Chapter 5 Discrete Probability Distributions Random … · Chapter 5 Discrete Probability Distributions ... = probability of 5 cars sold . ... Does it fulfill the basic requirements

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Chapter 5

Discrete Probability Distributions

Random Variables

x is a random variable which is a numerical description of the outcome of an experiment.

• Discrete: If the possible values change by steps or jumps.

Example: Suppose we flip a coin 5 times and count the number of tails. The number of tails

could be 0, 1, 2, 3, 4 or 5. Therefore, it can be any integer value between (and including) 0

and 5. However, it could not be any number between 0 and 5. We could not, for example, get

2.5 tails. Therefore, the number of tails must be discrete.

• Continuous: If the possible values can take any value within some range.

Example: The height of trees is an example of continuous data. Is it possible for a tree to be

2.105m tall? Sure. How about 2.10567m? Yes. How about 2.105679821014m? Definitely!

Discrete Random Variables

Consider the sales of cars at a car dealership over the past 300 days.

Frequency Distribution:

Number of cars sold per day Number of days

(frequency)

0 54

1 117

2 72

3 42

4 12

5 3

300

• Define the random variable:

Let x = the number of cars sold during a day.

• Note: We make the assumption that no more than 5 cars are sold per day.

• Sample Space: S = {0, 1, 2, 3, 4, 5}

• Notation:

P(X = 0) = f(0) = probability of 0 cars sold

P(X = 1) = f(1) = probability of 1 car sold






• Note: f(x) = probability function

• The probability function provides the probability for each value of the random variable

• Probability distribution for the number of cars sold per day at a car dealership

x Number of days

(frequency)

f(x)

0 54 54300 = 0.18

1 117 117300 = 0.39

2 72 72300 = 0.24

3 42 42300 = 0.14

4 12 12300 = 0.04

5 3 3300 = 0.01

300 1

• Question: Does the above mentioned probability function fulfill the required conditions for a

discrete probability function?

There are two requirements:

(i) 0 ≤ (�) ≤ 1 for all (�)

(ii) ∑ (�) = 1 Yes, both requirements are fulfilled.

0.18

0.39

0.24

0.14

0.04

0.01

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 1 2 3 4 5

Pro

ba

bil

ity

Number of cars sold per day

Graphical representation of the probability

distribution for the number of cars sold per day


Questions:

a) The probability that 2 cars are sold per day?

�(� = 2) = (2) = 0.24

b) The probability that, at most, 2 cars are sold per day?

�(� ≤ 2)

= �(� = 0) + �(� = 1) + �(� = 2)

= (0) + (1) + (2)

= 0.18 + 0.39 + 0.24 = 0.81

c) The probability that more than 2 cars are sold per day?

�(� > 2)

= �(� = 3) + �(� = 4) + �(� = 5)

= (3) + (4) + (5)

= 0.14 + 0.04 + 0.01 = 0.19

d) The probability that at least 2 cars are sold per day?

�(� ≥ 2)

= �(� = 2) + �(� = 3) + �(� = 4) + �(� = 5)

= (2) + (3) + (4) + (5)

= 0.24 + 0.14 + 0.04 + 0.01

= 0.43

e) The probability that more than 1 but less than 4 cars are sold per day?

�(1 < � < 4)

= �(� = 2) + �(� = 3)

= (2) + (3)

= 0.24 + 0.14

= 0.38


Discrete Uniform probability function:

(�) = 1�

where n = the number of values the random variable may assume

Example: Dice (�) = �� for x = 1, 2, 3, 4, 5, 6

x f(x)

1 ��

2 ��

3 ��

4 ��

5 ��

6 ��

Does the above mentioned probability function fulfill the required conditions for a discrete probability

function?


(i) 0 ≤ (�) ≤ 1 for all (�)


Another example of a random variable x with the following discrete probability distribution

(�) = �� for x = 1, 2, 3, 4

x f(x)

1

2

3

4

Does the above mentioned probability function fulfill the required conditions for a discrete probability

function?


(i) 0 ≤ (�) ≤ 1 for all (�)



Expected value, variance, standard deviation and median:

Expected Value

�(�) = � = ∑ � (�)

= (0)(0.18) + (1)(0.39) + (2)(0.24) + (3)(0.14) + (4)(0.04) + (5)(0.01)

= 1.5

Variance

� !(�) = "# = ∑(� − �)# (�)

= (0 − 1.5)#(0.18) + (1 − 1.5)#(0.39)+(2 − 1.5)#(0.24) + (3 − 1.5)#(0.14) + (4 − 1.5)#(0.04) +(5 − 1.5)#(0.01)

= 1.25

Standard deviation

" = √"# = √1.25 = 1.118

Median

0 → 0.18

0 and 1 → 0.18 + 0.39 = 0.57 > 0.5

Therefore, the median = 1

0.18

0.39

0.24

0.14

0.04

0.01

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0 1 2 3 4 5

Pro

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Number of cars sold per day


distribution for the number of cars sold per day


OR use a table to calculate the expected value, variance and standard deviation:

x f(x) x f(x) ' − ( (' − ()) (' − ())*(')

0 0.18 0 -1.5 2.25 0.4050

1 0.39 0.39 -0.5 0.25 0.0975

2 0.24 0.48 0.5 0.25 0.0600

3 0.14 0.42 1.5 2.25 0.3150

4 0.04 0.16 2.5 6.25 0.2500

5 0.01 0.05 3.5 12.25 0.1225

( = 1.5 +) = 1.25

+ = √,. )- = ,. ,,.

Example:

A psychologist has determined that the number of hours required to obtain the trust of a new patient is

either 1, 2 or 3 hours.

Let x = be a random variable indicating the time in hours required to gain the patient’s trust.

The following probability function has been proposed:

(�) = �� for x = 1, 2, 3

Questions:

a) Set up the probability function of x.

b) Is this a valid probability function? Explain.

c) Give a graphical representation of the probability function of x.

d) What is the probability that it takes exactly 2 hours to gain the patient’s trust?

e) What is the probability that it takes at least 2 hours to gain the patient’s trust?

f) Calculate the expected value, variance and standard deviation.


Answers:

a)

x f(x)

1 (1) = 1 60 = 0.161 2 (2) = 2 60 = 0. 31 3 (3) = 3 60 = 0.5

1

b) There are two requirements:

(i) 0 ≤ (�) ≤ 1 for all (�)


c)

d) (2) = 2 60 = 0. 31

e) �(� ≥ 2) = (2) + (3) = 2 60 + 3 60 = 5 60 = 0.831

f) � = ∑ � (�) = 2. 31

"# = ∑(� − �)# (�) = 0. 51

" = √"# = 20. 51 = 0.745

x f(x) x f(x) ' − ( (' − ()) (' − ())*(')

1 0.1667 0.1667 -1.3333 1.7778 0.2963

2 0.3333 0.6667 -0.3333 0.1111 0.0370

3 0.5 1.5000 0.6667 0.4444 0.2222

2.3333 0.5556

0

0.1

0.2

0.3

0.4

0.5

0.6

1 2 3

Pro

bab

ilit

y

x


distribution


Binomial distribution

1. The experiment consists of a sequence of n identical trials.

2. Two outcomes are possible on each trial. We refer to a

� Success

� Failure

3. The probability of a success, denoted by p does not change from trial to trial. Consequently, the

probability of a failure, denoted by 1 – p, does not change from trial to trial.

4. The trials are independent

In general:

� Let: x = number of successes

� Then x has a binomial distribution of n trials and the probability of a success of p.

The Binomial probability function is:

(�) = 3��4 5�(1 − 5)67�

Martin clothing store problem:

Let us consider the purchase decisions of the next 3 customers who enter the Martin clothing store. On the

basis of past experience, the store manager estimates the probability that any one customer will make a

purchase is 0.3.

Let: S = customer makes a purchase (success)

F = customer does not make a purchase (failure)

The above mentioned is a Binomial experiment, because:

1. n = 3 identical trials

2. Two possible outcomes

• customer makes a purchase (success)

• customer does not make a purchase (failure)

3. Probability of a success p = 0.3 and a failure 1 – p = 0.7

4. The trials are independent

Let x = number of customers that make a purchase

OR

x = number of successes


Tree Diagram:

1st 2nd 3rd

Outcomes

(S, S, S)

(S, S, F)

(S, F, S)

(S, F, F)

(F, S, S)

(F, S, F)

(F, F, S)

(F, F, F)

Value of x

3

2

2

1

2

1

1

0

Total number of experimental outcomes:

Using the tree diagram we count 8 experimental outcomes.

Using the counting rule for multiple-step experiments we get (n1)(n2)(n3) = (2)(2)(2) = 8.

Since the binomial distribution only as two possible outcomes on each step (success or failure), we can

use the formula 26 which in this case equals 28 = 8 where n denotes the number of trials in the binomial

experiment.

S

S

S

S

S

S

S

F

F

F

F

F

F

F


Calculating binomial probabilities: (�) = 3��4 5�(1 − 5)67�

Question 1: Calculate the probability that 2 out of the 3 customers make a purchase.

Answer 1:

(2) = 3324 0.3#0.7� = (3)(0.09)(0.7) = 0.189


Answer 2:

(1) = 3314 0.3�0.7# = (3)(0.3)(0.49) = 0.441


Answer 3:

(3) = 3334 0.380.7878 = (1)(0.027)(1) = 0.027


Answer 4:

(0) = 3304 0.3�0.787� = (1)(1)(0.343) = 0.343


The probability distribution for the number of customers making a purchase:

x f(x)

0 0.343

1 0.441

2 0.189

3 0.027

1

Does it fulfill the basic requirements for a discrete probability function?


(iii) 0 ≤ (�) ≤ 1 for all (�)

(iv) ∑ (�) = 1 Yes, both requirements are fulfilled.

Calculate the expected value, variance and standard deviation of x:

x f(x) x f(x) ' − ( (' − ()) (' − ())*(')

0 0.343 0 -0.9 0.81 0.27783

1 0.441 0.441 0.1 0.01 0.00441

2 0.189 0.378 1.1 1.21 0.22869

3 0.027 0.081 2.1 4.41 0.11907

0.9 0.63

� = ∑ � (�) = 0.9

"# = ∑(� − �)# (�) = 0.63

" = √"# = √0.63 = 0.794

Formulas of � and "# for the Binomial Probability Distribution:

�(�) = � = �5

� !(�) = "# = �5(1 − 5)

Test:

�(�) = � = �5 = (3)(0.3) = 0.9

� !(�) = "# = �5(1 − 5) = (3)(0.3)(0.7) = 0.63

9:;<=(�) = √0.63 = 0.794


EXCEL:

BINOMDIST(x, n, p, false) – normal probability

BINOMDIST(x, n, p, true) – cumulative probability

Formula Worksheet

Value Worksheet

Value Worksheet with explanations


Example: (Extension of the Martin-experiment)

• Suppose 10 customers go into the store.

• The probability of purchasing something is 0.3

• Let x = number of customers that make a purchase

Questions:

1. What is the distribution of x.

Binomial with n = 10 and p = 0.3

2. Calculate the expected value, variance and standard deviation of x.

�(�) = � = �5 = (10)(0.3) = 3

� !(�) = "# = �5(1 − 5) = (10)(0.3)(0.7) = 2.1

9:;<=(�) = " = √2.1 = 1.449

3. Calculate the probability distribution of x.

x f(x)

0 f(0) = 3100 4 (0.3)�(0.7)��7� = (1)(1)(0.028) = 0.0282

1 f(1) = 3101 4 (0.3)�(0.7)��7� = (10)(0.3)(0.040) = 0.1211

2 f(2) = 3102 4 (0.3)#(0.7)��7# = (45)(0.09)(0.058) = 0.2335

3 f(3) = 3103 4 (0.3)8(0.7)��78 = (120)(0.027)(0.082) = 0.2668

4 f(4) = 3104 4 (0.3)?(0.7)��7? = (210)(0.0081)(0.118) = 0.2001

5 f(5) = 3105 4 (0.3)@(0.7)��7@ = (252)(0.00243)(0.168) = 0.1029

6 f(6) = 3106 4 (0.3)�(0.7)��7� = (210)(0.0007)(0.2401) = 0.0368

7 f(7) = 3107 4 (0.3)A(0.7)��7A = (120)(0.00022)(0.343) = 0.0090

8 f(8) = 3108 4 (0.3)B(0.7)��7B = (45)(0.000066)(0.49) = 0.0014

9 f(9) = 3109 4 (0.3)C(0.7)��7C = (10)(0.0000197)(0.7) = 0.0001

10 f(10) = 310104 (0.3)��(0.7)��7�� = (1)(0.0000059)(1) = 0.0000


4. A graphical representation of the probability distribution of x.

5. Calculate the cumulative distribution of x.

Formula worksheet

0.028

0.121

0.233

0.267

0.200

0.103

0.037

0.0090.001 0.000 0.000

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 1 2 3 4 5 6 7 8 9 10

f(x)

x

Probability distribution of 10 customers


Value worksheet

Value worksheet with explanations


6. Calculate the probability that:

(a) At most 3 clients purchase something:

�(� ≤ 3) = 0.6496

(b) Only 3 clients purchase something:

�(� = 3) = 0.2668 or (3) = 3103 4 (0.3)8(0.7)��78 = (120)(0.027)(0.082) = 0.2668

(c) More than 1 client purchase something:

�(� > 1) = 1 − �(� ≤ 1) = 1 − 0.1493 = 0.8507

since �(� ≤ 1) + �(� > 1) = 1

(d) More than 2 but less than 5 clients purchase something:

�(2 < � < 5) = �(� = 3) + �(� = 4) = 0.2668 + 0.2001 = 0.4669

OR

�(2 < � < 5) = �(� = 3) + �(� = 4) = �(� ≤ 4) − �(� ≤ 2) = 0.8497 − 0.3828 =0.4669

(e) Less than 5 clients purchase something:

�(� < 5) = �(� ≤ 4) = 0.8497

(f) At least 4 clients purchase something:

�(� ≥ 4) = 1 − �(� ≤ 3) = 1 − 0.6496 = 0.3504

(g) Exactly 6 clients do not purchase anything:

If 6 clients do not purchase something, then 4 clients purchase something

�(� = 4) = 0.2001 or �(� ≤ 4) − �(� ≤ 3) = 0.8497 − 0.6496 = 0.2001

(h) Difficult question: Calculate the probability that the first three clients make a purchase:

555(1 − 5)(1 − 5)(1 − 5)(1 − 5)(1 − 5)(1 − 5)(1 − 5) = 58(1 − 5)A = 0.380.7A = 0.00222


Homework

A university found that 20% of its students withdraw without completing the introductory statistics

course.

Assume: That 20 students have registered for the course this quarter.

Let: x = number of students withdrawing from the course.

Binomial Experiment: n = 20 and p = 0.2

x f(x) Cumulative prob

0 0.0115 0.0115

1 0.0576 0.0692

2 0.1369 0.2061

3 0.2054 0.4114

4 0.2182 0.6296

5 0.1746 0.8042

6 0.1091 0.9133

7 0.0545 0.9679

8 0.0222 0.9900

9 0.0074 0.9974

10 0.0020 0.9994

11 0.0005 0.9999

12 0.0001 1.0000

13 0.0000 1.0000

14 0.0000 1.0000

15 0.0000 1.0000

16 0.0000 1.0000

17 0.0000 1.0000

18 0.0000 1.0000

19 0.0000 1.0000

20 0.0000 1.0000


Homework (work through this on your own):

a) Calculate the probability that 2 or less will withdraw.

D(E ≤ )) = F. )FG,

b) Calculate the probability that exactly 4 will withdraw.

D(E = H) = F. ),.) or D(E = H) = D(E ≤ H) − D(E ≤ I) = F. G)JG − F. H,,H = F. ),.,

c) Calculate the probability that more than 3 will withdraw.

D(E > I) = , − D(E ≤ I) = , − F. H,,H = F. -..G

d) What is the expected number of withdrawals?

K(E) = LM = ()F)(F. )) = H

e) What is the expected number of students that will not withdraw?

Suppose y = number of students not withdrawing from the course.

Then K(N) = L(, − M) = ()F)(F. .) = ,G

We use the probability , − M, since:

• M = 0.2 is the probability that a student will withdraw

• , − M = 1 – 0.2 = 0.8 is the probability that a student will not withdraw

f) Calculate the probability that 15 will not withdraw.

If 15 students from 20 students will not withdraw, then 5 students will withdraw.

�(� = 5) = �(� ≤ 5) − �(� ≤ 4) = 0.8042 − 0.6296 = 0.1746


Shape of the Binomial distribution:

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6 7 8 9 10 11

Pro

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ity

x

Binomial: n = 10 and p < 0.5

Skewed to the right

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4 5 6 7 8 9 10 11

Pro

ba

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ity

x

Binomial: n = 10 and p = 0.5

Symmetric

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

1 2 3 4 5 6 7 8 9 10 11

Pro

ba

bil

ity

x

Binomial: n = 10 and p > 0.5

Skewed to the left


Typical exam questions

Questions 1 to 4 are based on the following information:

30% of the tires of airplanes at an airport are faulty. Four tyres are randomly selected.

Let: x = Number of faulty tires found

Question 1

The standard deviation of x is:

Answer 1

x is binomially distributed with n = 4 and p = 0.3. Thus, O:;<=(�) = 2�5(1 − 5) = √4 × 0.3 × 0.7 =0.917.

Question 2

The probability that the first and third tyres in the sample of 4 selected tyres are faulty is:

Answer 2

(0.3)(0.7)(0.3)(0.7) = 0.3#0.7# = 0.0441.

Question 3

The probability that less than 3 tyres selected are faulty, is:

Answer 3

�(� < 3) = �(� = 0) + �(� = 1) + �(� = 2) = 3404 0.3�0.7? + 34

14 0.3�0.78 + 3424 0.3#0.7# =

0.2401 + 0.4116 + 0.2646 = 0.9163.

Question 4

The probability that 1 of the 4 randomly selected tyres is not faulty, is:

Answer 4

If 1 out of 4 is not faulty, it means that 3 out of 4 are faulty.

�(� = 3) = 3434 0.380.7� = 0.0756.



It is known that 60% of all South Africans will watch the opening match of the 2010 Soccer World Cup.

A random sample of 40 South Africans was asked whether they will watch the opening match.

Let � = the number of South Africans who will watch the opening match.

Consider the following results in Excel of two different binomial distributions:

Formula worksheet:

Value worksheet:


Question 5

Suppose a random sample of 7 South Africans is selected. The probability that the first 5 will watch the

opening match is:

Answer 5

55555(1 − 5)(1 − 5) = (0.6)@(0.4)# = 0.0124.

Question 6

The probability that exactly 23 South Africans will watch the opening match is:

Answer 6

�(� = 23) = (23) = 340234 (0.6)#8(0.4)�A = 0.1204.

Or using Excel we obtain: �(� = 23) = �(� ≤ 23) − �(� ≤ 22) = 0.4319 − 0.3115 = 0.1204.

Question 7

The probability that more than 18 but less than 24 South Africans will watch the opening match is:

Answer 7

�(18 < � < 24) = �(� ≤ 23) − �(� ≤ 18) = 0.4319 − 0.0392 = 0.3927

Question 8

The probability that more than 20 but less than 24 South Africans will not watch the opening match is:

Answer 8

Let y = the number of South Africans who will not watch the opening match.

The probability that someone will not watch the opening match is 1 – 0.6 = 0.4. These probabilities are

given in column C of the Excel spreadsheet [see cells C9, C10 and C11].

�(20 < Q < 24) = �(Q = 21) + �(Q = 22) + �(Q = 23) = 0.0352 + 0.0203 + 0.0106 = 0.0661.

Question 9

The standard deviation of � is:

Answer 9

9:;<=(�) = 2�5(1 − 5) = √40 × 0.6 × 0.4 = 3.10.


Consider the XYZ University. Five students writing the Accounting exam were selected at random. It is

known from previous experience that the probability that any one student will pass the exam is 0.4.

Let: x = the number of students who pass the exam.

Define: An experimental outcome is a sequence of successes and failures in a 5-trial binomial

experiment. (Tree diagram)


S

S

S

S

S

S

S

F

S

S

S

S

S

S

S

S

F

F

F

F

F

F

F

F

F

F

F

F

F

F

Question 10

The total number of experimental outcomes with 3 successes out of 5 is:

By drawing a tree diagram we find the answer = 10.

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

S

F

Outcome

(S, S, S, S, S)

(S, S, S, S, F)

(S, S, S, F, S)

(S, S, S, F, F)

(S, S, F, S, S)

(S, S, F, S, F)

(S, S, F, F, S)

(S, S, F, F, F)

(S, F, S, S, S)

(S, F, S, S, F)

(S, F, S, F, S)

(S, F, S, F, F)

(S, F, F, S, S)

(S, F, F, S, F)

(S, F, F, F, S)

(S, F, F, F, F)

(F, S, S, S, S)

(F, S, S, S, F)

(F, S, S, F, S)

(F, S, S, F, F)

(F, S, F, S, S)

(F, S, F, S, F)

(F, S, F, F, S)

(F, S, F, F, F)

(F, F, S, S, S)

(F, F, S, S, F)

(F, F, S, F, S)

(F, F, S, F, F)

(F, F, F, S, S)

(F, F, F, S, F)

(F, F, F, F, S)

(F, F, F, F, F)

Number of

successes

5

4

4

3

4

3

3

2

4

3

3

2

3

2

2

1

4

3

3

2

3

2

2

1

3

2

2

1

2

1

1

0


Question 11

The probability that only the first and last student will pass the exam is:

Answer 11

5(1 − 5)(1 − 5)(1 − 5)5 = 5#(1 − 5)8 = 0.4#(0.6)8 = 0.03456

Question 12

The probability that at most 4 of the 5 students will pass the exam is:

Answer 12

�(� ≤ 4)

= �(� = 0) + �(� = 1) + �(� = 2) + �(� = 3) + �(� = 4)

= 3504 0.4�0.6@ + 35

14 0.4�0.6? + 3524 0.4#0.68 + 35

34 0.480.6# + 3544 0.4?0.6�

= 0.07776 + 0.2592 + 0.3456 + 0.2304 + 0.0768 = 0.98976.

OR

�(� ≤ 4) = 1 − �(� = 5) = 1 − 3554 [email protected]� = 1 − 0.01024 = 0.98976.

Question 13

The probability that 2 of the 5 students will fail is:

Answer 13

If 2 out of 5 fail the exam, the 3 out of 5 pass the exam. Therefore,

�(� = 3) = 3534 0.480.6# = 0.2304.

OR

Let y = the number of students who fail the exam

The probability that any one student will fail the exam is 0.6.

�(Q = 2) = 3524 0.6#0.48 = 0.2304.

Documents

Chapter 5 Discrete Probability Distributions Random … · Chapter 5 Discrete Probability Distributions ... = probability of 5 cars sold . ... Does it fulfill the basic requirements