# ADDITION THEOREM OF PROBABILITY EXAMPLES

Example 1 :

The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of

(i) P(A U B) (ii) P(A n B') (iii) P(A' n B).

Solution :

(i) P(A U B)

P(A)  =  0.5, P(B)  =  0.3

Since A and B are mutually exclusive events, P(AnB) will be 0.

P(A U B)  =  P(A) + P(B)

=  0.5 + 0.3

P(A U B)  =  0.8

(ii) P(A n B')

P(A n B')  =  P(A) - P(AnB)

P(A n B')  =  0.5 - 0

P(A n B')  =  0.5

(iii) P(A' n B)

P(A' n B)  =  P(B) - P(AnB)

=  0.3 - 0

P(A' n B)  =  0.3

Example 2 :

A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.

(i) What is the probability that a fire engine is available when needed?

(ii) What is the probability that neither is available when needed?

Solution :

Let A and B be the chance of getting fire engines.

P(A)  =  0.96 and P(B) =  0.96

(i) What is the probability that a fire engine is available when needed?

= Probability of getting two fire engines at a time + Probability of getting A but not B + Probability of getting B not A.

=  P(AnB) + P(AnB') + P(A'nB)

P(A')  =  1 - 0.96  =  0.04

P(B') =  1 - 0.96  =  0.04

Note : A and B are independent events.

=  P(A) ⋅ P(B) + P(A) ⋅ P(B') + P(A') ⋅ P(B)

=  0.96 (0.96) + 0.96 (0.04) + 0.04(0.96)

=  0.9216 + 0.0384 + 0.0384

=  0.9984

(ii) What is the probability that neither is available when needed?

=  P(A' n B')

=  P(A')  P(B')

=  0.04 (0.04)

P(A' n B')  =  0.0016

Example 3 :

The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that (i) it will get at least one of the two awards (ii) it will get only one of the awards.

Solution :

Let A and B be the event of getting an award for its design and efficient use of materials respectively.

P(A)  =  0.48, P(B)  =  0.36, P(AnB)  =  0.2

(i) it will get at least one of the two awards

P(AUB)  =  P(A) + P(B) - P(AnB)

=  0.48 + 0.36 - 0.2

P(AUB)  =  0.64

(ii) it will get only one of the awards

He may get either an award A or award B.

=  P(A'nB) + P(AnB')

=  P(B) - P(AnB) + P(A) - P(AnB)

=  0.36 - 0.2 + 0.48 - 0.2

=  0.44

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