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Unit 27: Additive and Multiplicative Law of Probability

Similarly Notes

B= ( B ∪ )A ∩ ( A ∩ B )

∴ P (B) = ( PA ) ( ∩ ∩ B + PA B ) ... (b)

The events ( B and ( )A ∩ B being mutually exclusive. Thus, from (a) and (b), one gets
) A ∩
)
P (A) + P (B) = ( PA ) ( ∩ ∩ B + PA B +P ( )A ( ∩ ∩ B + P B ) A ... (c)
Now the last three terms on R.H.S. of (c), i.e., ( ) +PA ∩ B ) + P ( A ∩B B ) represent the probability
( ∩
PA
)
of occurrence of the events A or B or both A and B, i.e., ( PA ∪ B . Thus, replacing these three terms
by ( PA ∪ B , equation (c) can be written as
)
P (A) + P (B) = ( PA ∩ ) P (B + B )A ∪

(
)
or PA ∪ B = P (A) + P (B) – ( PA ∩ B ) ... (d)
The rule in (d) is called the addition of rule probability.
If A and B are mutually exclusive, ( PA ∩ B = 0 and the addition rule of probability becomes
)
)
(
PA ∪ B = P (A) + P (B) ... (e)
Example 1: What is the probability of getting an odd number in tossing a die ?
Solution: There are three odd numbers on a die, i.e., 1, 3 and 5. Let A, B and C be the respective
events of getting 1, 3 and 5. Thus, P (A) = 1/6, P (B) = 1/6 and P (C) = 1/6. Since A, B
and C are mutually exclusive therefore

1 1 1 3 1
P (A or B or C) = P (A) + P (B) + P (C) = + + = = .
6 6 6 6 2
Example 2: An urn contains 4 white and 2 red balls. Two balls are drawn randomly with
replacement. Find the probability that
(i) both balls are white
(ii) both balls will be of the same colour.
Solution: Here total number of balls = 6

6
Two balls can be drawn out of 6 in C ways i.e., 15 ways. Let A be the event that
2
4
both balls are white. The number (i) of ways of selecting 2 balls out of 4 is C i.e., 6
2
ways.
6
∴ P (A = both balls white) = 15

4
(ii) The number of ways of selecting 2 balls out of 4 white balls is C = 6 ways
2
2
The number of ways of selecting 2 both out of 2 red balls is C = 1 way
2
4 C 2 2 C 2
∴ P (both balls will be of the same colour) = 6 C 2 + 6 C 2

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