ADDITION THEOREMS OF PROBABILITY

Theorem 1 :

For any two events A and B, the probability that either event ‘A’ or event ‘B’ occurs or both occur is

P(AuB)  =  P(A) + P(B) – P(AnB)

Theorem 2 :

For any three events A, B and C, the probability that any one of the events occurs or any two of the events occur or all the three events occur is

P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) - P(BnC) - P(AnC) + P(AnBnC)

Theorem 3 :

For any two mutually exclusive events A and B, the probability that either A or B  occurs is given by the sum of individual probabilities of A and B.

P(AuB)  =  P(A) + P(B)

Note :

If two events A and B are mutually exclusive, then

AnB  =  Null set

P(AnB)  =  0

Theorem 4 :

For any three mutually exclusive events A, B and C, the probability that the event A or B or C  occurs is given by the sum of individual probabilities of A B and C.

P(AuBuC)  =  P(A) + P(B) + P(C)

Practice Problems

Problem 1 :

A number is selected from the first 25 natural numbers. What is the probability that it would be divisible by 4 or 7 ?

Solution :

Let A be the event that the number selected would be divisible by 4 and B, the event that the selected number would be divisible by 7.

Then AuB denotes the event that the number would be divisible by 4 or 7.

Next we note that

A = {4, 8, 12, 16, 20, 24}

and

B = {7, 14, 21}

whereas S  =  {1, 2, 3, ……... 25}.

Since AnB  =  Null set , the two events A and B are mutually exclusive and as such we have

P(AuB)  =  P(A) + P(B)

P(AuB)  =  (6/25) + (3/25)

P(AuB)  =  9/25

Problem 2 :

A number is selected from the first 30 natural numbers. What is the probability that it would be divisible by 4 or 7 ?

Solution :

Let A be the event that the number selected would be divisible by 4 and B, the event that the selected number would be divisible by 7.

Then AuB denotes the event that the number would be divisible by 4 or 7.

Next we note that

A = {4, 8, 12, 16, 20, 24, 28}

B = {7, 14, 21, 28}

and 

AnB = {28}

whereas S  =  {1, 2, 3, ……... 30}.

Here, the required probability is

P(AuB)  =  P(A) + P(B) - P(AnB)

P(AuB)  =  (6/25) + (3/25) - 1/25

P(AuB)  =  8/25

Problem 3 :

There are three persons A, B and C having different ages. The probability that A survives another 5 years is 0.80, B survives another 5 years is 0.60 and C survives another 5 years is 0.50. The probabilities that A and B survive another 5 years is 0.46, B and C survive another 5 years is 0.32 and A and C survive another 5 years 0.48. The probability that all these three persons survive another 5 years is 0.26. Find the probability that at least one of them survives another 5 years.

Solution :

From the given information, we have the following

P(A)  =  0.80

P(B)  =  0.60

P(C)  =  0.50,

P(AnB)  =  0.46

P(BnC)  =  0.32

P(AnC)  =  0.48

P(AnBnC)  =  0.26

The probability that at least one of them survives another 5 years in given by

=  P(AuBuC)

=  P(A) + P(B) + P(C) – P(AnB) - P(BnC) - P(AnC) + P(AnBnC)

= 0.80 + 0.60 + 0.50 – 0.46 – 0.32 – 0.48 + 0.26

P(AuBuC)  =  0.90

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