Chapter 5: Discrete Probability Distributions

5.1  Random Variable (RV): A RV is a numerical description of the outcome associated with

an experiment. This value will depend on the outcome of the experiment, thus it is variable.

 Discrete RV : A random variable with a finite number of values or an infinite sequence of values

such as 0, 1, 2, 3, … is referred to as discrete random variable. Table 5.1 (p.178) represents

examples of discrete random variable.

Continuous RV : A random variable which assumes any values in an interval or a number of 

intervals is called continuous random variable. Experimental outcomes that are based on

measurement scales such as time, weight, distance and temperature can be described by

continuous RVs. A non-discrete RV is continuous. Some examples are given in Table 5.2

(p.178).

5.2 Discrete Probability Distributions

 Probability distribution of a RV : It describes how probabilities are distributed over the values of 

the RV. For a discrete random variable x, the probability distribution is defined by a

probability function, denoted by f(x). It provides the probability for each value of the RV.

Table 5.3 (p.180) shows the probability distribution for the number of automobiles sold 

during a day at DicCarlo Motors.

Required conditions for a discrete probability function:

(a) f (x) ≥ 0, and (b) Σ  f (x) = 1

where the summation ranges over the values of RV x.

The probability in Table 5.3 satisfied the above conditions.

Fig.5.1 is a graphical representation of probability in Table 5.3.

Expected Value and Variance

 Expected value: The mean of a RV. For a discrete RV x, it is defined as

 E(x) =  µ = Σ x f(x)

From Table 5.5 the expected value is calculated as 1.5 automobiles per day. Thus the average

monthly sales is 30(1.5) = 45 automobiles.

Variance of a discrete RV : Var(x) = σ  2 = Σ (x- µ  )2 f(x).

From Table 5.6, the variance is calculated as 1.25 (p.186).

 Standard deviation: σ = √σ2. The Standard deviation is 1.118.

Example 1: A Doctor has determined that the number of hours required to obtain the trust of a

new patient is 1, 2, 3, 4 or, 5. 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.

 f ( x) = cx, x = 1, 2, 3, 4, 5. where c is a constant.

(a) Find the value of the constant c.

Answer:

∑ )( x  f   = c∑ x =c(1+2+3+4+5)=15c = 1

Therefore, c =15

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(b) What is the probability that it takes at least 2 hours to gain the patient's trust?

Answer: Here,  f ( x) =15

 x

P(x≥2) = f (2)+ f (3)+ f (4)+ f (5) =15

2+15

3+15

4+15

5=15

14=0.9333

(c) Find Var(x).

 Answer: Var(x)= E(x2) –[E(x)]2

E(x)=15

2∑ x=

15

2516941 ++++

=3

11, E(x2)=

15

3∑ x=

15

125642781 ++++

=15

Var(x)= E(x2) –[E(x)]2 = 15- (11/3)2 = 14/9 = 1.56

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Discrete Uniform Probability Function  f(x) = 1/n

E(x) = (n+1)/2, Var(x) = (n2-1)/12

Where n is the number of values the random variable may assume.

Example 2: For a die rolling experiment, the random variable x = 1, 2, 3, 4, 5, 6. Here n = 6. The

 probability function for x is f(x) = 1/6 

The possible values of the random variables and their associated probabilities are given belowx 1 2 3 4 5 6 Total

f( x) 1/6 1/6 1/6 1/6 1/6 1/6 1

E(x) =2

16+=3.5, Var(x) =

12

136−=12

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5.4 The Binomial Probability Distribution

When a distribution that consists of discrete probability function derived from a multi-step

experiment, it is called the binomial experiment.

 Binomial experiment : An experiment with the following properties:

1. The experiment consists of n identical trials.

2. Two outcomes, a success and a failure, are possible on each trial.

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

 probability of failure, 1-p, does not change either.

4. The trials are independent, or the outcome of a trial is not influenced by the outcome of a

 previous trial.

An experiment with properties 2-4 is called a Bernoulli process. Hence, a binomial experiment is

a sequence of identical (i.e., the success probability p remains constant) Bernoulli processes.

If we tossed a coin 5 times, then

1. The experiment consists of 5 identical trials.

2. Two outcomes are possible for each trial: a head (success), a tail (Failure).

3. The probability of head and the probability of a Tail are the same for each trial with p = .5

and q =(1-p) =.5

4. The trials or tosses are independent.

Thus the properties of a Binomial experiment are satisfied.

 Binomial RV : The variable x denotes the number of successes in n trials.

x = 0, 1, 2, 3, - - - n.

The binomial probability function:

Where  f (x) = probability of  x successes in n trials

   

  x

n= Number of exactly x successes in n trials =

)!(!

!

 xn x

n

− 

 p = probability of success on any trial

1-p = probability of failure on any trial.

The probability distribution associated with the binomial random variable is called Binomial

 probability distribution.

By knowing n and p, we will know the probability of each value of  x. n and p are said to be the

 parameters of the binomial distribution. The probabilities are calculated in Table 5.7 and

 presented graphically in Fig. 5.4. We can also find the probability value from Table 5 of 

Appendix B. (see Table 5.8).

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n x p p x

n x  f  

xn x,...,1,0,)1()( =−  

 

  

 = −

 

Expected value and variance for a binomial variable with parameters n and p:

 E (x) =  µ = np

Var (x) = σ 2 = np(1-p)

Example 3: Ten percent items produced by a machine are defective. If we randomly select 15

items,

(a) What is the probability that at least one item will be defective?

[Number of defective items follows binomial distribution]

Answer: n= 15 and p =0.1,  f (x)= x x

 x

−

15)9.0()1.0(15

P(x≥1) = 1-f(0) = 1-(0.9)15 =1- 0.2059 = 0.7941

(b) Find the mean and variance of this distribution.

Answer:

Mean =E(x)= np= 15 ×0.1=1.5

Var(x)=np(1-p)=15×0.1×0.9=1.35

5.5 Poisson Probability Distribution

This is used to describe the number of occurrences in an interval of time or space. It has a discrete

random variable that may assume an infinite sequence of values (x = 0, 1, 2, - - -). The Poisson

random variable has no upper limit The characteristics of the Poisson distribution are:

1. The occurrence of the events are independent

2. An infinite number of outcomes must be possible in an interval of times.

3. The probability of 2 or more occurrence of the event is negligible in any small proportion

of time.

Examples:

1. The arrival of telephone calls at the switch board of SQU in an interval of time.

2. The number of traffic accident in Muscat city.

3. The arrival of customers in a departmental store in a particular time of a day.

Poisson Probability Function

where

 f (x) = probability of  x occurrences in an interval

   µ = mean number of occurrences in an interval

  e = 2.71828, the base of natural logarithm

The mean and the variance of Poisson distribution are same (µ).

We can also find the probability value from Table 7 of Appendix B.

Example 4: Number of car accidents in a city follows a Poisson distribution with  µ =3. What is

the probability that there will be at most 2 accidents?

Answer:  f (x) =!

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 x

e x−

, x=0, 1, 2, 3, . . .

P(x≤2)= f( 0)+ f (1)+ f (2)= 0.0498+0.1494+0.2240=0.4232

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,...2,1,0,!

)( ==

−

 x x

e x  f  

 x µ  µ 

 

************

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