# Solution to homework 3 ```Homework 3 Solution
1. [&sect;3-5] An urn contains 6 white and 9 black balls. If 4 balls are to be
randomly selected without replacement, what is the probability that the
first 2 selected are white and the last 2 black?
Let A ={the first 2 selected are white} and B ={the last 2 selected are
black}. Then
9
6
2
15
2
P(A) =
and
P(B|A) =
2
13 .
2
Therefore,
9
2
13
2
P(A ∩ B) = P(B|A) &middot; P(A) =
9!
2!7!
13!
2!11!
6!
2!4!
15!
2!13!
6
2
15
2
&middot;
6!9!11!
9!6!2!11!2!13!
=
2!7!2!4!13!15!
4!7!15!
6&middot;5&middot;9&middot;8
6
=
=
.
15 &middot; 14 &middot; 13 &middot; 12
91
=
=
2. [&sect;3-8] A couple has 2 children. What is the probability that both are girls
if the older of the two is a girl?
Let A ={the older is girl} and B ={the younger is girl}. Then P(A) =
P(B) = 1/2 and P(A ∩ B) = 1/4. Thus
P(A ∩ B|A) =
P(A ∩ B)
1
= .
P(A)
2
3. [&sect;3-11] Two cards are randomly chosen without replacement from an ordinary deck of 52 cards. Let B be the event that both cards are aces, let
As be the event that the ace of spades is chosen, and let A be the event
that at least one ace is chosen. Find
(a) P(B|As )
(b) P(B|A)
Since
4
2
52 ,
2
P(B) =
51
1
52 ,
2
P(As ) =
1
48
2
52 ,
2
and,
P(A) = 1 −
and
3
1
52
2
P(B ∩ As ) =
we have
and
and
P(B ∩ A) = P(B),
(31)
P(B ∩ As )
3
1
(52
2)
P(B|As ) =
= 51
=
,
=
(1)
P(As )
51
17
52
(2)
(42)
1
P(B ∩ A)
6
(52
2)
=
=
.
P(B|A) =
=
48
(
)
P(A)
198
33
2
1 − 52
(2)
4. [&sect;3-12] A recent college graduate is planning to take the first three actuarial
examinations in the coming summer. She will take the first actuarial exam
in June. If she passes that exam, then she will take the second exam in
July, and if she also passes that one, then she will take the third exam
in September. If she fails an exam, then she is not allowed to take any
others. The probability that she passes the first exam is .9. If she passes
the first exam, then the conditional probability that she passes the second
one is .8, and if she passes both the first and the second exams, then the
conditional probability that she passes the third exam is .7.
(a) What is the probability that she passes all three exams?
(b) Given that she did not pass all three exams, what is the conditional
probability that she failed the second exam?
(a) Let A1 ={she passes exam 1}, A2 ={she passes exam 2}, A3 ={she
passes exam 3}, and A ={she passes all three exams}. Then
P(A1 ) = 0.9,
, P(A2 |A1 ) = 0.8,
and
P(A3 |A1 ∩A2 ) = 0.7.
Since A = A1 ∩ A2 ∩ A3 ,
P(A) = P(A3 |A1 ∩ A2 ) &middot; P(A2 |A1 ) &middot; P(A1 ) = 0.504.
(b) Since Ac2 ⊂ Ac , Ac2 ∩ Ac = Ac2 . According to the total probability
formula,
P(Ac2 ) = P(Ac2 |Ac1 ) &middot; P(Ac1 ) + P(Ac2 |A1 ) &middot; P(A1 )
= 0 &middot; 0.1 + 0.2 &middot; 0.9 = 0.18.
Thus
P(Ac2 |Ac ) =
P(Ac2 ∩ Ac )
0.18
=
= 0.363.
c
P(A )
0.496
2
5. [&sect;3-26] Suppose that 5 percent of men and .25 percent of women are color
blind. A color-blind person is chosen at random. What is the probability
of this person being male? Assume that there are an equal number of
males and females. What if the population consisted of twice as many
males as females?
Let M ={the selected person is a man}, W ={the selected person is a
woman}, and C ={the selected person is color-blind}. Then,
P(C|M ) = 0.05
and
P(C|W ) = 0.0025.
Furthermore, if there are an equal number of males and females, then
P(M ) = P(W ) = 0.5.
Thus if there are an equal number of males and females, then
P(M |C) =
(0.05)(0.5)
P(C|M ) &middot; P(M )
=
= 0.951.
P(C|M ) &middot; P(M ) + P(C|W ) &middot; P(W )
(0.05)(0.5) + (0.0025)(0.5)
On the other hand, if the population consisted of twice as many males as
females, then
P(M ) = 0.667
P(M |C) =
and
P(W ) = 0.333.
P(C|M ) &middot; P(M )
(0.05)(0.667)
=
= 0.975.
P(C|M ) &middot; P(M ) + P(C|W ) &middot; P(W )
(0.05)(0.667) + (0.0025)(0.333)
6. [&sect;3-50] Suppose that an insurance company classifies people into one of
three classes: good risks, average risks, and bad risks. The company’s
records indicate that the probabilities that good-, average-, and bad-risk
persons will be involved in an accident over a 1-year span are, respectively,
.05, .15, and .30. If 20 percent of the population is a good risk, 50 percent
an average risk, and 30 percent a bad risk, what proportion of people have
accidents in a fixed year? If policyholder A had no accidents in 1997, what
is the probability that he or she is a good or average risk?
Let G ={A person is of good-risk class}, V ={A person is of average-risk
class}, and B ={A person is of bad-risk class}. Furthermore, let C ={A
person has accidents in a fixed year}. Then
P(G) = 0.2, P(V ) = 0.5, P(B) = 0.3,
P(C|G) = 0.05, P(C|V ) = 0.15, P(C|B) = 0.30.
Thus
P(C) = P(C|G) &middot; P(G) + P(C|V ) &middot; P(V ) + P(C|B) &middot; P(B)
= (0.05)(0.2) + (0.15)(0.5) + (0.30)(0.3)
= 0.175.
3
Moreover,
P(C c |G) &middot; P(G)
(0.95)(0.2)
=
= 0.230, and
c
P(C )
0.825
P(C c |V ) &middot; P(V )
(0.85)(0.5)
P(V |C c ) =
=
= 0.515.
c
P(C )
0.825
P(G|C c ) =
7. [&sect;3-59] Independent flips of a coin that lands on heads with probability p
are made. What is the probability that the first four outcomes are
(a) H, H, H, H?
(b) T, H, H, H?
(c) What is the probability that the pattern T, H, H, H occurs before
the pattern H, H, H, H?
(a) According to independence, the probability for H, H, H, H is p4 .
(b) According to independence, the probability for T, H, H, H is p3 (1−p).
(c) The trick is that the only possible outcome such that a HHHH
pattern appears before a T HHH is a sequence start with HHHH
(think about it), and this probability is p4 . Therefore, the probability
that the pattern T, H, H, H occurs before the pattern H, H, H, H is
1 − p4 .
4
```