Module 3: Probability and Probability Distributions

Module 3: Probability and Probability Distributions

The concept of probability is used in everyday communication. For example, we may say that “there is a 50 – 50 chance that it will rain today”. On another day, one may say that “it is unlikely that it will rain today”. Events which are common have probabilities of occurrence that are large while events which are unlikely have small probabilities of occurrence. Whereas in everyday communication we use words like might, almost or certainly to indicate how likely or unlikely an event is, in probability we use a number lying between 0 and 1 to express the same idea.

Impossible event is an event that cannot occur and as such has a probability of 0. Sure event is an event that is certain to occur and as such has a probability of 1. Probability is a term which is used to describe the degree of belief which we have about something. It is relevant whenever the outcomes are uncertain.

Definitions:

(i) Experiment: is any process by which several chance observations are

obtained. If the process is done once, it is referred to as a trial. For example, rolling a die where we can obtain 6 chance observations i.e. 1, 2, 3, 4, 5, or 6.

(ii) Outcome: it is the result obtained when an experiment is carried out. It is

one of the possible results in a trial. For example, 2 is an outcome from rolling a die.

(iii) Event: is a set which consists of one or more possible outcomes of an

experiment. Example, the set of outcomes (2, 4, 6) is the event of obtaining an even number when a die is rolled. The set (1, 3, 5) is the event of obtaining an odd number when a die is rolled.

(iv) Sample Space: is the set of all possible outcomes of an experiment.

Example, the sample space when a die is rolled is {1, 2, 3, 4, 5, 6}. The sample space when a coin is tossed is {Head, Tail}.

(v) Mutually Exclusive Events: two or more events are said to be mutually

exclusive if they do not have any outcomes in common. In other words, two or more events are mutually exclusive if they cannot occur simultaneously.

Elementary Properties of Probability

(1) Given an experiment with n mutually exclusive outcomes (E1, E2, … , En),

say, the probability of an event Ei is non-negative. That is 𝑃 Ei ≥ 0.

(2) The sum of the probabilities of all mutually exclusive outcomes is equal to 1. That is,

Properties 1 and 2 imply that 0 ≤ 𝑃 Ei ≤ 1.

(3) (a) The probability of the occurrence of either 𝐸𝑖 or 𝐸𝑗 is equal to the sum of

their individual probabilities provided that the two events 𝐸𝑖 and 𝐸𝑗 are

mutually exclusive. That is,

𝑃 𝐸𝑖or𝐸𝑗 = 𝑃 Ei + 𝑃 Ej

(b) If the two events are not mutually exclusive, then 𝑃 𝐸𝑖or𝐸𝑗 = 𝑃 Ei + 𝑃 Ej − 𝑃 𝐸𝑖 ∩ 𝐸𝑗

(4) The probability of the occurrence of events 𝐸𝑖 and 𝐸𝑗 is equal to the product

of their individual probabilities provided the two events are independent. That is

𝑃 𝐸_𝑖and𝐸_𝑗 = 𝑃 𝐸_𝑖∩ 𝐸_𝑗 = 𝑃 E_i × 𝑃 E_j

Note:

 For mutually exclusive events:𝑃 𝐸𝑖∩ 𝐸𝑗 = 0 since 𝐸𝑖 ∩ 𝐸𝑗 = ∅ and 𝑃 ∅ = 0

 For independent events:𝑃 𝐸𝑖∩ 𝐸𝑗 = 𝑃 Ei × 𝑃 Ej

Probability Distributions

There are two main types of probability distributions. These are discrete and continuous distributions. We shall consider two (2) common discrete distributions: Binomial and Poisson and one (1) continuous distribution: Normal distribution.

Binomial Distribution

A discrete random variable, X is said to follow the Binomial distribution with parameters 𝑛 and 𝑝 if:

𝑃 𝑋 = 𝑥 = 𝑛 𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥; 𝑥 = 0, 1, 2, … , 𝑛; 0 < 𝑝 < 1 Or 𝑃 𝑋 = 𝑥 = 𝑛 𝑥 𝑝𝑥𝑞𝑛−𝑥 Where: 𝑞 = 1 − 𝑝

This can be written as 𝑋~𝐵𝑖 𝑛, 𝑝 . 𝑋 is said to be a Binomial random variable. 𝑛

Mean of Binomial Distribution

The mean of the Binomial distribution denoted as 𝐸(𝑋) is given as: 𝐸 𝑋 = 𝑛𝑝; 𝐸 represents the expected value (or mean)

Variance of Binomial Distribution

The variance of the Binomial distribution denoted Var (X) is given as: Var (X) = 𝑛𝑝 1 − 𝑝 = 𝑛𝑝𝑞

Note:

To get the 𝑛_𝑥 on your calculator, find the function, 𝑛𝑪𝑟.

Example 9:

In a family of 8, the chance that any of them will get a visa is 0.3. What is the probability that:

(i) Exactly four get a visa (ii) None gets a visa (iii) At most two get a visa (iv) At least three get a visa (v) Find the mean

(vi) Find the variance

Solution: 𝑋~𝐵𝑖 𝑛, 𝑝 ; 𝑋~𝐵𝑖 8, 0.3 . 𝑃 𝑋 = 𝑥 = 𝑛 𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥; 𝑥 = 0, 1, 2, … , 𝑛; 0 < 𝑝 < 1 𝑛 = 8, 𝑝 = 0.3, 1 − 𝑝 = 1 − 0.3 = 0.7 (i) 𝑃 𝑋 = 4 = 8₄ 0.3 4 _0.7 8−4 = 8 4 0.3 4 0.7 4 = 0.1361

Note: to get 8₄ in this example  Press 8

 then press the SHIFT key

 then press the key that has 𝑛𝑪𝑟 written on it (𝑪will appear after 8)  then press 4 (your answer will be displayed on the screen)

 In this case, 70

(ii) 𝑃 𝑋 = 0 = 8₀ 0.3 0 _0.7 8−0 = 8 0 0.3 0 0.7 8 = 0.0576 (iii) 𝑃 𝑋 ≤ 2 = 𝑃 𝑋 = 0 + 𝑃 𝑋 = 1 + 𝑃(𝑋 = 2) = 8 0 (0.3)0 0.7 8−0+ 8 1 (0.3)1 0.7 8−1+ 8 2 (0.3)2 0.7 8−2 = 8 0 (0.3)0 0.7 8+ 8 1 (0.3)1 0.7 7+ 8 2 (0.3)2 0.7 6 = 0.0576 + 0.1977 + 0.2965 = 0.5518

Simple Illustration to help answer (iv)

This illustration is essential in the understanding of probability and probability distributions.

Consider Elementary property 2:

The sum of the probabilities of all mutually exclusive outcomes is equal to 1. That is, 𝑃 E₁ + 𝑃 E₂ + ⋯ + 𝑃 E_n = 1

In our current Example our 𝒏 = 𝟖

𝑃 E₁ + 𝑃 E2 + 𝑃 E3 + 𝑃 E4 + 𝑃 E5 + 𝑃 E6 + 𝑃 E7 + 𝑃 E8 = 1

We will replace Ei’s with X’s.

That is

𝑃 X = 0 + 𝑃 X = 1 + 𝑃 X = 2 + 𝑃 X = 3 + 𝑃 X = 4 + 𝑃 X = 5 + 𝑃 X = 6 + 𝑃 X = 7 + 𝑃 X = 8 = 1

Dividing the above into two parts:

For instance, A and B

Where: A = 𝑃 X = 0 + 𝑃 X = 1 + 𝑃 X = 2 and

B = 𝑃 X = 3 + 𝑃 X = 4 + 𝑃 X = 5 + 𝑃 X = 6 + 𝑃 X = 7 + 𝑃 X = 8

A can be simplified as follows:

𝐴 = 𝑃 𝑋 < 3 = 𝑃 𝑋 ≤ 2 This is because 𝐴 contains values from 0 to 2.

B can be simplified as follows:

This is because 𝐵 contains values from 3 to 8 (in this case)

We therefore have:

𝐴 + 𝐵 = 1

That is:

𝑃 𝑋 < 3 + 𝑃 𝑋 ≥ 3 = 1

To get any part from the above, just make it the subject of the formula. For example,

𝐴 = 1 − 𝐵 that is:

𝑃 𝑋 < 3 = 1 − 𝑃 𝑋 ≥ 3 𝐵 = 1 − 𝐴 that is:

𝑃 𝑋 ≥ 3 = 1 − 𝑃 𝑋 < 3

This technique can also be applied to cases where the values of 𝒙 run to infinity, like the Poisson distribution.

(iv) 𝑃 𝑋 ≥ 3 = 1 − 𝑃 𝑋 < 3 = 1 − 𝑃 𝑋 ≤ 2 = 1 − 0.5518 = 0.4482 (v) 𝐸 𝑋 = 𝑛𝑝 = 8 0.3 = 2.4 (vi) Var (X) = 𝑛𝑝 1 − 𝑝 = 8 0.3 0.7 = 1.68 Poisson Distribution

A discrete random variable X is said to have a Poisson distribution with parameter lambda (𝜆) if

𝑃 𝑋 = 𝑥 =𝑒

−𝜆_𝜆𝑥

𝑥! ; 𝑥 = 0, 1, 2, … ; 𝜆 > 0 This can be written as 𝑋~𝑃(𝜆).

𝑥! is read as 𝑥 factorial. 0! = 1; 𝑒 is the exponential function

Mean of Poisson Distribution

Variance of Poisson Distribution

The variance of the Poisson distribution is given as: Var 𝑋 = 𝜆

Note:

An important characteristic of the Poisson distribution is that the parameter 𝜆 is the mean as well as the variance. That is

E 𝑋 = Var 𝑋 = 𝜆

Example 10:

If the number of accidents occurring on a highway follows a Poisson distribution with variance 3, compute the probability that within a period there will be:

(i) One accident (ii) No accident

(iii) Exactly five accidents (iv) At most three accidents (v) At least four accidents (vi) Find the mean

Solution: 𝑋~𝑃(𝜆); 𝑋~𝑃(3) 𝑃 𝑋 = 𝑥 =𝑒−𝜆𝜆𝑥 𝑥! ; 𝑥 = 0, 1, 2, … ; 𝜆 > 0

(i) 𝑃 𝑋 = 1

=

𝑒

3

1!

= 0.1494

Note: to solve the above:

 Press SHIFT function on your calculator

 Then press the key with 𝑒𝑥 _{on it (𝒆 will appear on the screen)}

 Press -3 after the 𝒆  Then press =

 It gives you 0.049787068

 Without clearing the screen, multiply by

3

 It gives you 0.149361205

 Without clearing the screen, divide by

1! (

(ii) 𝑃 𝑋 = 0

=

𝑒

3

0!

= 0.0498

(iii) 𝑃 𝑋 = 5

=

𝑒

3

5!

= 0.1008

(iv) 𝑃 𝑋 ≤ 3 = 𝑃 𝑋 = 0 + 𝑃 𝑋 = 1 + 𝑃 𝑋 = 2 + 𝑃 𝑋 = 3

=

𝑒

3

0!

+

𝑒

₃

1!

+

𝑒

₃

2!

+

𝑒

₃

3!

= 0.0498 + 0.1494 + 0.2240 + 0.2240

= 0.6472

(v) 𝑃 𝑋 ≥ 4 = 1 − 𝑃 𝑋 < 4 = 1 − 𝑃 𝑋 ≤ 3

= 1 − 0.6472

= 0.3528

(vi) E 𝑋 = 𝜆 = 3

Poisson Approximation to the Binomial Distribution

There are situations when for the Binomial distribution, 𝑛 is large and 𝑝 is small and calculations of probabilities become cumbersome. In such situations, the Binomial distribution may be approximated by the Poisson distribution. Calculations of probabilities are easier for the Poisson distribution. The approximation requires the following:

𝜆 = 𝑛𝑝

Example 11:

Assume that 0.2% of students eat shawarma. What is the probability that two or more students randomly chosen from 2000 students eat shawarma?

Solution:

= 1 − 0.0183 + 0.0733 = 1 − 0.0916

= 0.9084

Normal Distribution

A random variable X is said to have a normal distribution with parameters 𝜇 and 𝜎2

if: 𝑓 𝑥 = 1 2𝜋𝜎2𝑒𝑥𝑝 − 𝑥 − 𝜇 2 2𝜎2 , −∞ < 𝑥 < ∞ Where:

𝜋 and exp are the familiar mathematical constants 3.14159 and 2.71828 respectively. This can be written as 𝑋~𝑁 𝜇, 𝜎2 _.

Mean of Normal Distribution

The mean of the Normal distribution is given as: E 𝑋 = 𝜇

Variance of Normal Distribution

The variance of the Normal distribution is given as: Var 𝑋 = 𝜎2

Standard Normal Distribution

A random variable Z is said to have a standard normal distribution when 𝜇 = 0 and 𝜎2 _{= 1 such that:} 𝑓 𝑧 = 1 2𝜋𝑒𝑥𝑝 − 𝑧2 2 , −∞ < 𝑧 < ∞ Where: 𝑍 = 𝑋 − 𝜇 𝜎 Example 12:

(i) Less than 20 minutes (ii) Greater than 70 minutes (iii) Between 20 and 50 minutes

(iv) How many goods will take more than 80 minutes in packaging if there are 2500 goods in the factory?

Solution: 𝑋~𝑁 𝜇, 𝜎2 _{; 𝑋~𝑁 50, 17}2 _{; since 𝜎 = 17} 𝑍 = 𝑋 − 𝜇 𝜎 (i) 𝑃 𝑋 < 20 = 𝑃 𝑋 − 𝜇 𝜎 < 20 − 𝜇 𝜎 = 𝑃 𝑍 <20 − 50 17 = 𝑃 𝑍 <−30 17 = 𝑃 𝑍 < −1.76 = 0.0392 Note: To get 𝑃 𝑍 < −1.76 ,

 Go to the negative part of the normal table (See Table 1 below)

 The value of 𝑧 to first decimal is given in the column labelled 𝑧 while the second decimal is given in the top row.

 Locate where you have -1.7 under the column labelled 𝑧  trace that row to the column labelled 0.06

Note:

The normal table does not cater for greater than. As such we need to get the value for the other part and subtract from 1. (Applying the illustration under Binomial distribution). (iii) 𝑃 20 < 𝑋 < 50 = 𝑃 20 − 𝜇 𝜎 < 𝑋 − 𝜇 𝜎 < 50 − 𝜇 𝜎 = 𝑃 20 − 50 17 < 𝑍 < 50 − 50 17 = 𝑃 −30 17 < 𝑍 < 0 17 = 𝑃 −1.76 < 𝑍 < 0 = 𝑃 𝑍 < 0 − 𝑃(𝑍 < −1.76) = 0.5000 − 0.0392 = 0.4608 (iv) 𝑃 𝑋 > 80 = 𝑃 𝑋 − 𝜇 𝜎 > 80 − 𝜇 𝜎 = 𝑃 𝑍 >80 − 50 17 = 𝑃 𝑍 > 30 17 = 𝑃 𝑍 > 1.76 = 1 − 𝑃 𝑍 ≤ 1.76 = 1 − 0.9608 = 0.0392

Assignment 3:

(1) The probability of winning a football match for team A is 0.6. If team A plays 8 matches, what is the probability of winning:

(i) Not more than three matches (ii) At least four matches

(iii) Exactly six matches (iv) Find the mean

(2) If it is assumed that 0.09% of Bowen University students have type O negative blood. What is the probability that not less than five students randomly chosen from 5000 students have type O negative blood?

(3) If the number of daily customer complaints follows a Poisson distribution with 𝜆 = 2.4. What is the probability that there will be:

(i) Exactly seven complaints (ii) Not more than four complaints (iii) At least five complaints

(iv) Find the variance

(4) Suppose there are 7000 cows in the Bowen University cattle ranch and the cows have an average weight of 575kg with a standard deviation of 75kg. Assuming that the weights are normally distributed, find the probability that a randomly selected cow will weigh:

(i) Less than 500kg

(ii) Between 550kg and 650kg