The document discusses probability and random processes. It defines random variables as functions that assign real numbers to elements of a sample space from a random experiment. Random variables can be discrete or continuous. Discrete random variables take countable values while continuous random variables take all values in an interval. Probability mass functions define probabilities for discrete random variables and probability density functions define probabilities for continuous random variables. The cumulative distribution function gives the probability that a random variable is less than or equal to a value. Examples are provided to illustrate discrete and continuous random variables and their probability functions.

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