Untitled Lesson

Unit 4: Probability Distribution

Learning Material in Math 203 – Statistics and Probabilities

Graduate Studies, 1st Semester AY 2022 – 2023

Unit 4

Probability

Distribution

Prepared by

WILBERT F. MARCO

Unit 4: Probability Distribution

Unit 4

Probability

Distribution

______________________________________________

Introduction to Probability Distribution

Probability is widely used in mathematics, science, engineering, finance and philosophy to draw

conclusions about the likelihood of potential events and the underlying mechanics of complex systems

Probability is a measure of how likely it is for an event to happen
We measure probability with a number between 0 and 1
If an event is certain to happen, then the probability of the event is 1
If an event is certain not to happen, then the probability of the event is 0

Random Experiment

An experiment is called random if the outcome of the experiment is uncertain
For a random experiment:

—The set of all possible outcomes is known before the experiment
—The outcome of the experiment is not known in advance

Sample spaceΩ of an experiment is the set of all possible outcomes of the experiment
Example: Consider random experiment of tossing a coin twice. Sample space is:

Ω = *(𝐻, 𝐻), (𝐻, 𝑇), (𝑇, 𝐻), (𝑇, 𝑇)+

Random Variable

A Random variable is a numerical quantity that is assigned to the outcome of an

expirement.

We use capital letters to represent a random variable.

Unit 4: Probability Distribution

Example 1:

Suppose two coins are tossed and we are interested to determine the number of Heads
that will come out.

S = ( HH, HT, TH, TT)

Outcome

Number of Heads (Value of H)

HH

2

HT

1

TH

1

TT

0

The values of the random variable H (number of heads) in this experiment are 0, 1 and 2.

Example 2

A basket contains 10 ripe and 4 unripe bananas. If three bananas are taken from a basket one
after the other, determine the possible values of random variable R representing the number of
ripe bananas.

Let R represent the ripe bananas and let U represent the unripe bananas.

S = (RRR, RRU,RUR,URR,UUR,URU,RUU,UUU)

The values of the random variable R(number of ripe bananas) in this experiment are 0, 1, 2 and 3

Unit 4: Probability Distribution

Discrete and Continuous Random Variable

A random variable may be classified as discrete or continuous .

Discrete random variable is one that can assume only a countable number of values.

Continuous random variable can assume infinite number of values in one or more intervals.

Probability Distribution of the Discrete Random Variable

Probability Distribution of the Discrete Random Variable is sometimes called the probability
mass function.

It is the process of making frequency distribution of the values of the random variable and
determining the probability of each value of the random variable will occur.

Example 1:

A basket contains 10 ripe and 4 unripe bananas. If three bananas are taken from a basket one after the
other, determine the possible values of random variable R representing the number of ripe bananas.

Step 1: List the sample space of the experiment

S = (RRR, RRU,RUR,URR,UUR,URU,RUU,UUU)

Step 2: Count the number of ripe bananas R in each outcome and assign number to this outcome

Unit 4: Probability Distribution

Step 3: Construct the frequency distribution of the values of the random variable.

Step 4: Construct the probability distribution of the random variable R by getting the probability of
occurrence of each value of the random variable.

The probability of the distribution of the random variable R can be written as follows:

Properties of a Discrete Probability Distribution

Consider the probability distribution of the number of bananas given below:

Find the following.

1.P(R = 3)
2.P(R = 1)

Unit 4: Probability Distribution

3.P(R > 1)
4.P(R < 2)
5.∑P(R)

SOLUTION

1. P(R = 3) = 1/8

2. P(R = 1) = 3/8

3. P(R > 1) = P(2) + P(3) = 3/8 + 1/8 = 4/8 = ½

4. P(R < 2) = P(1) + P(0) = 3/8 + 1/8 = 4/8 = ½

5. ∑P(R) = P(3) + P(2) + P(1) + P(0)

= 1/8 + 3/8 + 3/8 + 1/8

= 1

Finding the Discrete Probability Distribution Described by a formula

Sometimes, discrete probability distributions are described by a formula.

Example

Determine whether the formula below describes a probability distribution.

P(X) = (𝑋 +1)/7 where x = 0,1,3 and find the following

P(X = 3)

P(X ≥ 1)

P(X ≤ 1)

Solution:

if X = 0

P(X) =

=

=

if X = 1

P(X) =

=

=

Unit 4: Probability Distribution

if X = 3

P(X) =

=

=

Finding the Mean of a Discrete Probability Distribution

To find the mean ( ) or expected value E(X) of a discrete Probability
distribution, we use the following formula.

= E(X) = ∑[X. P(X)] where = mean

X = the value of random variable

P(X) = the probability value of the random variable

Example 1: Find the mean of the discrete random variable X with the following
probability distribution.

Step 1: Multiply the value of X by its corresponding probability value P(X).

Step 2 : Find the mean or expected value of the probability distribution by getting
the sum of X. P(X).

Unit 4: Probability Distribution

To find the mean ( ) or expected value E(X) of a discrete Probability distribution, we use the following
formula.

= E(X) = ∑[X. P(X)]

= 3/8 + 6/8 + 3/8

= 12/8

= 3/2

= 1.5

Finding the variance and the standard deviation of a Discrete Probability Distribution

To find the variance and the standard deviation of a discrete Probability distribution, we use the
following formulas:

𝜎² = ∑[X². P(X)] - ² where = mean

X = the value of random variable

P(X) = the probability value of the random variable

𝜎²=𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒

𝜎 = standard deviation

Example: Find the variance and the standard deviation of the discrete random variable X with the
following probability distribution.

Unit 4: Probability Distribution

Step 1: Find the mean of the probability distribution.

Step 2: Square each value of the random variable and multiply by the corresponding probability value
[X². P(X)].

Step 3: Find the variance and the standard deviation by applying the formulas 𝜎² = ∑[X². P(X)] - ²

The standard deviation is computed by getting the square root of the variance.

𝜎 = ∑[X². P(X)] - ² = ( 1 + ½ + 0) - 1² = ½ = 0.5

𝜎 = √0 = 0. 707

The variance is 0.5 and the standard deviation is 0.707

Example 2: Find the mean, variance and the standard deviation of the following probability distribution.

Unit 4: Probability Distribution

𝜎² = ∑[X². P(X)] - ² = ( 27/8 +12/8 + 3/8 + 0) – 1.5² = 5.25 – 2.25 = 4

𝜎 = √(4 ) = 2

The mean is 1.5,variance is 4 and the standard deviation is 2


Applied Problems Involving the Mean and the Variance of a Discrete Probability
Distribution

The following data show the probability distribution of the number of computers sold daily in a
computer shop during the past several months.

Questions:

1.What is the probability that on a given day-
a.Fewer than 3 computers will be sold?
b.At least 4 computer will be sold?
c.Exactly 3 computer will be sold?

2. What is the expected number of computer will be sold on that given day?
3. What is the standard deviation of the number of computers that will be sold on that given day?

Solution

a.P(X < 3) = P(0) + P(1) +P(2)

0.10 + 0.20 + 0.45

0.75

b. P(X ≥ 4) = P(4) +P(5)

0.05 + 0.05
0.10

c. P(X = 3) = 0.15

Unit 4: Probability Distribution

= E(X) = ∑[X. P(X)] = 2.00

𝜎² = ∑[X². P(X)] - ² = 5.40 – 4.0 = 1.40

𝜎= √1.40 = 1.18

Binomial Probability Distribution

Binomial Probability Distribution is a very good approach for resolving probability involving random
experiment which has two possible outcomes. The outcome that the event:

(a)Will occur

(b)Will not occur

The typical examples are:

a.Tossing of coin ( Head or not a head)(tail or not a tail)

b. throwing of an unbiased die repeatedly.( the event of having 6 or not 6 or any choices of number.

Unit 4: Probability Distribution

If we consider the probability that in n number of trials, with r (successes) and n – r (failures) then,

(x = r) =

where p is probability of success and q is the probability of failure.

Recall:

=

( )

Note that the events are independent of one another for the number of trials.

Properties of binomial Probability distribution

If we denote the mean and standard deviation of the binomial distribution as and 𝜎 respectively, then

a. Mean ( ) = np

b. Standard Deviation (𝜎)= √𝑛

c. Variance = npq

Example1

A coin is tossed 6 times. Find the probability of obtaining:

a. Exactly 4 heads

b. At least 5 heads

c. At most 2 heads

Solution:

Probability of success (p) = ½

Probability of failure (q) = ½

Number of times a coin is tossed (n) = 6

_𝑟(x = r) = (_^𝑛)_𝑟^𝑟^(𝑛−𝑟)

a. Exactly 4 heads

(p) = ½ (q) = ½ (n) = 6 r = 4

Solution:

Unit 4: Probability Distribution

(x = r) =

(x = 4) =

(

)(

)

(x = 4) = 15(

)(

)

(x = 4) = 15(

)(

)

(x = 4) =

𝟓
𝟔𝟒

b. At least 5 heads

(p) = ½ (q) = ½ (n) = 6 r = 4

Solution:

_𝑟(x ≥ 5) = _𝑟(x = 5) + _𝑟(x = 6)

_𝑟(x ≥ 5) = (_^6)_5 〖(1/2)〗^5 〖(1/2)〗^1 + (_^6)_6 〖(1/2)〗^6 〖(1/2)〗^0

_𝑟(x ≥ 5) = 6( 1/32)(1/2) + 1( 1/64)(1)

_𝑟(x ≥ 5) = ( 6/64) + ( 1/64)

_𝑟(x ≥ 5) = /𝟔

c. At most 2 heads

(p) = ½ (q) = ½ (n) = 6 r = 4

Solution:

_𝑟(x ≤ 2) = _𝑟(x = 0) + _𝑟(x = 1) + _𝑟(x = 2)

_𝑟(x ≤ 2) = (_^6)_0 〖(1/2)〗^0 〖(1/2)〗^6 + (_^6)_1 〖(1/2)〗^1 〖(1/2)〗^5 + (_^6)_2 〖
(1/2)〗^2 〖(1/2)〗^4

_𝑟(x ≤ 2) = 1(1)(1/64) + 6( 1/2)(( 1/32) + 15( 1/4)( 1/16)

_𝑟(x ≤ 2) = ( 1/64) + ( 6/64) + ( 15/64)

Unit 4: Probability Distribution

_𝑟(x ≤ 2) = 𝟐/𝟔𝟒 = /𝟑𝟐

Example 2

An unbiased die with 6 faces is thrown 5 times. Find the probability that:

Factor of 6 appears exactly 3 times;

Perfect square appears at most 4 times.

(a)S = {1,2,3,4,5,6} and n(S) = 6

F = {1,2,3,6} and n(F) = 4

(p) = 4/6 = 𝟐/𝟑 (q) = 2/6= /𝟑 (n) = 5 r = 3

Solution:
_𝑟(x = 3) = (_^5)_3 〖(2/3)〗^3 〖(1/3)〗^2
_𝑟(x = 3) = 10( 8/27)( 1/9)
_𝑟(x = 3) = 𝟎/𝟐𝟒𝟑

a)S = {1,2,3,4,5,6} and n(S) = 6
b) E = {1,4} and n(E) = 2

(p) = 2/6 = /𝟑 (q) = 4/6= 𝟐/𝟑 (n) = 5 r = 0,1,2,3 and 4

Solution:

_𝑟(x ≤ 4) =1- _𝑟(x = 5 ) = (_^5)_5 〖(1/3)〗^5 〖(2/3)〗^0

= 1 – 1 (1/243)(1)

= 𝟐𝟒𝟐/𝟐𝟒𝟑

Example 3

A test contain 10 multiple choice questions comprising of 4 options in which only one option is correct.
Find the probability that a candidate can get 7 correct answers out of the 10 questions.

Unit 4: Probability Distribution

(p) = ¼ (q) = 3/4 (n) = 10 r = 7
Solution:
_𝑟(x = 7) = (_^10)_7 〖(1/4)〗^7 〖(3/4)〗^3
= 120 (1/16384)(27/64)
= 0.0030899

Hypergeometric Distribution :

Suppose there is a population with 2 types of elements:

1-st Type = success

2-nd Type = failure

· N= population size

· K= number of elements of the 1-st type

· N -K = number of elements of the 2-nd type

We select a sample of n elements at random from the population

· Let X = number of elements of 1-st type (number of successes) in the sample

· We need to find the probability distribution of X.

There are to two methods of selection:

Unit 4: Probability Distribution

1. selection with replacement

2. selection without replacement

(1) If we select the elements of the sample at random and with replacement, then

X ~ Binomial(n,p); where

(2) Now, suppose we select the elements of the sample at random and without replacement. When the
selection is made without replacement, the random variable X has a hyper geometric distribution with
parameters N, n, and K. and we write X~h(x;N,n,K).

Note that the values of X must satisfy:

0xK and 0nxNK 0xK and nN+K xn

Example

Lots of 40 components each are called acceptable if they contain no more than 3 defectives. The
procedure for sampling the lot is to select 5 components at random (without replacement) and to reject
the lot if a defective is found. What is the probability that exactly one defective is found in the sample if
there are 3 defectives in the entire lot.

Let X= number of defectives in the sample

· N=40, K=3, and n=5

· X has a hypergeometric distribution with parameters N=40, n=5, and K=3.

· X~h(x;N,n,K)=h(x;40,5,3).

· The probability distribution of X is given by:

N
K
p 

Unit 4: Probability Distribution