Chapter １ 習題答案
1.7 Prove Bonferroni’s inequality :
P( A  B)  P(A) + P(B)
–1
SOL :
P(A  B) = P(A) + P(B) – P(A  B)  1

P( A  B)  P(A) + P(B) – 1
1.8 Prove that
n
n
 P( A )
P(  Ai ) 
i
i 1
i 1
SOL :
n
prove that P(  Ai ) 
i 1
n
 P( A )
i
i 1
2
n = 2 P(  Ai ) = P(A 1  A 2 ) = P(A 1 ) + P(A 2 ) - P(A 1  A 2 ) 
i 1
k
設 n=k 成立 即 P(  Ai ) 
i 1
k 1
k
i 1
i 1
2
 P( A )
i 1
i
k
 P( A )
i 1
i
k
n = k + 1 時 P(  Ai ) = P((  Ai )  A k 1 )  P(  Ai ) + P(A k 1 )
i 1
k
 [  P( Ai ) ] + P(A k 1 )
i 1
k 1
=
 P( A )
i 1
n
由數學歸納法得知 P(  Ai ) 
i 1
i
n
 P( A )
i 1
i
1.9 The weather forecaster says that the probability of rain on Saturday is 25 % and
that the probability of rain on Sunday is 25 %. Is the probability of rain during the
weekend 50%? Why or why not?
SOL :
3
No, the probability of rain during the weekend is １ –  
4
1.34 Prove the following identity :
n
n m  n
m
  
 =  

k 0  k 
nk 
n
SOL :
let Χ be a hypergeometric random variable
2
M 
 
x
P(X = x | N , M , K) =  
N  M 


K

x

 x = 0、1、2  k
N
 
K
M  N  M 
N

 =  
x 0 
  Kx 
K
let N = m , M = n , k = n
n
n m  n
m
 =  
we have    
x 0  x 
nx
n
n
 n m  n
m
 =  
即    
k 0  k 
nk 
n

k
  x 
1.59 A box has three coins. One has two heads,one has two tails,and the other is a fair
coin with one head and one tail. A coin is chosen at random,si flipped,and comes up
heads.
a. What is the probability that the coin chosen is the two-headed coin?
b. What is the probability that if is is thrown another time it will come up
heads?
c. Answer part(a) again, supposing that the coin is throw a second time and
comes up heads again.
©. SOL :
A : two headed coin
B : one head and one tail coin
C : two tailed coin
P( A, head and head)
P(A| head and head) =
P(head and head)
=
P( head and head | A)P(A)
P(head and head | A)P(A)  P(head and head | B)P(B)  P(head and head | C)P(C)
1
=
1
3
1
1 1
1  0  
3
4 3
=
4
5
1.62 Show that if P(A|E)  P(B|E) and P(A|E c )  P(B|E c ),then P(A)  P(B).
SOL :
P(A|E)  P(B|E)  P(A  E)  P(B  E)
P(A|E c )  P(B|E c )  P(A  E c )  P(B  E c )

P(A  E) + P(A  E c )  P(B  E) + P(B  E c )

P(A)  P(B)
1.69 If A and B are disjoint , can they be independent?
SOL :
IF A   , B   ( P(A) > 0 , P(B) > 0)
 IF A and B are disjoint , then A and B must be dependent.
IF suppose A and B are independent,
0 = P(A  B) = P(A)  P(B)  0 
1.79 Many human diseases are genetically transmitted(for example, hemophilia or
Tay-Sachs disease). Here is a simple model for such a disease. The genotype aa is
diseased and dies before it mates. The genotype Aa is a carrier but is not diseased. The
genotype AA is not a carrier and is not diseased.
a. If two carriers mate, what are the probabilities that their offspring are of each
of the three genotypes?
b. If the male offspring of two carriers is not diseased,what is the probability
that he is a carrier?
c. Suppose that the nondiseased offspring of part (b) mates with a member of
the population for whom no family history is available and who is thus
assumed to have probability p of being a carrier(p is a very small number).
What are the probabilities that their first offspring has the genotypes
AA,Aa,and aa?
d. Suppose that the first offspring of part © is not diseased.What is the
probability that the father is a carrier in light of this evidence?
SOL :
a.
A
A
A
a
AA
Aa
Aa
aa
P(AA) = P(aa) =
P(Aa) =
1
4
1
2
1
 
2
2
b. P(二代 Aa | Aa  Aa , and 二代未死) =   =
 3 3
 
4
c.
♂ : P(Aa) =
2
1
, P(AA) =
3
3
♀ :P(Aa) = p , P(AA) = 1=p
三代 : P(AA) = P(♂Aa  ♀Aa & get AA) + P(♂Aa  ♀AA & get AA)
+ P(♂AA  ♀Aa & get AA) + P(♂AA  ♀AA & get AA)
=
2
1
2
1
1
1
1
P
+ (1  p ) 
+  p  +  (1  p )  1
3
4
3
2
3
2
3
=
2 p

3 3
Similarly P(Aa) =
1 p
p

, P(aa) =
3 6
6
d.
P(♂Aa|三代沒病) =
P(三代沒病 | Aa ) P( Aa )
P(三代沒病)
1
1 2 1
2
 2 1

2
 (1  p)   2      p       p  1
(1  p ) 
4
3
2
3
2
2
3
2
 
 
 =
3 = 4 p
=
p
6 p
 2 p 1 p
1
    
6
3 3  3 6 
1.75 This problem deals with an elementary aspect of a simple branching process.A
population starts with one member,at time t = 1, it either divides with probability p
ordies with probability 1 – p . If it divides ,then both of its children behave
independently with the same two alternatives at time t = 2. What is the probability that
there are no members in the third generation? For what value of p is this probability
equal to 0.5?
SOL:
(1-P) + P(1-P) 2 + P 3 (1-P) 4 + 2P 2 (1-P) 3
1.74 What is the probability that the following system works if each unit fails
independently with probability p (see figure 1.4)?
□ figure 1.4
甲
乙
丙
SOL :
P(甲 works) = (1-P) 2
P(乙 works) = (1-p)
P(丙 works) = (1-P) 2
P(system works) = P(甲  乙  丙 )
= P(甲) + P(乙) + P(丙) – P(甲  乙 ) – P(乙  丙) - P(甲  丙)
+ P(甲  乙  丙)
= 2(1-p) 2 + (1-p) – 2(1-p) 3 + (1-p) 4 + (1-p) 5