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2-6 INDEPENDENCE 47
Table 2-4 Parts Classified
Surface Flaws
Yes (event F) No Total
Defective Yes (event D) 2 18 20
No 38 342 380
Total 40 360 400
EXAMPLE 2-24 The information in Table 2-3 related surface flaws to functionally defective parts. In that case,
we determined that P1D ƒ F2 10
40 0.25 and P1D2 28
400 0.07. Suppose that the
situation is different and follows Table 2-4. Then,
P1D ƒ F2 2
40 0.05 and P1D2 20
400 0.05
That is, the probability that the part is defective does not depend on whether it has surface
flaws. Also,
P1F ƒ D2 2
20 0.10 and P1F2 40
400 0.10
so the probability of a surface flaw does not depend on whether the part is defective.
Furthermore, the definition of conditional probability implies that
P1F ¨ D2 P1D ƒ F2P1F2
but in the special case of this problem
2 2 1
P1F ¨ D2 P1D2P1F2
40 20 200
The preceding example illustrates the following conclusions. In the special case that
P1B 0 A2 P1B2, we obtain
P1A ¨ B2 P1B ƒ A2P1A2 P1B2P1A2
and
P1A ¨ B2 P1A2P1B2
P1A ƒ B2 P1A2
P1B2 P1B2
These conclusions lead to an important definition.
Definition
Two events are independent if any one of the following equivalent statements is true:
(1) P1A ƒ B2 P1A2
(2) P1B ƒ A2 P1B2
(3) P1A ¨ B2 P1A2P1B2 (2-9)