CHAPTER IV
PROBABILITY AND COUNTING RULES
Probability can be defined as a chance of an event occurring.
Sample spaces and probability
A probability experiment is a chance process that leads to well defined results called outcomes.
An outcome is a result of a single trial of a probability experiment.
A sample space is the set of all possible outcomes of a probability experiment.
Example: Find the sample space for rolling two dice.
Solution:
Die 1
Die 2
1
2
3
4
5
6
1
2
3
4
5
6
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
Example: Find the sample space of drawing a card from an ordinary deck of cards.
Solution: There are four suits (hearts, diamonds, spades and clubs)
Example: Find the sample space of the gender of the children if a family has three children.
Solution: BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG
An event consists of a set of outcomes of a probability experiment.
Classical Probability assumes that all outcomes in the sample space are equally likely to occur.
Equally likely events are events that have the same probability to occur.
The probability of any event E, denoted by P(E) and is given by
P(E) =
𝒏(𝑬)
𝒏(𝑺)
=
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒊𝒏 𝑬
𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒑𝒂𝒄𝒆
Examples: Tossing a coin, rolling a die, taking a card from a deck of cards.
Rules of probability
1. If E is any event, then 0 ≤ 𝑃(𝐸) ≤ 1
2. If an event E cannot occur, then 𝑃(𝐸) = 0
3. If an event E cannot certain, then 𝑃(𝐸) = 1
4. ∑ 𝑃(𝐸) = 1, that is sum of the probabilities of all the outcomes in the sample space is 1
Example: If a die is rolled one time, find these probabilities.
(a) Getting a 2 (b) Getting a number greater than 6 ( c) Getting an odd number (d) Getting a 4 or an odd
number (e) Getting a number less than 7 (f) Getting a number greater than or equal to 3 (g) Getting a
number greater than 2 and an even number.
Solution:
(a) P(Getting a 2) =
1
6
(b) P(Getting a number greater than 6) = 0
3
1
4
2
(c) P(Getting an odd number) = 6 = 2 (d) P(Getting an 4 or an odd number) = 6 = 3
(e) P(Getting a number less than 7) =
6
6
=1
(f) P(Getting a number greater than or equal to 3) =
(g) P(Getting a number greater than 2 and an even number) =
2
6
=
4
6
=
2
3
1
3
Example: If two dice are rolled one time, find the probability of getting these results.
(a) A sum of 9 (b) A sum of 7 or 11 (c) Doubles (d) A sum less than 9 (e) A sum greater than or equal to 10
Solution:
4
1
(a) P(A sum of 9) = 36 = 9
(b) P(A sum of 7 or 11) =
(c) P(Doubles) =
6
=
36
2
9
1
=
36
8
6
26
=
Sample space
= {(3,6), (4,5),(5,4),(6,3)}
Sample space
= {(1,6),(2,5),(3,4), (4,3),(5,2),(6,1),(5,6),(6,5)}
Sample space
= {(1,1),(2,2),(3,3), (4,4),(5,5),(6,6)}
13
Sample space = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2)
36
18
(2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (4,1) (4,2) (4,3) (4,4) (5,1) (5,2) (5,3) (6,1) (6,2) }
(d) P(A sum less than 9) =
(e) P(A sum greater than or equal to 10) =
6
36
=
1
6
Sample space
= { (4,6)
(5,5) (5,6) (6,4) (6,5) (6,6) }
Example: If one card is drawn from a deck , find the probability of getting these results.
(a) A queen (b) A club (c) A queen of clubs (d) A 3 or an 8 (e) A 6 or a spade (f) A black king (g) A red card
and a seven (h) A diamond or a heart (i) A black card.
Solution:
(a) P(A queen) =
(b) P(A club) =
4
52
13
52
1
=
1
=
8
(e) P(A 6 or a spade) =
1
16
52
2
52
(There is 1 queen of club cards)
52
=
52
(f) P(A black king) =
(There are 13 club cards)
4
(c) P(A queen of clubs) =
(d) P(A 3 or an 8) =
(There are 4 queen cards)
13
2
(There are 4 cards with number 3 and 4 cards with 8)
13
=
=
4
13
1
26
(There are 3 cards with number 6 and 13 cards of spade)
(There are 2 black king cards)
(g) P(A red card and a seven) =
(h) P(A diamond or a heart) =
(i) P(A black card) =
26
52
=
2
52
26
52
=
=
1
(There are 2 red cards with number 7)
26
1
(There are 13 cards of diamond and 13 cards of heart)
2
1
(There are 26 black cards)
2
Example: If a family has three children, find the probability that two of the three children are girls.
Solution: P(Two girls) =
3
8
(There are 8 outcomes BBB,BBG,BGB,GBB, BGG,GBG,GGB and GGG and 3
ways to have 2 girls)
Example: A couple has three children, find each probability.
(a) All boys (b) All girls or all boys
Solution: (a) P(All boys ) =
(c) Exactly two boys or two girls (d) At least one child of each gender
1
2
6 3
(b) P(All boys or all girls) =
(c) P(Exactly two boys or two girls) = 
8
8
8 4
(d) P(At least one child of each gender) =
6 3

8 4
Example: A card is drawn from an ordinary deck. Find the probabilities of
(a) getting a jack (b) getting a 6 of clubs (c) getting a three or a diamond (d) getting a 3 or a 6.
Solution: See the text book.
Example: When a single die is rolled, find the probabilities of
(a) getting a 9 (b) getting a number less than 7
Solution: See the text book.
The complement of an event is the set of outcomes in the sample space that are not included in the outcomes of
event E. This is denoted by 𝑃(𝐸̅ ).
Rules for complementary events
𝑃(𝐸̅ ) = 1 − 𝑃(𝐸) , 𝑃(𝐸) = 1 − 𝑃(𝐸̅ ) , 𝑃(𝐸) + 𝑃(𝐸̅ ) = 1
P(S) =1
Example: When a single die is rolled, find the probabilities of (a) getting an odd number (b) getting an even
number.
Solution: (a) P(Getting an odd number) =
3
6
=
1
2
1
1
(b) P(Getting an even number) = 1 − 2 = 2
Example: When two coins are tossed, find the probabilities of (a) getting all heads (b) getting at least one head.
Solution: (a) P(Getting all heads) =
1
4
1
3
(b) P(E) = P(Getting at least one head) = 1 − 4 = 4
Empirical Probability relies on actual experience to determine the likelihood of outcomes.
Given a frequency distribution, the probability of an event being in a given class is
P(E) =
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠
𝑇𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
=
𝑓
𝑛
This probability is called empirical probability and is based on observation
Example: In a sample of 50 people, 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 probabilities.
(a) A person 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 O blood
Solution:
Type
A
B
AB
O
(a) P(Type O blood) =
Frequency
22
5
2
21
n = 50
27
21
7
(b) P(Type A or B blood) =
(c) P(Neither type A nor O blood) =
50
50
50
(d) P(Does not have Type AB blood) = 1 
2 48

.
50 50
Example: Hospital records indicated that knee replacement patients stayed in the hospital for the number of
days shown in the distribution.
No.of
days
stayed
3
4
5
6
7
Frequency
15
32
56
19
5
127
Find the probabilities.
(a) A patient stayed exactly 5 days (b) A patient stayed at most 4 days (c) A patient stayed less than 6 days
(d) A patient stayed at least 5 days .
Solution: (a) P(Exactly 5 days) =
56
127
(c) P(Less than 6 days) =
(b) P(At most 4 days) =
47
127
(3 or 4)
103
80
(3 or 4 or 5) (d) P(At least 5 days) =
127
127
(5 or 6 or 7)
Example: Rural speed limit for all 50 states is indicated below.
60 mph
1
65 mph
18
70 mph
18
75 mph
13
Choose one state at random. Find the probability that its speed limit is
(a) 60 or 70 miles per hour (b) greater than 65 miles per hour (c) 70 miles per hour or less
Solution: (a) P(60 or 70 mph) =
(c) P(75 mph or less) =
19
31
 0.38 (b) P(greater than 65 mph) =
 0.62 ( 70 or 75)
50
50
37
 0.74
50
THE ADDITION RULES FOR PROBABILITY
Two events are mutually exclusive events if they cannot occur at the same time
(i.e., they have no outcomes in common)
Examples:
1) When a single die is rolled,
Getting an odd number and getting an even number (mutually exclusive)
Getting a number less than 4 and getting a number greater than 4(mutually exclusive)
Getting a 3 and getting an odd number (not mutually exclusive)
Getting an odd and getting a number less than 4(not mutually exclusive)
2) When a single card is drawn from a deck,
Getting a 7 and getting a jack (mutually exclusive)
Getting a club and getting a king (not mutually exclusive)
Getting a face card and getting an ace (mutually exclusive)
Getting a face card and getting a spade (not mutually exclusive)
Addition Rules
1.If two events A and B are mutually exclusive , the probability that A or B will occur is
P(A or B) = P(A) + P(B)
2. If two events A and B are not mutually exclusive , the probability that A or B will occur is
P(A or B) = P(A) + P(B) – P(A and B)
Venn diagram for addition rules
Example: A single card is drawn from a deck. Find the probability of selecting the following.
(a) A 4 or a diamond (b) A club or a diamond (c) A jack or a black card
Solution:
(a) P ( Selecting a 4) =
4
13
1
, P ( Selecting a diamond) =
, P ( Selecting a 4 and a diamond) =
52
52
52
P ( Selecting a 4 or a diamond) =
4 13 1 16 4
 


52 52 52 52 13
13
13
, P(A diamond) =
, P (A club and a diamond) = 0 ( mutually exclusive)
52
52
(b) P(A club) =
P (A club or a diamond) =
(c) P(A jack) =
13 13 26 1



52 52 52 2
26
4
2
, P(A black) =
, P(A Jack and a black) =
52
52
52
P(A Jack and a black) =
4 26 2 28 7




52 52 52 52 13
Example: A city has 9 coffee shops: 3 Starbuck’s, 2 Caribou Coffees and 4 Crazy Mocho Coffees. If a person
selects one shop at random to buy a cup of coffee, find the probability that it either a Starbuck’s or Crazy
Mocho Coffees.
Solution:
P(Starbuck’s or Crazy Mocho Coffees) = P(Starbuck’s ) + P(Crazy Mocho Coffees)
=
3 4 7
 
(Events are mutually exclusive)
9 9 9
Example: A corporate research and development centres for three local companies have the following number
of employees:
U.S Steel
110
Alcoa
750
Bayer Material Science 250
If a research employee is selected at random, find the probability that the employee is employed by U.S Steel or
Alcoa.
Solution: See the text book.
Example: At a particular school with 200 male students, 58 play foot ball, 40 play basket ball and 8 play both.
What is the probability that the randomly selected male student plays neither sport?
Solution:
P(foot ball or basket ball) = P(foot ball ) P(basket ball) P(foot ball and basket ball)
=
58
40
8
90
9




 0.45
200 200 200 200 20
P(Neither sport) = 1  0.45  0.55
Example: 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 probability of selecting the following.
(a) A junior or a female (b) A senior or a female (c) A junior or a senior
Solution:
Male
Female
Total
Senior
4
6
10
Junior
12
6
18
(a) P(A junior or a female ) = P(A junior ) + P(A female ) - P(A junior and a female )
=
18 12 6 24 6
 


28 28 28 28 7
(b) P(A senior or a female ) = P(A senior ) + P(A female ) - P(A senior and a female )
=
10 12 6 16 4
 


28 28 28 28 7
(c) P(A junior or a senior ) = P(A junior ) + P(A senior )
=
18 10 28


1
28 28 28
Example: In a hospital unit there are 8 nurses are 5 physicians: 7 nurses and 3 physicians are females. If a staff
person is selected at random, find the probability that the subject is a nurse or female.
Solution: See the text book
MULTIPLICATION RULES AND CONDITIONAL PROBABILITY
Two events A and B are independent events if the occurrence of
A does not affect the the occurrence of B
Example: Rolling two dice, drawing a card and replacing it and drawing a second card.
MULTIPLICATION RULES
1.When two events are independent, the probability of both occurring is
P(A and B) = P(A) . P(B)
2.When two events are dependent, the probability of both occurring is
P(A and B) = P(A) . P(B/A)
where P(B/A) is the conditional probability of B given that A has
occurred and is given by
P(B/A) =
P(A and B)
P(A and B)
. Also P(A/B) =
.
P(A)
P(B)
Example: A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 in the
die.
Solution: A: getting head B: getting 4
P(A and B) = P(A) . P(B) =
1 1 1
. 
(Events are independent)
2 6 12
Example: A card is drawn from a deck and replaced; then the second card is drawn. Find the probability of
getting a queen and then an ace.
Solution: A: Queen B: Ace
P(A and B) = P(A) . P(B) =
4 4
1
.

(Events are independent)
52 52 169
Example: An urn contains 3 red balls,2 blue balls and 5 white balls.A ball is selected and its colour is noted.
Then it is replaced. A second ball is selected and its colour is noted. Find the probability of each of these.
(a) Selecting 2 blue balls (b) Selecting 1 blue ball and then 1 white ball (c) Selecting 1 red ball and then 1
blue ball.
Solution:
(a) P(2 blue) =
2 2
1
. 
10 10 25
(b) P(1 blue and 1 white) =
2 5
1
.

10 10 10
(c) P(1 red and 1blue) =
3 2
3
. 
10 10 50
Example: In a pizza restaurant , 95% of the customers order pizza. If 65% of the customers order pizza and a
salad, find the probability that a customer who orders pizza will also order a salad.
Solution: P(order pizza) = 95% = 0.95, P(order pizza and salad) = 65% = 0.65
P (order pizza also order a salad) =
0.65
P(order pizza and salad)
 0.684  68.4%

0.95
P(order pizza)
Example: If 2 cards are selected from a standard deck of 52 cards without replacement, find the probabilities.
(a) Both are spades
(b) both are the same suit (c) Both are kings
Solution:
(a) P( Both are spades ) =
13 12 1
. 
(Events are dependent)
52 51 17
(b) P( Both are the same suit ) =
13 12 1
. 
(Events are dependent)
52 51 17
(c) P( Both are kings ) =
4 3
1
. 
(Events are dependent)
52 51 221
Example: A box contains black chips and white chips. A person selects two chips without replacement. If the
15
probability of selecting a black chip and a white chip is
and probability of selecting a black chip on the first
56
3
draw is , find the probability of selecting a white chip on the second draw ,given that the first chip selected
8
was a black chip.
Solution: A-Black chip B-White chip
15
P(A and B)
15 8 5
 56 
. 
P(B/A) =
3
P(A)
56 3 7
8
Example: A recent survey asked 100 people if they thought women in the armed forces should be permitted to
participate in combat.The result of the survey are shown.
Gender
Male
Female
Total
Yes
32
8
40
No
18
42
60
Total
50
50
100
Find the probabilities.
a. The respondent answered yes given that the respondent is a female
b. the respondent is a male given that the respondent answered no
Solution: (Respondent) M – Male , F- Female , (Answer) Y-Yes , N- No
a. P(Y/F) =
8
P(Y and F)
8
4
 100 

50
P(F)
50 25
100
b. P(M/N) =
18
P(M and N)
18 3
 100 

60
P(N)
60 10
100
PROBABILITIES FOR “At least’’
The multiplication rules can be used with the complementary rule to simplify solving probability problems
involving “at least”.
Example: A coin is tossed 5 times. Find the probability of getting at least 1 tail.
̅̅̅
Solution: P(E) = 1- P(𝑬)
5
1
1
1

P(At least 1 tail) = 1- P(All heads) = 1     1 
32 31
2
Example: A game is played by drawing 4 cards from an ordinary deck and replacing each card after it is drawn.
Find the probability that at least one ace is drawn.
̅̅̅
Solution: P(E) = 1- P(𝑬)
4
4
 48 
 12 
P(At least one ace) = 1- P(No aces) = 1     1     0.27
 52 
 13 
Example: It is reported 72% working women use computers at work. Choose 5 working women at random.
Find (i) the probability that at least one doesn’t use a computer at work
(ii) the probability that all the 5 use a computer in their jobs.
Solution: P(E) =
72
̅̅̅= 1 - P(E) = 1  0.72  0.38
 0.72 P(𝑬)
100
(i) P(At least 1 doesn’t use) = 1- P(All 5 use) = 1  0.72  1  0.193  0.807
5
(ii) P(All 5 use) = 0.72  0.193
5
Example: In 2006, 86% of US households had cable TV. Choose 3 households at random. Find the probability
that
a) None of the 3 households had cable TV b) All 3 households had Cable TV c) At least 1 of the 3 households
had Cable TV .
Solution: P(E) =
86
̅̅̅= 1 - P(E) = 1  0.86  0.14
 0.86 P(𝑬)
100
P(None of the 3 households had cable TV) = (0.14)3 = 0.0027 = 0.003
P(All 3 households had cable TV) = 0.86  0.636
3
P(At least 1 of the 3 households had Cable TV) = 1- P(None of the 3 households had cable TV)
= 1- 0.003= 0.997.