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Statistics

Notes We have already seen that if A, and’A2 are independent, then

c
c
c
c
A and A or A and A or A and A are independent. We now give a similar remark about
1 2 1 2 1 2
n independent events.
Remark 4 : If A , A , . . . . A are n independent events, then we may replace some or all of them
1 2 n
by their complements without losing independence. In particular, when A , A , . . . , A are
1 2 n
independent, the product rule (17) holds even with some or all of A , . . . , A are replaced by
1 i r i
their complements.
We shall not prove this assertion, but shall use it in the following examples.
Example 22: Suppose A , A , A are three independent events, with P(A ) = P and we want
1 2 3 j j
to obtain the probability that at least one of them occurs.
We want to find P(A  A  A ). Recall that (Example 8)
1 2 3
P(A A A ) = P(A ) + P(A ) + P(A ) – P(A  A ) – P(A A ) – P(A  A) + P(A  A A )
1 2 3 1 2 3 1 2 2 3 3 l 2 3
= P + P +P – P P – P P – P P + P P P
1 2 3 1 2 2 3 3 1 1 2 3
= 1 – (1 – P ) (1 – P ) (1 – P3).
1 2
We could have amved at this expression more easily by using Remark 4. This is how we can
proceed.
c
(A  A  A ) = 1 – P((A  A  A ) )
1 2 3 1 2 3
c
= 1 – P(A  A  A )
c
c
3
2
1
c
c
c
= 1 – P(A )P(A )P(A )
3
1
2
Example 23: If A , A and A are independent events, then can we say that A  A and A are
1 2 3 1 2 3
independent? Let’s see.
We have
P(A  A ) = P(A ) + P(A ) – P(A  A )
1 2 1 2 1 2
= P(A ) + P(A ) – P(A ) P(A )
1 2 1 2
and P((A  A )  A ) = P((A  A )  (A  A ))
1 2 3 1 3 2 3
= P(A  A ) + P(A  A ) – P(A  A  A )
1 3 2 3 1 2 3
= {P(A ) + P(A ) – P(A ) P(A )} P(A )
1 2 1 2 3
= P(A  A ) P(A ).
1 2 3
implying the independence of A  A and A .
1 2 3
Example 24: An automatic machine produces bolts. Each bolt has probability 1/10 of being
defective. Assuming that a bolt is defective independently of all other bolts, let’s find
(i) the probability that a good bolt is followed by two defective ones.
(ii) the probability of getting one good and two defective bolts, not necessarily in that order.
Let A denote the event that the j-th inspected bolt is defective, j = 1,2,3. The assumption of
j
independence implies that A , A and A are independent.
1 2 3

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