LESSON 2: PROBABILITY

Introduction
Probability is simply how likely an event is to happen. “The chance of rain today is
50%” is a statement that enumerates our thoughts on the possibility of rain. The
likelihood of an outcome is measured by assigning a number from the interval [0, 1]
or as percentage from 0 to 100%. The higher the number means the event is more
likely to happen than the lower number. A zero (0) probability indicates that the
outcome is impossible to happen while a probability of one (1) indicates that the
outcome will occur inevitably.

This module intends to discuss the concept of probability for discrete sample spaces,
its application, and ways of solving the probabilities of different statistical data.

Intended Learning Outcomes

At the end of this module, it is expected that the students will be able to:

1. Understand and describe sample spaces and events for random experiments

2. Explain the concept of probability and its application to different situations

3. Define and illustrate the different probability rules

4. Solve for the probability of different statistical data.

PROBABILITY

Probability is the likelihood or chance of an event occurring.

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑎𝑐ℎ𝑖𝑒𝑣𝑖𝑛𝑔 𝑠𝑢𝑐𝑐𝑒𝑠𝑠

/ 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

For example, the probability of flipping a coin and it being heads is 1⁄2, because
there is 1 way of getting a head and the total number of possible outcomes is 2 (a
head or tail). We write P(heads) = 1⁄2 .

• The probability of something which is certain to happen is 1.

• The probability of something which is impossible to happen is 0.
• The probability of something not happening is 1 minus the probability that it
will happen.

Experiment – is used to describe any process that generates a set of data.

Event – consists of a set of possible outcomes of a probability experiment. Can be

one outcome or more than one outcome.

Simple event – an event with one outcome.

Compound event – an event with more than one outcome.

2.1 Sample Space and Relationships among Events
Sample space is the set of all possible outcomes or results of a random experiment.
Sample space is represented by letter S. Each outcome in the sample space is called
an element of that set. An event is the subset of this sample space, and it is
represented by letter E. This can be illustrated in a Venn Diagram. In Figure 2.1, the
sample space is represented by the rectangle and the events by the circles inside
the rectangle.

The events A and B (in a to c) and A, B and C (in d and e) are all subsets of the
sample space S.

Figure 2.1 Venn diagrams of sample space with events (adapted from Montgomery
et al., 2003)

For example, if a dice is rolled we have {1, 2, 3, 4, 5, and 6} as sample space. The
event can be {1, 3, and 5} which means set of odd numbers. Similarly, when a coin
is tossed twice the sample space is {HH, HT, TH, and TT}.

Difference between Sample Space and Events

As discussed in the beginning sample space is set of all possible outcomes of an

experiment and event is the subset of sample space. Let us try to understand this
with few examples. What happens when we toss a coin thrice? If a coin is tossed
three times we get following combinations,

HHH, HHT, HTH,THH, TTH, THT, HTT and TTT

All these are the outcomes of the experiment of tossing a coin three times. Hence,
we can say the sample space is the set given by,

𝑆 = {𝐻𝐻𝐻, 𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝑇𝐻𝐻, 𝑇𝑇𝐻, 𝑇𝐻𝑇, 𝐻𝑇𝑇, 𝑇𝑇𝑇}

Now, suppose the event be the set of outcomes in which there are only two heads.
The outcomes in which we have only two heads are HHT, HTH and THH hence the
event is given by,

𝐸 = {𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝑇𝐻𝐻}

 We can clearly see that each element of set E is in set S, so E is a subset of S. There
can be more than one event. In this case, we can have an event as getting only one
tail or event of getting only one head. If we have more than one event we can
represent these events by E1, E2, E3 etc. We can have more than one event for a
Sample space but there will be one and only one Sample space for an Event. If we
have Events E1, E2, E3, ...... En as all the possible subset of sample space then we
have,

𝑆 = 𝐸1 ∪ 𝐸2 ∪ 𝐸3 ∪ . . . . . . .∪ 𝐸𝑛

We can understand this with the help of a simple example. Consider an experiment
of rolling a dice. We have sample space,

𝑆 = {1, 2, 3, 4, 5, 6}

Now if we have Event E1 as getting odd number as outcome and E2 as getting even
number as outcome for this experiment then we can represent E1 and E2 as the
following set,

𝐸1 = {1, 3, 5}

𝐸2 = {2, 4, 6}

So we have,

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

𝑂𝑟 𝑆 = 𝐸1 ∪ 𝐸2

Hence, we can say union of Events E1 and E2 is S. Null space – is a subset of the
sample space that contains no elements and is denoted by the symbol ∅. It is also
called empty space.

Operations with Events

Intersection of events

The intersection of two events A and B is denoted by the symbol A ꓵ B. It is the

event containing all elements that are common to A and B. This is illustrated as
the shaded region in Figure 2.1 (c).

For example,

Let 𝐴 = {3,6,9,12,15} and 𝐵 = {1,3,5,8,12,15,17}; then 𝐴 ꓵ 𝐵 = {3,12,15}

Let 𝑋 = {𝑞, 𝑤, 𝑒, 𝑟, 𝑡, } and 𝑌 = {𝑎, 𝑠, 𝑑, 𝑓}; then 𝑋 ꓵ 𝑌 = ∅, since X and Y have no

elements in common.

Mutually Exclusive Events

We can say that an event is mutually exclusive if they have no elements in common.
This is illustrated in Figure 2.1 (b) where we can see that A ꓵ B =∅.
Union of Events

The union of events A and B is the event containing all the elements that belong to
A or to B or to both and is denoted by the symbol A B. The elements A ∪ B
maybe listed or defined by the rule 𝐴 ∪ 𝐵 = {𝑥 | 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}.

For example,

Let 𝐴 = {𝑎, 𝑒, 𝑖, 𝑜, 𝑢} and 𝐵 = {𝑏, 𝑐, 𝑑, 𝑒, 𝑓}; then 𝐴 ∪ 𝐵 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑖, 𝑜, 𝑢}

Let 𝑋 = {1,2,3,4} and 𝑌 = {3,4,5,6}; then 𝐴 ∪ 𝐵 = {1,2,3,4,5,6}

Compliment of an Event

The complement of an event A with respect to S is the set of all elements of S that
are not in A and is denoted by A’. The shaded region in

Figure 2.1 (e) shows (A ∪ C)’.

For example,

Consider the sample space 𝑆 = {𝑑𝑜𝑔, 𝑐𝑜𝑤, 𝑏𝑖𝑟𝑑, 𝑠𝑛𝑎𝑘𝑒, 𝑝𝑖𝑔}

Let 𝐴 = {𝑑𝑜𝑔, 𝑏𝑖𝑟𝑑, 𝑝𝑖𝑔}; then 𝐴’ = {𝑐𝑜𝑤, 𝑠𝑛𝑎𝑘𝑒}

Probability of an Event

Sample space and events play important roles in probability. Once we have sample
space and event, we can easily find the probability of that event. We have following
formula to find the probability of an event.

Probability of an event = number of elements in the event set / number of elements

in sample space of an experiment

𝑃(𝐸) = 𝑛(𝐸) / 𝑛(𝑆)

Where,

n (S) represents number of elements in a sample space of an experiment;

n (E) represents a number of elements in the event set; and

P (E) represents the probability of an event.

When probabilities are assigned to the outcomes in a sample space, each

probability must lie between 0 and 1 inclusive, and the sum of all probabilities
assigned must be equal to 1. Therefore,

0 ≤ 𝑃 (𝐸) ≤ 1 𝑎𝑛𝑑 𝑃(𝑆) = 1

Let us try to understand this with the help of an example. If a die is tossed, the
sample space is {1, 2, 3, 4, 5, 6}. In this set, we have a number of elements equal
to 6. Now, if the event is the set of odd numbers in a dice, then we have {1, 3, and
5} as an event. In this set, we have 3 elements. So, the probability of getting odd
numbers in a single throw of dice is given by

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 3/6 = 1/2

2.2 Counting Rules Useful in Probability

Multiplicative Rule / Fundamental Counting Rules

Suppose you have j sets of elements, n1 in the first set, n2 in the second set, ...
and nj in the jth set. Suppose you wish to form a sample of j elements by taking
one element from each of the j sets. The number of possible sets is then defined
by:

𝑛1 ∙ 𝑛2 ∙ . . .∙ 𝑛𝑗

Permutation Rule

The arrangement of elements in a distinct order is called permutation. Given a

single set of n distinctively different elements, you wish to select k elements from
the n and arrange them within k positions. The number of different permutations
of the n elements taken k at a time is denoted nPc and is equal to

𝑛𝑃𝑘 = 𝑛!/(𝑛 − 𝑘)!

Partitions rule

Suppose a single set of n distinctively different elements exists. You wish to

partition them into k sets, with the first set containing n1 elements, the second
containing n2 elements, ..., and the kth set containing nk elements. The number
of different partitions

is

𝑛!/ 𝑛1! 𝑛2! . . . 𝑛𝑘!

Where,

𝑛1 + 𝑛2 + . . . + 𝑛𝑘 = 𝑛

The numerator gives the permutations of the n elements. The terms in the
denominator remove the duplicates due to the same assignments in the k sets
(multinomial coefficients).

Combinations Rule

A sample of k elements is to be chosen from a set of n elements. The number of

different samples of k samples that can be selected from n is equal to

(𝑛𝐶𝑘) = 𝑛! / 𝑘! (𝑛 − 𝑘)!

2.3 Rules of Probability

Before discussing the rules of probability, we state the following definitions:

Two events are mutually exclusive or disjoint if they cannot occur at the same
time.

The probability that Event A occurs, given that Event B has occurred, is called a
conditional probability. The conditional probability of Event A, given Event B, is
denoted by the symbol P (A|B).

The complement of an event is the event not occurring. The probability that Event
A will not occur is denoted by P (A').
 The probability that Events A and B both occur is the probability of the intersection
of A and B. The probability of the intersection of Events A and B is denoted by P
(A ∩ B). If Events A and B are mutually exclusive, P (A ∩ B) = 0.

The probability that Events A or B occur is the probability of the union of A and B.
The probability of the union of Events A and B is denoted by P(A ∪ B).

If the occurrence of Event A changes the probability of Event B, then Events A and
B are dependent. On the other hand, if the occurrence of Event A does not change
the probability of Event B, then Events A and B are independent.

Rule of Addition

Rule 1: If two events A and B are mutually exclusive, then:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵)

Rule 2: If events A and B are not mutually exclusive events, then:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

Rule of Multiplication

Rule 1: When two events A and B are independent, then:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)

Dependent - Two outcomes are said to be dependent if knowing that one of the
outcomes has occurred affects the probability that the other occurs.

Conditional Probability - an event B in relationship to an event A is the probability

that event B occurs after event A has already occurred. The probability is
denoted by P(B|A).

Rule 2: When two events are dependent, the probability of both occurring is:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴)

Where 𝑃(𝐵|𝐴) = 𝑃(𝐴 ∩ 𝐵)/𝑃(𝐴), provided that 𝑃 (𝐴) ≠ 0.

Example 1. A day’s production of 850 manufactured parts contains 50 parts that do

not meet customer requirements. Two parts are selected randomly without
replacement from the batch. What is the probability that the second part is
defective given that the first part is defective?

Solution:

Let A = event that the first part selected is defective

Let B = event that the second part selected is defective.

𝑃 (𝐵|𝐴) =?

If the first part is defective, prior to selecting the second part, the batch contains
849 parts, of which 49 are defective, therefore

𝑃 (𝐵|𝐴) = 49/849
 Example 1. A student goes to the library. The probability that she checks out (a) a
work of fiction is 0.40, (b) a work of non-fiction is 0.30, and (c) both fiction and
non-fiction is 0.20. What is the probability that the student checks out a work of
fiction, non-fiction, or both?

Solution:

Let F = the event that the student checks out fiction.

Let N = the event that the student checks out non-fiction.

Then, based on the rule of addition:

𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐹) + 𝑃(𝑁) − 𝑃(𝐹 ∩ 𝑁)

𝑃(𝐴 ∪ 𝐵) = 0.4 + 0.3 − 0.2 = 0. 5

Rule of Multiplication

Rule 1: When two events A and B are independent, then:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)

Dependent - Two outcomes are said to be dependent if knowing that one of the
outcomes has occurred affects the probability that the other occurs.

Conditional Probability - an event B in relationship to an event A is the probability

that event B occurs after event A has already occurred. The probability is
denoted by P(B|A).

Rule 2: When two events are dependent, the probability of both occurring is:

𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴)

Where 𝑃(𝐵|𝐴) = 𝑃(𝐴 ∩ 𝐵)/𝑃(𝐴), provided that 𝑃 (𝐴) ≠ 0

Example 1. A day’s production of 850 manufactured parts contains 50 parts that do not
meet customer requirements. Two parts are selected randomly without replacement
from the batch. What is the probability that the second part is defective given that the
first part is defective?

Solution:

Let A = event that the first part selected is defective

Let B = event that the second part selected is defective.

𝑃 (𝐵|𝐴) =?

If the first part is defective, prior to selecting the second part, the batch contains
849 parts, of which 49 are defective, therefore

𝑃 (𝐵|𝐴) = 49/849

Example 2. An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn
without replacement from the urn. What is the probability that both marbles are black?

Solution:

Let A = the event that the first marble is black;

and let B = the event that the second marble is black.

We know the following:

In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore,
𝑃(𝐴) = 4/10.

After the first selection, there are 9 marbles in the urn, 3 of which are black.
Therefore, 𝑃 (𝐵|𝐴) = 3/9.

𝑃(𝐴 ∩ 𝐵) = (4/10) (3/9) = 0. 133

Example 3. Two cards are selected from a pack of cards. What is the probability that
they are both queen?

Solution:

Let A = First card which is a queen

Let B = Second card which is also a queen

We require P (A ∩ B). Notice that these events are dependent because the probability
that the second card is a queen depends on whether or not the first card is a queen.

𝑃 (𝐴 ∩ 𝐵) = 𝑃 (𝐴) 𝑃 (𝐵|𝐴)

𝑃 (𝐴) = 1/13 𝑎𝑛𝑑 𝑃 (𝐵|𝐴) = 3/51

𝑃 (𝐴 ∩ 𝐵) = (1/13) (3/51) = 1/221 = 0.004525

Rule of Subtraction

The probability that event A will occur is equal to 1 minus the probability that event A
will not occur.

𝑃(𝐴) = 1 − 𝑃(𝐴′)

Example 1. The probability of Bill not graduating in college is 0.8. What is the
probability that Bill will not graduate from college?

Solution:

𝑃(𝐴) = 1 − 0.8 = 0. 2

Prepared by:

Engr. Loara Adyz Acosta-Villa

 PROBLEM SET

1. A menu offers a choice of 3 salads, 8 main dishes, and 5 desserts. How many
different meals consisting of one salad, one main dish, and one dessert are
possible?
2. How many 4-digit debit card personal identification numbers (PIN) can be
made?
3. How many ways can the three letters a, b, and c be arranged with no letters
repeating?
4. How many ways can you arrange five people standing in line?
5. Critical Miss, PSU's Tabletop Gaming Club, has 15 members this term. How many
ways can a slate of 3 officers consisting of a president, vice-president, and
treasurer be chosen?
6. Critical Miss, PSU's Tabletop Gaming Club, has 15 members this term. They
need to select 3 members to have keys to the game office. How many ways can
the 3 members be chosen?
7. What is the probability of winning the jackpot in the Pick-4 Lottery? To play
Pick-4, you choose 4 numbers from 0 to 9. This will give you a number between
0000 and 9999. You win the jackpot if you match your 4 numbers in the exact
order they are drawn.
8. What is the probability of winning the Powerball jackpot? As of 2021, the
Powerball lottery consists of drawing five white balls in any order numbered 1
through 69, and one red Powerball numbered 1 through 26.
9. Three events are shown on the Venn diagram in the following figure:

Reproduce the figure and shade the region that corresponds to each of the
following events.

a. A’ b. A ∩ B c. (A ∩ B) ∪ C d. (B ∪ C)’ e. (A ∩ B)’ ∪ C

10. Each of the possible five outcomes of a random experiment is equally likely.
The sample space is {a, b, c, d, e}. Let A denote the event {a, b}, and let B
denote the event {c, d, e}. Determine the following:
a. P(A) b. P(B) c. P(A’) d. P(A ∩ B) e. P(A ∪ B)
11. If A, B, and C are mutually exclusive events with P (A) = 0.2, P(B) = 0.3, and
P(C) = 0.4, determine the following probabilities:
a. P(A ∪ B ∪ C) c. P(A ∩ B) e. P(A’ ∩ B’ ∩ C’)

b. P(A ∩ B ∩ C) d. P[(A ∩ B) ∪ C]