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Addition Theorem

Theorem1: If A and B are two mutually exclusive events, then
                   P(A ∪B)=P(A)+P(B)

Proof: Let the n=total number of exhaustive cases
                      n1= number of cases favorable to A.
                      n2= number of cases favorable to B.

Now, we have A and B two mutually exclusive events. Therefore, n1+n2 is the number of cases favorable to A or B.

Addition Theorem

Example: Two dice are tossed once. Find the probability of getting an even number on first dice or a total of 8.

Solution: An even number can be got on a die in 3 ways because any one of 2, 4, 6, can come. The other die can have any number. This can happen in 6 ways.
        ∴ P (an even number on Ist die) = Addition Theorem

A total of 8 can be obtained in the following cases:

                {(2,6),(3,5),(4,4),(5,3),(6,2)}
        ∴     P (a total of 8) = Addition Theorem
            ∴     Total Probability =Addition Theorem

Theorem2: If A and B are two events that are not mutually exclusive, then
                 P(A ∪B)=P(A)+P(B)- P (A∩B).

Proof: Let n = total number of exhaustive cases
                      n1=number of cases favorable to A
                      n2= number of cases favorable to B
                      n3= number of cases favorable to both A and B

But A and B are not mutually exclusive. Therefore, A and B can occur simultaneously. So,n1+n2-n3 is the number of cases favorable to A or B.

Therefore, P(A ∪B)=Addition Theorem
But we have, P(A)=Addition Theorem, P(B) =Addition Theoremand P (A∩B)=Addition Theorem

Hence,       P(A ∪B)=P(A)+P(B)- P (A∩B).

Example1: Two dice are tossed once. Find the probability of getting an even number on first dice or a total of 8.

Solution: P(even number on Ist die or a total of 8) = P (even number on Ist die)+P (total of 8)= P(even number on Ist die and a total of 8)
∴    Now, P(even number on Ist die)= Addition Theorem

Ordered Pairs showing a total of 8 = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)} = 5
∴       Probability; P(total of 8) = Addition Theorem

P(even number on Ist die and total of 8) = Addition Theorem

∴         Required Probability = Addition Theorem

Example2: Two dice are thrown. The events A, B, C, D, E, F

A = getting even number on first die.
B= getting an odd number on the first die.
C = getting a sum of the number on dice ≤ 5
D = getting a sum of the number on dice > 5 but less than 10.
E = getting sum of the number on dice ≥ 10.
F = getting odd number on one of the dice.

Show the following:

1. A, B are a mutually exclusive event and Exhaustive Event.
2. A, C are not mutually exclusive.
3. C, D are a mutually exclusive event but not Exhaustive Event.
4. C, D, E are a mutually exclusive and exhaustive event.
5. A'∩B' are a mutually exclusive and exhaustive event.
6. A, B, F are not a mutually exclusive event.

Solution:

Addition Theorem

A: (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
     (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
     (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

B: (1,1), (1,2),(1,3),(1,4),(1,5),(1,6)
     (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
     (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

C: (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)

D: (1,5),(1,6),(2,4),(2,5),(2,6)
     (3,3),(3,4),(3,5),(3,6)
     (4,2),(4,3),(4,4),(4,5)
     (5,1),(5,2),(5,3),(5,4)
     (6,1),(6,2),(6,3)

E: (4,6),(5,5),(5,6),(6,5),(6,6),(6,4)

F: (1,2),(1,4),(1,6)
     (2,1),(2,3),(2,5)
     (3,2),(3,4),(3,6)
     (4,1),(4,3),(4,5)
     (5,2),(5,4),(5,6)
     (6,1),(6,3),(6,5)

1. (A∩B) =∅ and (A∪B)=S
     A, B are a mutually exclusive and exhaustive event.

2. (A∩C) are not mutually exclusive
     (2,1),(2,3),(4,1)≠ ∅

3. C∩D are a mutually exclusive but not exhaustive event.
     C∩D=∅       C∪ D≠S

4. C∩D=∅,D∩E=∅, C∩E=∅ are mutually exclusive and exhaustive event.

5. A'∩B' =(A∪B)' are a mutually exclusive and exhaustive event.

6. (A∩B) =∅ are a mutually exclusive
     A, B, F are not mutually exclusive events.






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