PROBABILITY AND QUEUEING THEORY all unit notes

Subject Name & Code: MA 8402 - Probability and Queuing Theory 2020-21


St. JOSEPH’S COLLEGE OF ENGINEERING, CHENNAI-119
DEPARTMENT OF MATHEMATICS
CLASS NOTES
MA 8402 – PROBABILITY AND QUEUEING THEORY
UNIT-1 RANDOM VARIABLES
Random experiment: An experiment whose all possible outcomes are known, but it is not possible to predict the
outcome.
Probability:

Let A be a event and B be a its sample space then its probability on the occurrence on events is defined as
No. of favourable Cases
P  A  .
Total no. of exhaustic Cases

Axioms of Probability:
n n
(i) 0  P( E)  1 (ii) P(S )  1 (iii) P(  Ei )   P( Ei ) if Ei’s are mutually exclusive events.
i 1 i 1

Example: (i) A fair coin is “tossed” (ii) A die is “rolled” are random experiments, since we cannot predict the
outcome of the experiment in any trial.
Mutually exclusive:
Two events are said to be mutually exclusive if the occurrence of any one of them excludes the occurrence of other in
a single experiment.
Example: Tossing of Coin.
Independent events:
Two (or) more events are independent if the occurrence of one does not affect the occurrence of the other.
Example: If coin is tossed twice; result of second throw is not affected by the result of first throw.
Addition Law of Probability:

If A and B are two events in a sample space “S” then P  A  B   P  A  P  B   P  A  B  .

Conditional Probability:

The conditional probability of an event B assuming that the event A has happened, is defined as
P  A  B
P  B A  , P  A  0
P  A

P  A  B
Similarly, P  A B   , P  B  0 .
P  B




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, Subject Name & Code: MA 8402 - Probability and Queuing Theory 2020-21


1. If A and B are independent events then a) A and B b) A and B are also independent.

Solution:
Since A and B are independent,

P  A  B   P  A P  B    1

 
a) P A  B  P  A  P  A  B 

 P  A  P  A P  B  [u sin g (1)]

= P  A 1  P  B  

P  A  B   P  A P  B   A & B are independent events

 
b) P A  B  P A  B  
 1 P  A  B

 1   P  A  P  B   P  A  B  [ By addition theorem]

 1  P  A  P  B   P  A  B 

 1  P  A  P  B   P  A P  B  [u sin g 1]

 1  P  A  P  B  1  P  A

 1  P  A 1  P  B 

P  A  B   P  A P  B 

 A & B are indepentent events .

1 1 1
2. A Problem in statistics is given to three students. A, B and C whose chances of solving it are , and
2 3 4
respectively. What is the probability that the problem will be solved?
Solution:
Let A, B, C Denote the events that the problem is solved by the students A, B, C respectively.

1 1 1
Then P  A   , P  B  , P C  
2 2 4



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, Subject Name & Code: MA 8402 - Probability and Queuing Theory 2020-21



P  A  1
1 1

2 2

P  B   1
1 2

3 3

P C   1 
1 3

4 4

     
P(all the three students will not solve the problem)  P A P B P C
1 2 3 1
 . . 
2 3 4 4

P(all the three students will solve the problem)  P  A  B  C   1  P  A  P  B  P  C   1 
1 3

4 4

, P  AB   and P  A   find P  B  .
3 1 2
3. Event A and B are such that P  A  B  
4 4 3
Solution:

, P  AB   , P  A  
3 1 2
Given P  A  B  
4 4 3

 
i.e. P  A   1  P A  1 
2 1

3 3
By addition theorem

P  A  B   P  A  P  B   P  A  B 

i.e. P  B   P  A  B   P  A  P  A  B 

3 1 1 943 8 2
P  B      
4 3 4 12 12 3
4. An integer is chosen at random from two hundred digits. What is the probability that the integer is divisible
by 6 or 8?
Solution:

The sample space  1,2,3......199,200

n  S   100

Let the event A be an integer chosen that is divisible by 6,

i.e. A  6,12,18........198




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, Subject Name & Code: MA 8402 - Probability and Queuing Theory 2020-21


198
n  A   33
6

n  A 33
 n  A  
n  S  200

Let the event B be an integer chosen that is divisible by 8

i.e. B  8,16, 24.....200

200
n B   25
8

n  B  25
 P  B  
n  S  200

The L.C.M of 6 & 8 is 24.
Hence, a number that is divisible by both 6 & 8 is divisible by 24.

 A  B  24,48,72,.....192

192
n A  B  8
24

n  A  B 8
 P  A  B  
nS  200

Hence by addition theorem on probability

33 25 8 58  8 50 1
P  A  B   P  A  P  B   P  A  B       
200 200 200 200 200 4
5. A and B throw alternatively with a pair of dice. A wins if he throws 6 before B throws 7 and 8 wins if he
throws 7 before a throws 6.If A begins, show that their respective chances of winning are in the ratio 30:61.
Solution:

Let Ai denote the event of A’s throwing 6 in the ith thrown i  1, 2,3,...

‘6’ can be obtained with two dice in the following ways

1,55,1 2, 4 4, 23,3
i.e. 5 distinct ways

P  Ai  
5
36
 
, P Ai  1  P  Ai  
31
36
, i  1, 2,...



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