEORY
CN303/1/ 47
7. A radio uses six batteries, two of which are defective. If two are selected at
random without replacement, find the probability that the first battery tests
good and the second one is defective.
8. Out of 120 students, 90 of them put on white t-shirts. If five students are
selected at random, one by one, find the probability that all will put on white tshirts.
9. Urn 1 contains five red balls and three black balls. Urn 2 contains three red
balls and one black ball. Urn 3 contains four red balls and two black balls. If
an urn is selected at random and a ball is drawn, find the probability it will be
red.
10. If a die is rolled three times, find the probability of getting at least one even
number.
48.
PROBABILITY THEORY
CN303/1/ 48
FEEDBACK TO SELF-ASSESMENT 1
Have you tried all the questions?? If “YES”, check your answer now.
1. a. Independent
b. Dependent
c. Dependent
d. Dependent
2. 0.462
3. 0.003
4. 1/12
5. 1/1728
6. 1/133,225
7. 4/15
8. 243/1024
9. 49/72
10. 7/8
e. Independent
f. Dependent
g. dependent
h. Independent
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