2. Conditional Probability 5
If w e appl y (1.9) wit h A {woman} an d B = {employed} , the n w e find
that P(A | B) = 15/35 « 0.4285 . Thi s agree s wit h ou r earlie r bare-hand s
calculation.
In general , A \— P(A \ B) i s a probability , an d ha s th e propertie s o f
Proposition 1.2. I n othe r words , w e hav e th e following .
P r o p o s i t i o n 1.10. For any event B that has positive probability, P(B \ B)
1 and P ( 0 | B) 0. In addition, for all events A, A\,Ai,
(1) P(A i UA 2\B) = P(A
±
| B) + P(A
2
| B) - P(A
X
nA 2\B).
(2) P(A
C\B)
= 1-P(A\B).
(3) 7 / ^ e AiDB's are disjoint, then P ( U g
1
A
i
| B) = J2Zi F (Ai I B )-
(4) / / A i C ^
2 ;
thenP{A x\B) P(A 2\B).
(5) (Boole' s Inequality ) P ( U g
1
A
i
| B) ^ ) £ i P ( ^ I B).
(6) (Inclusion-Exclusio n Formula )
(
n I \ n
\JAAB\=J2P(A\B)-
X X p ^ n A J B )
i = l I / z= l lziZ2 n
+ ... + (-i)
n-1
p(A1\B).nn---r\A
The followin g i s a particularl y usefu l corollary .
T h e La w o f Tota l Probability . Consider events B , A\,A
2
,..., A
n
where
the Ai 's are disjoint, Uf
=1
Ai = ft, and P(Aj) 0 for all j = 1 , . . . , n. Then,
n
(1.10) P(B) = Y,V(B\A J)P(A3).
3=1
The la w o f tota l probabilit y provide s u s wit h a metho d fo r computin g
weighted averages , viz. ,
E x a m p l e 1.11Dichromatism . i s a for m o f colo r blindness . I t i s a geneti c
disorder tha t i s cause d b y a defec t i n th e X-chromosome . Roughly , abou t
5% o f al l me n an d 1% o f al l wome n suffe r fro m dichromatism . Suppos e
tha t 60 % o f a certai n populatio n ar e women . I f w e sampl e a n individua l a t
random fro m thi s population , the n wha t i s th e probabilit y tha t thi s perso n
has dichromatism ?
Define th e event s D = {dichromatism } an d F {female} i n th e obviou s
way. Becaus e P(D \ F c) « 0.05 , P(D | F) « 0.01, an d P(F) « 0.6 ,
(1.11) P(D) = P(D | F)P(F) + P(D |
FC)P(FC)
« 0.026 .
Tha t is , abou t 2.6 % o f thi s populatio n suffer s fro m dichromatism .
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