PROBABILITY AND COUNTING RULES

Sample space

Some sample spaces for various probability experiments are shown here

Ex. Rolling Dice

Find the sample space for rolling two dice.

Solution:
Since each die can land in six different ways, and two dice are rolled, the sample space can be presented
by a rectangular array, as shown below. The sample space is the list of pairs of numbers in the chart.
 Ex. Drawing Cards
Find the sample space for drawing one card from an ordinary deck of cards.

Solution
Since there are 4 suits (hearts, clubs, diamonds, and spades) and 13 cards for each suit
(ace through king), there are 52 outcomes in the sample space. See Figure below

Ex. Gender of Children

Find the sample space for the gender of the children if a family has three children. Use
B for boy and G for girl.

Solution
There are two genders, male and female, and each child could be either gender. Hence,
there are eight possibilities, as shown here.
Other way to get the sample space : Tree diagram

Tree diagram is a device consisting of line segments emanating from a starting point and from the outcome point. It is used to
determine all possible outcomes of a probability experiment.

Ex. Gender of Children

Use a tree diagram to find the sample space for the gender of the children if a family has three children. Use B for boy and G for girl.

Solution

Since there are two possibilities (boy or girl) for the first child, draw two branches from a starting point and label one B and the other G.
Then if the first child is a boy, there are two possibilities for the second child (boy or girl), so draw two branches from B and
label one B and the other G. Do the same if the first child is a girl. Follow the same procedure for the third child. The completed tree
diagram is below. To find the outcomes for the sample space, trace through all the possible branches, beginning at the starting point for
each one.
 Events

Event - consists of a set of outcomes of a probability experiment.

1. Compound events – are the outcomes with several expected results.

2. Simple event – is an event which cannot be decomposed.

Example on Compound Event:

1. Toss a die and observe the number appearing on the upper face.
Event A: Observe an Odd number
Solution:
A = { 1, 3,5}
Event B: Observe a number less than 4.
Solution:
B = { 1, 2, 3}
2. Toss two coins and observe the outcomes
Solution:
Coin 1 Coin 2
Events 1 H H
Event 2 H T
Event 3 T H
Event 4 T T

Example on Simple Events:

1.Toss a die once and observe the number appearing on the upper face.
Solution:
Event 1 Observe a 1
Event 2 Observe a 2
Event 3 Observe a 3
Event 4 Observe a 4
Event 5 Observe a 5
Event 6 Observe a 6

2. Toss a coin once

Solution:
Event 1 Observe a head
Event 2 Observe a tail
 Set Operations

1. Complement(‘) – denoted by A’
-Complement of an event is the set of all outcomes in a sample space that rae not
contained in that event.
Ex. Roll a die once.
Solution:
S ={ 1,2,3,4,5,6}
Let A be the event that an even number appears on the die
A = { 2,4,6}
A’ = { 1,3,5}
2. Intersection ( Ո )
- Intersection of two events is the event consisting of all outcomes that are in both events.
Example:
A = { 1, 2,3,4} ; B ={ 1,3,5}
A Ո B = {1,3}

3. Union ( Ս )
- Union of two events consisting of all outcomes that are either in A or in B or in both events.
Example:
A ={ 1,2,3,4} ; B ={4,5,6}
A Ս B = { 1,2,3,4,5,6}
Joint Sets – are those events with common outcome/outcomes
Ex.
A ={ 2,4,6} ; B= { 1, 2,3,5}
A Ո B = { 2}
Disjoint sets (mutually exclusive) are sets where two events have no common outcome.

Probability
Definition:
- If an experiment is repeated a large number of time N and A is observed n times, the
probability of E is

P( E) = n/N

1. Probability of an event can be greater or equal to zero; or less than or equal to 1

2. The sum of the probability over the entire sample space is 1.
Ex.
Calulate the probability of Event E.
A die is tossed once. Let E be the probability of getting an odd number on the upper face of a
die.
Solution:
S = { 1,2,3,4,5,6}
E ={ 1,3,5}
P( E ) = 3/6 = 1/2

Ex. Drawing Cards

Find the probability of getting a black 10 when drawing a card from a deck .

Solution:
 There are four basic probability rules

Rule 1 states that probabilities cannot be negative or greater than 1.

Ex. Rolling a Die

When a single die is rolled, find the probability of getting a 9.

Solution

S = { 1,2,3,4,5,6}

Since it is impossible to get 9

P(9) = 0
Ex. Rolling a Die
When a single die is rolled, what is the probability of getting a number less than 7?

Solution
 Counting Sample Points

Theorem:
- with m elements a1,a2,a3,...am and n elements b1,b2,b3,...bn it is possible to form mn
pairs containing one element from each group.
Ex.
Two dice are tossed. How many sample points are associated with the experiment?
Solution:
The first die can fall I one of the 6 ways, m= 6, likewise the second die can fall in n=6 ways.
The total number,N of sample points is

N = mn
N = 6(6)
N = 36
 Ways of Determining the Sample Points

1. Enumeration
ex. Toss a die once, what are elements in sample space?

2. Tree diagram Diagram can be be used to represent pictorally all possible events.
Ex. Calculate the probability of observing exactly one head in a toss of two coins.
Solution:
A tree diagram can be drawn to come up with th epossible outcomes

S ={ HH,HT,TT,TH}
H
P(HH) = ¼
H P(HT) = ¼
T P(TH) = ¼
P(TT) = ¼
H
T Let E be the event that exactly one head
appears
T
P(E) = P(HT)+P(TH)

P(E) = (¼) + (1/4) = ½

3. Multiplication Rule
If the first element or object of an ordered pair can be selected in n1 ways anad for each of
these n1 ways the second element of the pair can be selected in n2 ways, then the number of
pair n1(n2)=N.
Ex.
Toss a coin once then roll a die once. Find the number of possibilities.
Solution:
Let n1 represent the number of outcomes in tossing a coin n1= 2 (iether head or tail)
Let n2 represnet the number of outcomess in rolling a die n2 = 6 (yhe number of face of a die)

N = n1(n2)
N = 2 (6)
N = 12 possibilities
 PERMUTATION

-Is an ordered arrangement of r distinct objects. The number of ways of ordering n distinct objects r
at a time will be designated by the symbol, nPr.
nPr = n(n-1)(n-2)...(n-r+1)

Expressed in terms of factorial

nPr = n!___
(n-r)!

Ex. Three lottery tickets are drawn from a total of fifty. Assume that order is of importance?
Solution:
To solve for total number of sample points
𝟓𝟎!
50P3 =
𝟓𝟎−𝟑 !
𝟓𝟎!
=
𝟒𝟕!
𝟓𝟎 𝟒𝟗 𝟒𝟖 𝟒𝟕!
=
𝟒𝟕!

50P3 = 117,600 samaple points

Ex. A piece of equipment i s composed of five parts which may in any order. A test is to be
conducted to determine the lenght of time necessary fro each order of assembly. If each order
is to be tested once, how many test must be conducted?
Solution:
𝟓!
5P5 =
𝟓−𝟓 !

𝟓!
=
𝟎!

5P5 = 120
 Circular Permutation Of Objects nPn = (n-1)!

Ex.
1. In how many ways can eight guest be seated in around table with eight chairs?
nPn = (n-1)!
8P8 = (8-1)!
= 7!
8P8 = 5,040 ways

2. Suppose there are 50 beads of different colors to form a necklace, in how many ways can they be
joined?
nPn = (n-1)!
50P50 = (50-1)!
= 49!
50P50 = 6.08 x1062
 Permutation of objects some of which are of the same kind

n!
nPn =
n1!n2!...nk!
Ex.
1. How many permutations can be made with word TETRAMETER?
n = 10 letters ; T = 3 ; E = 3 ; R = 2; A = 1; M = 1
10!
10P10 =
3!3!2!1!1!
10P10 = 50,400 ways

2. In how many ways can these lights be arranged on a string of christmas lights, if there are
2 green, 5 blue, 3 red and 2 yellow light?
12!
1. 12P12 =
2!5!3!2!
2. 12P12 = 166,320 ways
 Permutation of objects that are grouped into cells

n!
nPn =
n1!n2! n3!...nr!

Where: n1!n2! n3!...nr! =n

Ex. A college plays 12 football games during a season. In how many ways can the team end the
season with 7 wins, 3 losses and 2 ties?

N = 12 ; wins = 7 ; losses = 3 ; ties = 2

12!
12P12 =
7!3! 2!
= 7,920 ways
 Combination

The number of combinations of n objects taken r at a time will be denoted by the symbol, nCr.

nPr
nCr =
r!
n!
nCr =
r!(n-r)!

Ex.
A radio tube may be purchased from five suppliers. In how many ways can
three suppliers be chosen from the five?

5!
5 C3 =
3! 5−3 !

5(4)(3!)
=
3!2!

5 C3 = 10 ways
Ex. Five manufacturer, of varying but unknown quality, produce a certain type of electrical tube. If we are to select three
manufacturers at random, what is the chance that the selectrion would contain exactly two of the best three.
Solution:

Let n be the number of points in which two of the best three manufacturers are selected.
𝑛
P=
𝑁
- Our problem is to use the counting rules to find n and N.
- Since the selection is a combination

5!
N = 5 C3 =
3!2!
N = 10

Let a be the number of ways of selecting two from the best three
3!
3 C2 = =3
2!1!

Let b be the number of ways of selecting the remaining manufacturer from the two poorest
2!
2 C1 = =2
1!1!

The total number of ways of choosing two of the best three in a selection of three is
n = ab = 3(2) = 6
𝑛
P=𝑁
6
P = 10
 PROBABILITY OF AN EVENT

The probability of an event A is the sum of the weight (probabilities) of all sample pointa in A,
therefore
0≤P(A) ≤1, P(Ф) = 0 , P(S) =1
P(Ф) – sure not to happen
P(S) – sure to happen
1. Toss a balanced coin.
Solution:
S = { H, T}
1 probability (weight) that it will be a head + 1 probability(weight) that it will be a tail = 1
1W + 1W =1
2W = 1
W = 1/2
2. Toss a balanced coin 3 times, find the probability of getting at least two heads.
H
H
T
H H
T
T
H H
T T
H
T
T
 S = { HHH, HHT, HTT, HTH,TTH,THH,THT,TTT}
no. of possible outcomes = 8
8W =1
w = 1/8
Let A be the event that at least two heads appears on three losses of coin
A = {HHH, HHT, HTH, THH}
1 1 1 1
p(A) = 8 + 8+ 8+ 8

= 4/8
= 1/2
 Theorem:

If an experiment can result in one of N different equally likely outcomes, and if exactly n of these
outcomes correspondence to event A, then the probability of event A is ,
𝑛
P(A) =
𝑁
Ex. In a poker hand consisting of five cards, find the probability of holding,
a. 3 aces
Solution:
a) Let A be the probability of getting 3 aces.
- Number of ways of being dealt with 3 aces from 4 is :
4 4!
4 C3 =( )= =4
3 3!1!

- Number of ways of being dealt with 2 cards, 48 of which are not aces
48 48!
- 48C2 =( )= = 1128
2 2!46!

4 48
n= ( ) ( ) = 4512
3 2
52 52!
N=( )= = 2598960
5 5!47!

4512 94
P(A) = =
2598960 54145
Ex. Given: 7 green, 5 yellow and 3 blue bulbs.
a.) Find the probability of getting a green bulb.
b.) Probability of getting 2 yellow bulbs and a blue bulb.
c.) Find the probability of getting 2 green and 2 yellow when 5 light bulbs are drawn.
d.) find the probability of getting 3 light bulbs of different colors.
For:
a.) Find the probability of getting a green bulb.
Solution:
N = 7+5+3 = 15

7 7!
n=( ) = =7
1 1!6!

7
P(G) =
15

b.) Probability of getting 2 yellow bulbs and a blue bulb.

5 3
n= ( )( )
2 1
5! 3!
n = (2!3!) (1!2!)

n = 10 (3)
n = 30
15!
N = (3!12!) = 455

Let A be the event of getting 2 yellow and a blue bulb

30 6
P(A) = = 91
455
c.) Find the probability of getting 2 green and 2 yellow when 5 light bulbs are drawn.
Solution:
7 5 3
n= ( )( )( )
2 2 1
7! 5! 3!
=( )( )( )
2!5! 2!3! 1!2!

n = 630
15
N=( )
5
15!
= ( )
5!10!

N = 3003
Let B the event of getting 2 green and 2 yellow when 5 light bulb are drawn
630
P(B) =
3003
30
=
143

d.) find the probability of getting 3 light bulbs of different colors.

Let C be the event of getting 3 light bulbs of different colors
( 72) ( 51) ( 31)
P(C) =
( 15
3
)
(1!6!
7!
) (1!4!
5!
) (1!2!
3!
) 105
= =
(3!12!)
15! 455

3
P(C) =
13