*88*

# Addition Theorem

**Theorem1:** If A and B are two mutually exclusive events, then

Â Â Â Â Â Â Â Â Â Â P(A âˆªB)=P(A)+P(B)

**Proof:** Let the n=total number of exhaustive cases

Â Â Â Â Â Â Â Â Â Â Â n_{1}= number of cases favorable to A.

Â Â Â Â Â Â Â Â Â Â Â n_{2}= number of cases favorable to B.

Now, we have A and B two mutually exclusive events. Therefore, n_{1}+n_{2} is the number of cases favorable to A or B.

**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.

Â Â Â Â âˆ´ P (an even number on Ist die) =

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

Â Â Â Â Â Â Â Â {(2,6),(3,5),(4,4),(5,3),(6,2)}

Â Â Â Â âˆ´ Â Â P (a total of 8) =

Â Â Â Â Â Â âˆ´ Â Â Total Probability =

**Theorem2:** If A and B are two events that are not mutually exclusive, then

Â Â Â Â Â Â Â Â Â P(A âˆªB)=P(A)+P(B)- P (Aâˆ©B).

**Proof:** Let n = total number of exhaustive cases

Â Â Â Â Â Â Â Â Â Â Â n_{1}=number of cases favorable to A

Â Â Â Â Â Â Â Â Â Â Â n_{2}= number of cases favorable to B

Â Â Â Â Â Â Â Â Â Â Â n_{3}= 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,n_{1}+n_{2}-n_{3} is the number of cases favorable to A or B.

Therefore, P(A âˆªB)=

But we have, P(A)=, P(B) =and P (Aâˆ©B)=

Hence, Â Â Â P(A âˆªB)=P(A)+P(B)- P (Aâˆ©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)

âˆ´ Â Â Now, P(even number on Ist die)=

Ordered Pairs showing a total of 8 = {(6, 2), (5, 3), (4, 4), (3, 5), (2, 6)} = 5

âˆ´ Â Â Â Probability; P(total of 8) =

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

âˆ´ Â Â Â Â Required Probability =

**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 â‰¤ 5

D = getting a sum of the number on dice > 5 but less than 10.

E = getting sum of the number on dice â‰¥ 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. Aâ€™âˆ©Bâ€™ are a mutually exclusive and exhaustive event.

6. A, B, F are not a mutually exclusive event.

**Solution:**

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

Â Â Â (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

Â Â Â (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)

Â Â Â (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

Â Â Â (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)

Â Â Â (3,3),(3,4),(3,5),(3,6)

Â Â Â (4,2),(4,3),(4,4),(4,5)

Â Â Â (5,1),(5,2),(5,3),(5,4)

Â Â Â (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)

Â Â Â (2,1),(2,3),(2,5)

Â Â Â (3,2),(3,4),(3,6)

Â Â Â (4,1),(4,3),(4,5)

Â Â Â (5,2),(5,4),(5,6)

Â Â Â (6,1),(6,3),(6,5)

**1.** (Aâˆ©B) =âˆ… and (AâˆªB)=S

Â Â Â A, B are a mutually exclusive and exhaustive event.

**2.** (Aâˆ©C) are not mutually exclusive

Â Â Â (2,1),(2,3),(4,1)â‰ âˆ…

**3.** Câˆ©D are a mutually exclusive but not exhaustive event.

Â Â Â Câˆ©D=âˆ… Â Â Â Câˆª Dâ‰ S

**4.** Câˆ©D=âˆ…,Dâˆ©E=âˆ…, Câˆ©E=âˆ… are mutually exclusive and exhaustive event.

**5.** Aâ€™âˆ©Bâ€™ =(AâˆªB)â€™ are a mutually exclusive and exhaustive event.

**6.** (Aâˆ©B) =âˆ… are a mutually exclusive

Â Â Â A, B, F are not mutually exclusive events.