778 Combinatorics and Probability
P (B/A)=
P(B)
P (A)
·P(A/B).
For our problemAis the event that the mammogram is positive andBthe event that
the woman has breast cancer. ThenP(B)= 0 .01, whileP(A/B)= 0 .60. We compute
P (A)from the formula
P (A)=P(A/B)P(B)+P(A/notB)P(notB)= 0. 6 · 0. 01 + 0. 07 · 0. 99 = 0. 0753.The answer to the question is therefore
P(B/A)=
0. 01
0. 0753
· 0. 6 = 0. 0795 ≈ 0. 08
The chance that the woman has breast cancer is only 8%!
917.We call asuccessful stringa sequence ofH’s andT’s in whichHHHHHappears
beforeTTdoes. Each successful string must belong to one of the following three types:
(i) those that begin withT, followed by a successful string that begins withH;
(ii) those that begin withH,HH,HHH,orHHHH, followed by a successful string
that begins withT;
(iii) the stringHHHHH.
LetPHdenote the probability of obtaining a successful string that begins withH, and
letPTdenote the probability of obtaining a successful string that begins withT. Then
PT=
1
2
PH,
PH=
(
1
2
+
1
4
+
1
8
+
1
16
)
PT+
1
32
.
Solving these equations simultaneously, we find that
PH=
1
17
and PT=1
34
.
Hence the probability of obtaining five heads before obtaining two tails is 343.
(American Invitational Mathematics Examination, 1995)
918.Let us denote the eventsx= 70 ◦,y= 70 ◦, max(x◦,y◦)= 70 ◦, min(x◦,y◦)= 70 ◦
byA, B, C, D, respectively. We see thatA∪B=C∪DandA∩B=C∩D. Hence
P (A)+P(B)=P(A∪B)+P(A∩B)=P(C∪D)+P(C∩D)=P(C)+P(D).
Therefore,P(D)=P (A)+P(B)−P(C), that is,