1. PROBABILITY AND RANDOM PROCESS

2. Random Variables
A random variable is a function that assigns a real number X(s) to every elements s in
S, where S is the sample space corresponding to a random experiment.
Random variables
Discrete Random Variables Continuous Random variables
Dr. S. Kalyani PRP 2

3. Discrete Random variable:
If X is a random variable which can take a finite number or countably infinite
number of values, X is called a Discrete Random variable.
Eg:
1. The marks obtained by the student in an exam.
2. Number of students absent for a particular class.
Continuous Random variable:
If X is a random variable which can take all values in an interval, then is called
a Continuous Random variable.
Eg:
The length of time during which a vacuum tube installed in a circuit functions is
a continuous RV.
Dr. S. Kalyani PRP 3

4. Probability Mass Function (or) Probability Function:
If X is a discrete RV with distinct values x1 x2 x3 ,… xn…
then P ( X = xi) = pi, then the function p(x) is called the
Probability Mass Function.
Where p(i = 1, 2, 3, …) satisfy the following conditions:
1. for all i, and
2.
0,
pi
1
i
i
p 

Dr. S. Kalyani PRP 4

5. PROBABILITY DENSITY FUNCTION FOR CONTINUOUS CASE:
If X is a continuous r.v., then f(x) is defined the probability density
function of X.
Provided f(x) satisfies the following conditions,
1.
2.
3.
( ) 0
f x 
( ) 1
f x dx




( ) ( )
b
a
P a x b f x dx
   
Dr. S. Kalyani PRP 5

6. CUMULATIVE DISTRIBUTION FUNCTION OF C.R.V.:
The cumulative distribution function of a continuous r.v. X is
( ) ( ) ( ) ,
x
F x P X x f x dx for x

       

CUMULATIVE DISTRIBUTION FUNCTION OF D.R.V.:
The cumulative distribution function F(x) of a discrete r.v. X with
probability distribution p(x) is given by
( ) ( ) ( )
i
i
X x
F x P X x p x for x

      

6
Dr. S. Kalyani PRP

7. Dr. S. Kalyani PRP 7
Relationship between probability density function and distribution function
𝑖) 𝑓 𝑥 =
𝑑
𝑑𝑥
𝐹 𝑥
𝑖𝑖) 𝐹 𝑥 =
−∞
𝑥
𝑓 𝑥 𝑑𝑥

8. PROBLEM 1:
The number of telephone calls received in an office during lunch time has the
following probability function given below
No. of calls 0 1 2 3 4 5 6
Probability
p(x)
0.05 0.2 0.25 0.2 0.15 0.1 0.05
i) Verify that it is really a probability function.
ii) Find the probability that there will be 3 or more calls.
iii)Find the probability that there will be odd number of calls.
Dr. S. Kalyani PRP 8

9. ii) Let X denote the no. of telephone calls
P(X>=3) = P(X=3) + P(X=4) + P(X=5) +P(X=6)
= 0.2 + 0.15 + 0.1 + 0.05
= 0.5
iii) Let X denote the no. of telephone calls
P(X is odd) = P(X=1) + P(X=3) + P(X=5)
= 0.2 + 0.2 + 0.1
= 0.5
i) Probability function p(x) follows 2 properties
i) all probability values lies between 0 and 1
ii) total probability value is 1.
In the given data all probability values are between 0 and 1
And 0.05+0.2+0.25+0.15+0.1+0.05 = 1
So it is a probability function.
Solution:
Dr. S. Kalyani PRP 9

10. PROBLEM 2:
A r. v. X has the following probability functions
X 0 1 2 3 4 5 6 7
p(x) 0 k 2k 2k 3k k2 2k2 7k2 + k
Find (i) value of k
(ii) P(1.5 < X < 4.5 / X > 2)
(iii) if P(X ≤ a) > ½, find the minimum value of a.
Dr. S. Kalyani PRP 10

11. Solution:
7
0
2
( ) . . . ( ) 1
10 9 1
(10 1)( 1) 0
1
, 1
10
( ) cannot be negative,
1 is neglected.
1
10
x
i w k t p x
k k
k k
k
p x
k
k


  
   
  
 
 

Dr. S. Kalyani PRP 11

12. 7
3
( ) (1.5 4.5 / 2)
( )
( / )
( )
[(1.5 4.5) ( 2)]
(1.5 4.5 / 2)
( 2)
( 3) ( 4)
( )
5
5
10
=
7 7
10
r
ii P X x
P A B
P A B
P B
P X X
P X X
P X
P X P X
P X r

  


   
   

  




Dr. S. Kalyani PRP 12
0 1 2 3 4 5

13. ( ) By trials,
( 0) 0
1
( 1)
10
3
( 2)
10
5
( 3)
10
8
( 4)
10
1
the minimum value of 'a' satisfying the condition ( ) 4
2
iii
P X
P X
P X
P X
P X
P X a is
 
 
 
 
 
  
X 0 1 2 3 4 5 6 7
p(x) 0 1/10 2/10 2/10 3/10 1/100 2/100 17/100
Dr. S. Kalyani PRP 13

14. PROBLEM 3
A random Variable X has the following probability distribution.
Find
(i) the value of k
(ii) Evaluate P[X<2] and P[-2<X<2]
(iii)Find the cumulative Distribution of X.
x -2 -1 0 1 2 3
p(x) 0.1 k 0.2 2k 0.3 3k
Dr. S. Kalyani PRP 14

15. Solution:
Dr. S. Kalyani PRP 15
6
0
( ) . . . ( ) 1
6 0.6 1
6 0.4
1
15
x
i w k t p x
k
k
k


  
 
 


16.  
     
   
2 2 1
0 1
1 2
0.1 0.2
15 15
3 1
0.3
15 2
ii
P X P X P X
P X P X
      
   
   
  
Dr. S. Kalyani PRP 16

17. Dr. S. Kalyani PRP 17
( )
[ ]
[ ] [ ] [ ]
2 2
1 0 1
1 2
0.2
15 15
3
= 0.2
15
2
5
ii
P X
P X P X P X
- < <
= = - + = + =
= + +
+
=

18. (iii) The Cumulative Distribution of X:
X F(x) = P(X ≤ x)
-2 F(-2)=P(X≤-2)=P(X=-2)=0.1=1/10
-1 F(-1)=P(X≤-1)=F(-2)+P(-1)=1/10+k=1/10+1/15=0.17
0 F(0)=P(X≤0)=F(-1)+P(0)=1/6+2/10=1/6+1/5=0.37
1 F(1)=P(X≤1)=F(0)+P(1)=11/30+2/15=15/30=1/2
2 F(2)=P(X≤2)=F(1)+P(2)=1/2+3/10=(5+3)/10=8/10=4/5
3 F(3)=P(X≤3)=F(2)+P(3)=4/5+3/15=(12+3)/15=1
Dr. S. Kalyani PRP 18

19. Dr. S. Kalyani PRP 19
Problem
If the random variable X takes the values 1,2,3 and 4 such that
2P(X=1) = 3P(X=2) = P(X=3) = 5P(X=4), find the probability distribution and
cumulative distribution function.
Let P(X=3) = k
So P(X=1) = k/2
P(X=2) = k/3
P(X=4) = k/5
By property, 𝑘 +
𝑘
2
+
𝑘
3
+
𝑘
5
= 1
So 𝑘 =
30
61

20. X 1 2 3 4
P(X) 15/61 10/61 30/61 6/61
Dr. S. Kalyani PRP 20
F(X)= 1
𝑥
𝑝(𝑋)
When X < 1 , F(X) = 0
X 1 2 3 4
F(X) 15/61 25/61 55/61 61/61 = 1

21. Problem 3:
The diameter, say X of an electric cable, is assumed to be a continuous r.v.
with p.d.f. :
f(x) = 6x(1-x), 0 ≤ x ≤ 1
(i) Check that the above is a p.d.f.
(ii) Compute P(X ≤ ½ / ⅓ ≤ X ≤ ⅔)
(iii) Determine the number k such that P(X< k) = P(X > k)
(iv) Find the cumulative distribution function.
Dr. S. Kalyani PRP 21

22. Solution:
1 1
0 0
1
2 3
0
( )
( ) 6 (1 )
6
2 3
1
i
f x dx x x dx
x x
 
 
 
 
 

 
Dr. S. Kalyani PRP 22

23. 1
2
1
3
2
3
1
3
( )
1 1
( )
1 1 2 3 2
( | )
1 2
2 3 3 ( )
3 3
6 (1 )
=
6 (1 )
11
11
54
13 26
27
ii
P X
P X X
P X
x x dx
x x dx
 
   
 


 


Dr. S. Kalyani PRP 23

24. 1
0
2 3 2 3
3 2
( )
( ) ( )
6 (1 ) 6 (1 )
3 2 3(1 ) 2(1 )
4 6 1 0
1 1 3
,
2 2
1
The only possible value of k in the given range is .
2
1
2
k
k
iii
We have P X k P X k
x x dx x x dx
k k k k
k k
k
k
  
   
     
   

 
 
 
Dr. S. Kalyani PRP 24

25. Dr. S. Kalyani PRP 25
𝑖𝑣) 𝐹 𝑥 =
−∞
𝑥
𝑓 𝑥 𝑑𝑥
𝐹 𝑥 =
0
𝑥
6𝑥 1 − 𝑥 𝑑𝑥
=
0
𝑥
(6𝑥 − 6𝑥2
)𝑑𝑥
= 6
𝑥2
2
− 6
𝑥3
3 0
𝑥
𝐹(𝑥) = 3𝑥2
− 2𝑥3 is the cumulative distribution function

26. Dr. S. Kalyani PRP 26
Problem
The distribution function of a random variable X is given by 𝐹 𝑋 = 1 − 1 + 𝑥 𝑒−𝑥
, 𝑥 ≥ 0
Find the density function.
Solution:
We know that 𝑓 𝑥 =
𝑑
𝑑𝑥
𝐹 𝑥
=
𝑑
𝑑𝑥
{1 − 𝑒−𝑥 − 𝑥𝑒−𝑥}
= 0 − −𝑒−𝑥
− (−𝑥𝑒−𝑥
+ 𝑒−𝑥
)
= 𝑒−𝑥
+ 𝑥𝑒−𝑥
− 𝑒−𝑥
= 𝑥𝑒−𝑥, 𝑥 ≥ 0. is the density function