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Addition of multiple events

L 1

K 1 L 2 K 2
M N

L 3

Figure 2.1 – An electric circuit to Example 2.5

Solution. Denote by A j (j = 1, 2) the event meaning that an element K j is out of order, by A
that at least one element K j is out of order and by B that all three elements L i (i = 1, 2, 3) are
out of order. Then, the required probability is

p = P(A + B) = P(A) + P(B) − P(A)P(B).
Since
P(A) = P(A 1 ) + P(A 2 ) − P(A 1 )P(A 2 ) = 0.8,
P(B) = P(L 1 )P(L 2 )P(L 3 ) = 0.252,

we get p ≈ 0.85.

1
Example 2.6. A biased six-sided die has probabilities p, p, p, p, p, 2p of showing
2
1, 2, 3, 4, 5, 6 respectively. Compute p. ,
Solution. Given that the individual events are mutually exclusive, (2.10) can be applied to give

1 13
P(1 ∪ 2 ∪ 3 ∪ 4 ∪ 5 ∪ 6) = p + p + p + p + p + 2p = p.
2 2
The union of all possible outcomes on the LHS of this equation is clearly the sample space, S,
and so P(S) = 13 p. Now using (2.5), 13 p = P(S) = 1 ⇒ p = 2 .
2 2 13

Addition of multiple events

When the possible outcomes of a trial correspond to more than two events, and those events are
notmutuallyexclusive, thecalculationoftheprobabilityoftheunionofanumberofeventsismore
complicated, and the generalization of the addition law (2.6) requires further work. Let us begin
by considering the union of three events A 1 , A 2 and A 3 , which need not be mutually exclusive. We
first define the event B = A 2 ∪ A 3 and, using the addition law (2.6), we obtain

P(A 1 ∪ A 2 ∪ A 3 ) = P(A 1 ∪ B) = P(A 1 ) + P(B) − P(A 1 ∩ B). (2.12)
However, we may write P(A 1 ∩ B) as

P(A 1 ∩ B) = P[A 1 ∩ (A 2 ∪ A 3 )] = P[(A 1 ∩ A 2 ) ∪ (A 1 ∩ A 3 )] =

= P(A 1 ∩ A 2 ) + P(A 1 ∩ A 3 ) − P(A 1 ∩ A 2 ∩ A 3 ).
Substituting this expression, and that for P(B) obtained from (2.6), into (2.12) we obtain the
probability addition law for three general events,

P(A 1 ∪ A 2 ∪ A 3 ) = P(A 1 ) + P(A 2 ) + P(A 3 ) − P(A 2 ∩ A3) − P(A 1 ∩ A 3 )−
−P(A 1 ∩ A 2 ) + P(A 1 ∩ A 2 ∩ A 3 ). (2.13)

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