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Unit 27: Additive and Multiplicative Law of Probability

(ii) Now the event that both children are of the same sex is equivalent to the event Notes
that both are either boys or girls and these are mutually exclusive events. Thus
1 1 1
the required probability = + = .
4 4 2
Mutually exclusive events and independent events

These ideas are not equivalent ideas. We discuss them to bring out the difference between the two.
When the happening of one event precludes the happening of the other event, the two events are
mutually exclusive (or (disjoint). For two mutually exclusive events A and B
P (AB) = 0.
When the happening of one event has no effect on the probability of occurrence of happening of the
other event, the two events are independent. For two independent events A and B,
P (AB) = P (A) P (B).
Two events can be mutually exclusive and not independent. Again two events can be independent
and not mutually exclusive.
Suppose two coins are tossed. The events {H, H} ≡ A (head on both coins) and the event {T, T} ≡ B
are mutually exclusive (because if A happens B cannot happen) and
P (AB) = 0

1 1
But P (A) = , P (B) = and so P (AB) = 0 ≠ P (A) P (B) and hence A and B are not independent.
4 4
Consider the events A, B that 6 appears in the first and the second die respectively in throwing 2 dice
together (example 11). The events A and B are independent, as
1 1 1
P (AB) = = P (A) P (B) = × .
36 6 6
Here P (AB) ≠ 0 and hence the events are not mutually exclusive.
The only way that two mutually exclusive events A, B can be independent is when both
P (AB) = 0 and P (AB) = P (A) P (B)
hold simultaneously. Both of the above can hold simultaneously if at least one of P (A) or P (B) is
zero. In case at least one of P (A) or P (B) is zero then the two events A and B mutually exclusive
events as well as independent events.
Number of sample points in a combination of events or sets
Let N (A) denote the number of points in the set A.
Then, using Venn diagrams, it can be easily seen that
(
NA U B ) = N (A) + N (B) – N (AB). ... (7)
Example 14: Suppose that students in an Institution can enrol for one, two or none of the language
courses, French (A), German (B). If 30% are enrolled for French, 20% for German and
10% for both French and German, then the number (in percentage) enrolled for at
least one of the courses is given by
(
NA U B ) = N (A) + N (B) – N (AB)
= 30 + 20 – 10
= 40
and the percentage not enrolled for any of the courses is
100 – 40 = 60
Thus the probability that a student selected at random is (i) enrolled for at least one of
the courses is 0.4 and (ii) not enrolled for any of the courses is 1 – 0.4 = 0.6. Given that

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