chapter 5: discrete probability...

FAROUQ ALAM, Ph.D.

Department of Statistics, KAU

Textbook:

Bluman, A. G. (2016). Elementary Statistics a Step by Step Approach. McGraw-Hill Education. Customized edition for the Department of Statistics at King Abdulaziz University, pp. 258 – 309

Chapter 5: Discrete Probability Distributions

5 – 1: Probability Distributions

A random variable is a variable whose values are

determined by chance.

A discrete probability distribution consists of the

values a random variable can assume and the

corresponding probabilities of the values. The

probabilities are determined theoretically or by

observation.

Example 5 – 1: Rolling a Die

The probability distribution for rolling a single

die is:

Outcomes

X

1 2 3 4 5 6

Probability

P(X)

1/6 1/6 1/6 1/6 1/6 1/6

The random variable and its values

The probabilities of the values of the

random variable

Example 5 – 2: Tossing Coins

Find the probability distribution of the number

of heads when tossing three coins.

Solution: If X represents the number of heads,

then X = 0 (no heads), 1 (one head), 2 (two

heads), 3 (three heads).

Example 5 – 2 (cont.)

Sample space:

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

X = 0 means TTT; hence, P(X = 0) = 1/8.

X = 1 means HTT, THT, or TTH; hence, P(X = 1)

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

X = 2 means HHT, HTH, or THH; hence, P(X =

2) = 1/8 + 1/8 + 1/8 = 3/8.

X = 3 means HHH; hence, P(X = 3) = 1/8.


The probability distribution of the number of

heads when tossing three coins is:

Number of

heads X

0 1 2 3 Total

Probability

P(X)

1/8 3/8 3/8 1/8 8/8

= 1

Extra example:Example 5 – 3 (Page 260-261)

Two Requirements for a Probability

Distribution

Requirement #1:

0 ≤ 𝑃 𝑋 ≤ 1, ∀𝑋

Requirement #2:

∑𝑃 𝑋 = 1

Example 5 – 4: Probability Distributions

Determine whether each distribution is a probability

distribution.

(a)

(b)

X 5 8 11 14

P(X) 0.2 0.6 0.1 0.3

Not a probability

distribution!

X 1 2 3 4 5

P(X) 1/4 1/8 3/8 1/8 1/8

A probability

distribution!


(c)

(d)

X -1 2 3 4

P(X) ¼ ¼ ¼ ¼

A probability

distribution!

X 4 8 12

P(X) -0.5 0.6 0.4

Not a probability

distribution!

5 – 2: Mean, Variance, Standard

Deviation, and Expectation

Mean (expected value) of X:

𝜇 = 𝐸 𝑋 = ∑𝑋 ⋅ 𝑃(𝑋)

Variance of X:

𝜎2 = ∑𝑋2 ⋅ 𝑃 𝑋 − 𝜇2

Standard Deviation of X:

𝜎 = 𝜎2 = ∑𝑋2 ⋅ 𝑃 𝑋 − 𝜇2

Example 5 – 5 and 5 – 9: Rolling a Die

Find the missing value K, mean, variance, and

standard deviation of the number of spots that

appear when a die is tossed.

𝑿 1 2 3 4 5 6

𝑃(𝑋) 1/6 K 1/6 1/6 1/6 1/6

Example 5 – 5 and 5 – 9 (cont.)

𝑿 1 2 3 4 5 6 ∑

𝑃(𝑋) 1/6 1/6 1/6 1/6 1/6 1/6 1

𝑋 ⋅ 𝑃(𝑋) 1/6 2/6 3/6 4/6 5/6 6/6 21/6

𝑋2 ⋅ 𝑃(𝑋) 1/6 4/6 9/6 16/6 25/6 36/6 91/6

The mean (expected value):

𝜇 = 𝐸 𝑋 = ∑𝑋 ⋅ 𝑃 𝑋 =𝟐𝟏

𝟔= 𝟑. 𝟓

Example 5 – 5 and 5 – 9 (cont.)

The variance:

𝜎2 = ∑𝑋2 ⋅ 𝑃 𝑋 − 𝜇2 =𝟗𝟏

𝟔−

𝟐𝟏

𝟔

2

=91

6−441

36

=546 − 441

36=𝟏𝟎𝟓

𝟑𝟔= 𝟐. 𝟗𝟏𝟕

The standard deviation:

𝜎 = 2.917 ≈ 𝟏. 𝟕𝟎𝟖

The Expected value of X

The expected value E(X) of a discrete random

variable X of a probability distribution is the

theoretical average of the variable.

Example 5 – 14: Bond Investment

A financial adviser suggests that his client select one

of two types of bonds in which to invest $5000.

Bond A pays a return of 4% and has a default

rate of 2%. Bond B has a return 2.5% and a

default rate of 1%. Find the expected rate of

return and decide which bond would be a better

investment. When the bond defaults, the investor

loses all the investment.


Bond A:

E(X) = $96

Win Lose

X $200 -$5000

𝑃(𝑋) 0.98 0.02

𝑋 ⋅ 𝑃(𝑋) 196 -100


Bond B:

E(X) = $73.75

Win Lose

X $125 -$5000

𝑃(𝑋) 0.99 0.01

𝑋 ⋅ 𝑃(𝑋) 123.75 -50

Other examples

Mean: Examples 5 – 6, 5 – 7, and 5 – 8.

Variance and standard deviation: Examples 5 –

10, and 5 – 11.

Expected value: Example 5 – 12.

5 – 3: The binomial distribution

A binomial experiment is a probability experiment that

satisfies the following four requirements:

1. There must be a fixed number of trials.

2. Each trial can have only two outcomes or outcomes

that can be reduced to two outcomes. These

outcomes can be considered as either success or

failure.

3. The outcomes of each trial must be independent of

one another.

4. The probability of a success must remain the same

for each trial.

Example 5 – 15

Decide whether each experiment is a binomial

experiment:

Selecting 20 university students and recording their class

rank.

This is not a binomial experiment because we have

more than two outcomes.


Selecting 20 university students and recording their

gender.

This is a binomial experiment since all four

requirements are met.

Selecting 5 cards without replacement and recording

whether they are red or black cards.

This is not a binomial experiment because the

outcomes of each trial are not independent.


Selecting five students from a large school and asking if

they are on the dean’s list.

This is a binomial experiment since all four

requirements are met.

Recording the number of children in 50 randomly selected

families.

This is not a binomial experiment because we have more

than two outcomes.

The binomial distribution

The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

In a binomial experiment, the probability of exactly X successes in n trials is:

𝑃 𝑋 =𝑛!

𝑛 − 𝑋 ! 𝑋!𝑝𝑋𝑞𝑛−𝑋

Important note: 𝑝 + 𝑞 = 1, 0 ≤ 𝑝, 𝑞 ≤ 1

The binomial distribution (cont.)

Notation Meaning

𝑛 The number of trials

𝑋The number of successes in 𝑛 trials

𝑋 = 0,1, … , 𝑛

𝑝 The probability of success

𝑞 The probability of failure

Example 5 – 18: Survey on Employment

A survey from Teenage Research Unlimited (Northbrook, Illinois) found that 30% of teenage consumers receive their spending money from part-time jobs. If 5 teenagers are selected at random, find the probability that:

Exactly 3 of them will have part-time jobs.

At most 3 of them will have part-time jobs.

At least 3 of them will have part-time jobs.

Solution: Notice that n = 5, p = 0.30, and q = 0.70.


𝑃 0 =5!

5−0 !0!0.3 0 0.7 5−0 ≈ 0.168

𝑃 1 =5!

5−1 !1!0.3 1 0.7 5−1 ≈ 0.360

𝑃 2 =5!

5−2 !2!0.3 2 0.7 5−2 ≈ 0.309

𝑃 3 =5!

5−3 !3!0.3 3 0.7 5−3 ≈ 0.132

𝑃 4 =5!

5−4 !4!0.3 4 0.7 5−4 ≈ 0.028

𝑃 5 =5!

5−5 !5!0.3 5 0.7 5−5 ≈ 0.002


Exactly 3 of them will have part-time jobs:

𝑃 3 ≈ 0.132

At most 3 of them will have part-time jobs:

𝑃 At most 3 of them will have part − time jobs= 𝑃 3 +⋯+ 𝑃 0 ≈ 0.132 + 0.309 + 0.360 + 0.168= 0.969

At least 3 of them will have part-time jobs.

𝑃 At least 3 of them will have part − time jobs= 𝑃 3 + 𝑃 4 + 𝑃 5 ≈ 0.132 + 0.028 + 0.002 = 0.162

Mean, Variance, and Standard Deviation

for the binomial distribution

The mean:

𝜇 = 𝑛 ⋅ 𝑝

The variance:

𝜎2 = 𝑛 ⋅ 𝑝 ⋅ 𝑞

The standard deviation:

𝜎 = 𝜎2 = 𝑛 ⋅ 𝑝 ⋅ 𝑞

Example 5 – 22: Tossing a Coin

A coin is tossed 4 times. Find the mean, variance, and standard deviation of the number of heads that will be obtained.

Solution: First, the probability of getting a head is ½; therefore, p = ½ and q = ½. Second, the coin is tossed four times, so n = 4. Hence,

𝜇 = 4 ⋅1

2= 2, 𝜎2 = 4 ⋅

1

2⋅1

2= 1, 𝜎 = 1 = 1

Other examples

The binomial distribution: Examples 5 – 16, and 5

– 17.

Mean, variance and standard deviation for the

binomial distribution: Examples 5 – 23, and 5 –

24.

chapter 5: discrete probability...

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