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Probability

4. However, if A and B are mutually exclusive events (A ∩ B = ∅) then P(A ∩ B) = 0 and we
obtain the special case
P(A ∪ B) = P(A) + P(B). (2.7)

¯
¯
5. If A is the complement of A then A and A are mutually exclusive events.
Thus, from (2.5) and (2.7) we have

¯
¯
1 = P(S) = P(A ∪ A) = P(A) + P(A), (2.8)
from which we obtain the complement law

¯
P(A) = 1 − P(A). (2.9)

6. If A 1 , A 2 , . . . , A n are mutually exclusive events then (2.7) becomes

P(A 1 ∪ A 2 ∪ . . . ∪ A n ) = P(A 1 ) + P(A 2 ) + . . . + P(A n ). (2.10)

7. The sum of probabilities of events A 1 , A 2 , . . . , A n , which form a total set is equal to 1, that
is
P(A 1 ) + P(A 2 ) + . . . + P(A n ) = 1. (2.11)

∑ n
Since the events A 1 , A 2 , . . . , A n form a total set, their sum is a certain event A k = S.
∑ n k=1
On using the formula (2.10) and P( A k ) = P(S) we can show that is P(A 1 ) + P(A 2 ) +
k=1
. . . + P(A n ) = 1, since P(S) = 1.
The formula (2.8) is very important for practice. In many problems it is difficult to compute the
probability of an event while the probability of the complementary event may be easily computed.
In such cases formula (2.8) is useful.

Example 2.4. Compute the probability of drawing an ace or a spade from a pack
of cards. ,

Solution. Let A be the event that an ace is drawn and B the event that a spade is drawn. It
1
immediately follows that P(A) = 4 = 1 and P(B) = 13 = . The intersection of A and B
52 13 52 4
consists of only the ace of spades and so P(A ∩ B) = 1 . Thus, from (2.6)
52
1 1 1 4
P(A ∪ B) = + − = .
13 4 52 13

In this case it is just as simple to recognize that there are 16 cards in the pack that satisfy
the required condition (13 spades plus three other aces) and so the probability is 16 .
52

Example 2.5. The scheme of the electric circuit between two points M and N is
given in (Fig. 2.1). Malfunctions during an interval of time T of different
elements of the circuit represent independent events with the following
probabilities
Element K 1 K 2 L 1 L 2 L 3
Probability 0.6 0.5 0.4 0.7 0.9
Find the probability of a break in the circuit during the indicated interval
of time. ,

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