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Axiomatic Approach To Probability (Part II)

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Axiomatic Approach To Probability (Part II) - Lesson Summary

Theorem: If A and B are any two events of sample space S, then P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Proof:
 

 
(B - A) ∪ (A ∩ B) = B

(B - A) ∩ (A ∩ B) = ∅

⇒ (B – A) and (A ∩ B) are mutually exclusive events

P(B) = P[(B - A) ∪ (A ∩ B)]

⇒ P(B) = P(B - A) + P(A ∩ B)

⇒ P(B - A) = P(B) - P(A ∩ B)                    …(1)

From the figure, we have
A ∪ (B - A) = A ∪ B

Also, A ∩ (B - A) = ∅
⇒ A and (B - A) are mutually exclusive events

P[A ∪ (B - A)] = P(A ∪ B)

⇒ P(A ∪ B) = P(A) + P(B - A)

⇒ P(A ∪ B) = P(A) + P(B) - P(A ∩ B)  [from (1)]

If A and B are two mutually exclusive events in sample space S, then A ∩ B = ∅, and hence, P(A ∪ B) = P(A) + P(B).

If A and B are any two events in sample space S, then P(B - A) = P(B) - P(A ∩ B) and P(A - B) = P(A) - P(A ∩ B).

Example:
Find the probability of getting a prime number or an odd number, when a cubical die is rolled.
Let A be the event of the occurrence of a prime number.

A = {2, 3, 5}

Let B be the event of the occurrence of an odd number.
B= {1, 3, 5}

Thus, A ∩ B is the event of the occurrence of an odd prime number.
A ∩ B = {3,5}

P(A) = P(2) + P(3) + P(5)  = 1 6 + 1 6 + 1 6 = 3 6 = 1 2

P(B) = P(1) + P(3) + P(5)  =  1 6 + 1 6 + 1 6 = 3 6 = 1 2

P(A ∩ B) = P(3) + P(5) = 1 6 + 1 6 = 2 6 = 1 3

A ∪ B is the event of the occurrence of a prime number or an odd number.
A ∪ B = {1, 2, 3, 5}

The total number of possible outcomes is 6.

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

1 2 + 1 2 - 1 3

2 3

The required probability is  2 3 .
 
Theorem: If A, B and C are any three events in a sample space, then P(A ∪ B ∪ C) = P(A) + P(B) + P(C) -P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C).

Proof:
P(A ∪ B ∪ C) = P[A ∪( B ∪ C)]

= P(A) + P(B ∪ C) - P[A ∩( B ∪ C)]

= P(A) + P(B) + P(C) - P(B ∩ C) – P[(A ∩ B) ∪ (A ∩ C)]

= P(A) + P(B) + P(C) - P(B ∩ C) – { P(A ∩ B) + P(A ∩ C) - P[(A ∩ B) ∩ (A ∩ C)]}

= P(A) + P(B) + P(C) - P(B ∩ C) – P(A ∩ B) - P(A ∩ C) + P[(A ∩ B) ∩ (A ∩ C)]}

But (A∩B) ∩ (A∩C) = A∩B∩C

Hence, P(A ∪ B  ∪  C)= P(A) + P(B) + P(C) -P(A ∩ B) -P(B∩C)-P(C∩A) + P(A∩B∩C).
 
Theorem: If A is an event of a sample space, then A’ is the complement of event A, and P (not A) = 1 – P (A).

Proof:

We know that events A and A¢ are mutually exclusive and exhaustive.

⇒ A ∪ A' = S and A ∩ A' = ∅

Now, P(A ∪ A') = P(S)

⇒ P(A) + P(A') = 1  ...(by the axiomatic approach to probability)

⇒ P(not A) = 1 - P(A)
 
Example: The probability of event A = P(A) = 3 8

The probability of event “not A” = 1 – P(A) = 1 - 3 8 = 5 8 .

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