Introduction to Probability

Probability is used to quantify the likelihood, or chance, that an outcome of a random

experiment will occur. “The chance of rain today is 30%’’ is a statement that quantifies our
feeling about the possibility of rain. The likelihood of an outcome is quantified by assigning a
number from the interval [0, 1] to the outcome (or a percentage from 0 to 100%). Higher
numbers indicate that the outcome is more likely than lower numbers. A 0 indicates an
outcome will not occur. A probability of 1 indicates an outcome will occur with certainty.
The probability of an outcome is interpreted as the limiting value of the proportion of times the
outcome occurs in a large repetition of the random experiment. For example, if we assign
probability 0.10 to the outcome that there is a defective item in a production process, we might
interpret this as, if we inspect many items that have been produced, approximately 10% of
them will be defective.

Random Experiment: An experiment that can result in different outcomes, even though it is
repeated in the same manner every time, is called a random experiment. Example: flipping a
coin, rolling a die, measures the life time of same electronic component.
Sample Space: The set of all possible outcomes of a random experiment is called the sample
space of the experiment. The sample space is denoted as S. Sample space is discrete if it
consists of a finite or countable infinite set of outcomes. A sample space is continuous if it
contains an interval (either finite or infinite) of real numbers. Following are the examples
i. Consider an experiment in which you select a molded plastic part, such as a connector,
and measure its thickness. The possible values for thickness depend on the resolution of
the measuring instrument, and they also depend on upper and lower bounds for
thickness. However, it might be convenient to define the sample space as simply the
positive real {𝑆 = 𝑅+ = {𝑥|𝑥 > 0} line because a negative value for thickness cannot
occur. If it is known that all connectors will be between 10 and 11 millimeters thick, the
sample space could be 𝑆 = {𝑥|10 < 𝑥 < 11} .
ii. If the objective of the analysis is to consider only whether a particular part is low,
medium, or high for thickness, the sample space might be taken to be the set of three
outcomes: 𝑆 ={low, medium, high}

iii. If the objective of the analysis is to consider only whether or not a particular part
conforms to the manufacturing specifications, the sample space might be simplified to
the set of two outcomes 𝑆 = {𝑦𝑒𝑠, 𝑛𝑜} that indicates whether or not the part
conforms.

iv. If two parts (connectors) are selected and the objective of the analysis is to consider
only whether or not the parts conform to the manufacturing specifications, either part
 may or may not conform. We abbreviate 𝑦𝑒𝑠 and 𝑛𝑜 as 𝑦 and 𝑛. If the ordered pair 𝑦𝑛
indicates that the first connector conforms and the second does not, the sample space
can be represented by the four outcomes: 𝑆 = {𝑦𝑦, 𝑦𝑛, 𝑛𝑦, 𝑛𝑛}
v. If we are only interested in the number of conforming parts in the sample, we might
summarize the sample space as 𝑆 = {0, 1, 2}.
vi. As another example, consider an experiment in which the thickness is measured until a
connector fails to meet the specifications. The sample space can be represented as 𝑆 =
{𝑛, 𝑦𝑛, 𝑦𝑦𝑛, 𝑦𝑦𝑦𝑛, 𝑦𝑦𝑦𝑦𝑛, … }.

Event: An event is a subset of the sample space of a random experiment. For instance, if we
consider the previous example (1), the event 𝐴 might be “a connector will have thickness
between 10 and 10.5 millimeters” i.e., 𝐴 = {𝑥|10 < 𝑥 < 10.5} ; if we consider the previous
example (2), the event 𝐴 might be “a part will be medium or high thickness” i.e., 𝐴 ={ medium,
high}.

Tree diagrams: Sample spaces can also be described graphically with tree diagrams. When a
sample space can be constructed in several steps or stages, we can represent each of the n 1
ways of completing the first step as a branch of a tree. Each of the ways of completing the
second step can be represented as n2 branches starting from the ends of the original branches,
and so forth.

Suppose three coins will be tossed. The possible outcomes for three tosses can be displayed by
eight branches in the tree diagram shown in the following figure
Items selected with replacement and without replacement
In random experiments in which items are selected from a batch, we will indicate whether or
not a selected item is replaced before the next one is selected.
Counting rules
As sample spaces become larger, complete enumeration is difficult. Instead, counts of the
number outcomes in the sample space and in various events are often used to analyze the
random experiment. These methods are referred to as counting techniques as described below
Number of possible ways if we take r objects from n objects
Without replacement With replacement
(WOR) (WR)

Ordered 𝑛!
𝑛𝑃𝑟 = 𝑛𝑟
(permutation) (𝑛 − 𝑟)!

Unordered 𝑛!
𝑛𝐶𝑟 = (𝑛 + 𝑟 − 1)𝐶 𝑟
(Combination) 𝑟! (𝑛 − 𝑟)!

Where,
𝑛 = number of items which can be selected.
𝑟 = number of items which are selected.

Choices: With or without replacement?

When you are choosing several things, can you choose the same thing more than once? If so,
you are choosing with replacement.
 Example: You buy bagels at a bakery. You can buy more than one blueberry bagel if you
want. This is choosing with replacement.
 Example: You select five players for a basketball team from a pool of 20 candidates. You
can not pick the same person more than once. This is choosing without replacement.

Choices: Ordered or unordered?

When you are choosing several things, does the order in which you select them matter? If so,
you are making an ordered choice.
 Example: You select five players for a basketball team for a pickup game in the park. All
five will run on the court together, without organizing positions. This is an unordered
choice.
 Example: You select five players for a basketball team for a formal game with fixed
positions: center, power forward, small forward, shooting guard, and point guard. This
is an ordered choice.
Example 1: We have five different candy bars to distribute to 20 children. We do not want to
give any child more than one candy bar. Is this selection with replacement or without? Is it
unordered or ordered? How many ways can we distribute the candy?

Solution: We are choosing r = 5 children from n = 20 possibilities. No child gets more than one,
so this is without replacement. The candy bars are different, so this is ordered. There are
𝑛! 20!
𝑛𝑃𝑟 = (𝑛−𝑟)! = 15! = 20 × 19 × 18 × 17 × 16 = 1860480 ways.

Example 2: We have five identical candy bars to distribute to 20 children. We do not want to
give any child more than one candy bar. Is this selection with replacement or without? Is it
unordered or ordered? How many ways can we distribute the candy?

Solution: We are choosing r = 5 children from n = 20 possibilities. No child gets more than one,
so this is without replacement. The candy bars are identical, so this is unordered. There are
𝑛! 20!
𝑛𝐶𝑟 = 𝑟!(𝑛−𝑟)! = 5!×15! = 15504 ways.

Example 3: We have five identical candy bars to distribute to 20 children. We are willing to give
some children more than one candy bar. Is this selection with replacement or without? Is it
unordered or ordered? How many ways can we distribute the candy?
Solution: We are choosing r = 5 children from n = 20 possibilities. Children can get more than
one, so this is with replacement. The candy bars are identical, so this is unordered. There are
24!
(𝑛 + 𝑟 − 1)𝐶 𝑟 = 5!×19! = 42504 ways.

Example 4: We have five different candy bars to distribute to 20 children. We are willing to give
some children more than one candy bar. Is this selection with replacement or without? Is it
unordered or ordered? How many ways can we distribute the candy?

Solution: We are choosing r = 5 children from n = 20 possibilities. Children can get more than
one, so this is with replacement. The candy bars are different, so this is ordered. There are
205 = 3200000 ways.

Example 5: There are 20 computers in a store. Among them, 15 are brand new and 5 are
refurbished. How many ways are there to choose 7 computers that includes 3 refurbished
computers?
Solutions: For the number of favorable outcomes, 3 refurbished computers are selected from a
total of 5, and the remaining 4 new ones are selected from a total of 15.
There are 5𝐶3 × 15𝐶4 = 10 × 1365 = 13650 different ways.

Example 6: In a library, there are 4 books on fairy tales, 5 books are novels and 3 books are on
plays. In how many ways can you arrange these so that books on the fairy tales are together in one
place, the novels are together, and plays are also together. The requirement is that these books
should be in a specific order i.e., books on fairy tales, before novels, before plays.

Solution: There are 4 books on fairy tales and they have to be put together. They can be arranged
in 4! ways. Similarly, there are 5 novels. They can be arranged in 5! ways. Also there are 3 books on
plays. They can be arranged in 3! ways. So, by the counting principle all of them together can be
arranged in 4! × 5! × 3! ways = 17280 ways.

Example 7: In the above example what is the number of permutations if the books are not to be
kept in order?
Solution: Whenever you are asked to keep a particular class of objects together, a convenient trick
is to sort of glue them together in your head and treat them as one object. First, we consider the
books on fairy tales, novels and plays as single objects. These three objects i.e the one group of
fairy tale books, the one group of novels and the one group of plays can be arranged in 3! ways = 6
ways. Let us fix one of these 6 arrangements. This may give us a specific order, say, novels → fairy
tales → plays. Given this order, the books on the same subject can be arranged as follows. In other
words, now we have to count the internal permutations. The 4 books on fairy tales can be
arranged among themselves in 4! = 24 ways.
The 5 novels can be arranged in 5! = 120 ways. The 3 plays can be arranged in 3! = 6 ways. For a
given order, the books can be arranged in 24×120×6 = 17280 ways.
Therefore, for all the 6 possible orders the books can be arranged in 6×17280 = 103680 ways.

Exercise:
1. Electronic device usually require a personal code to operate. This particular device uses
4-digits code. Calculate how many codes are possible from 10 digits. Ans. 10000.
2. In how many ways can 3 students be seated in a classroom with 5 chairs? Ans. 5𝑃3 .
3. In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are
together? Ans. 4! × 6! = 17280.
4. There are 6 boys who enter a boat with 8 seats, 4 on each side. In how many ways can they
sit anywhere on the boat? Ans. 8𝑃6 = 20160.
5. How many different 7-place license plates are possible if first three places are to be
occupied by letters and final 4places are numbers? Ans. 263 × 104 .
6. There are four girls and six boys in a class. They are arranged randomly at 10 seats.
i. How many different arrangements are possible? Write five of them. Ans. 10! .
ii. How many arrangements are possible where girls will be occupied first four
seats? Write three of them. Ans. 4! × 6! .
iii. How many ways are there where a particular boy will be occupied the 7th seat?
Ans. 9! .
Assignment
1. Suppose that after 10 years of service, 40% of computers have problems with motherboards
(MB), 30% have problems with hard drives (HD), and 15% have problems with both MB and
HD. What is the probability that a 10-year old computer still has fully functioning MB and
HD? Ans: 0.45
2. A new computer virus can enter the system through e-mail or through the internet. There is a
30% chance of receiving this virus through e-mail. There is a 40% chance of receiving it
through the internet. Also, the virus enters the system simultaneously through e-mail and the
internet with probability 0.15. What is the probability that the virus does not enter the system
at all? Ans: 0.45
3. In a college of 675 students, 320 studied mathematics, 385 studied history, and 150 studied
both mathematics and history. If one of these students is selected at random, find
(i). the probability that the student took mathematics or history. Ans: 0.82
(ii). the student did not take either of these subjects. Ans: 0.18
(iii). the student took history but not mathematics. Ans: 0.35
4. A group contains 3 students. If we want to take 2 students in order to form a queue, how
many ways are possible? Ans: 6
5. A group contains 4 students: 2 male and 2 female. Suppose we want to take 3 students at
random in order to form a queue. What is the probability that a female is in the middle
position? Ans: (2 × 3𝑃2 )⁄4𝑃3 = 0.5
6. A group contains 20 students. If we want to take 4 students in order to participate a
campaign, how many ways are possible? Ans: 4845
7. A group contains 6 students: 2 male and 4 female. Suppose we want to take 3 students at
random in order to participate a campaign. How many ways we can select 3 students from 6?
List all the outcomes in the sample space.
8. Refer to problem 7, find the probability that
4
(a) all female students are selected? Ans: (2𝐶0 × 4𝐶3 )⁄6𝐶3 = 20
(b) all male students are selected? Ans: 0
12
(c) 2 females are selected? Ans: (2𝐶1 × 4𝐶2 )⁄6𝐶3 = 20
4
(d) 2 males are selected? Ans: (2𝐶2 × 4𝐶1 )⁄6𝐶3 =
20
9. Suppose 2 coins are flips simultaneously. How many outcomes are possible? Ans: 4
10. Suppose 3 fair coins are flips simultaneously. What is the probability of getting exactly 2
heads? Ans: 3𝐶2 ⁄23
11. Fifty-two percent of the students at a certain college are females. Five percent of the students
in this college are majoring in computer science. Two percent students are female and
majoring in computer science. If a student is selected at random, find the conditional
probability that
(a) this student is female, given that the student is majoring in computer science;
(b) this student is majoring in computer science, given that the student is female.
12. Armco, a manufacturer of traffic light systems, found that under accelerated-life tests, 95
percent of the newly developed systems lasted 3 years before failing to change signals
properly.
(a) If a city purchased four of these systems, what is the probability all four systems would
operate properly for at least 3 years?
(b) Which rule of probability does this illustrate?