Math 151 HW 2 Solutions
Problems
4.
Let
A
be the probability that at least on of a pair of dice lands on 6. Easy computaion gives:
P
(
A

i
= 2) = 0,
P
(
A

i
= 3) = 0,
P
(
A

i
= 4) = 0,
P
(
A

i
= 5) = 0,
P
(
A

i
= 6) = 0,
P
(
A

i
= 7) = 2
/
6,
P
(
A

i
= 8) = 2
/
5,
P
(
A

i
= 9) = 2
/
4,
P
(
A

i
= 10) = 2
/
3,
P
(
A

i
= 11) = 2
/
2,
P
(
A

i
= 2) = 1
/
1.
18.
Let
E, I, L, C
be the event that a person voted in the local election, is an independent, is a liberal, is
a conservative, respectively.
(a) Using Bayes Formula,
P
(
I

E
) =
P
(
IE
)
P
(
E
)
=
P
(
I
)
P
(
E

I
)
P
(
I
)
P
(
E

I
)+
P
(
L
)
P
(
E

L
)+
P
(
C
)
P
(
E

C
)
=
(0
.
46)(0
.
35)
(0
.
46)(0
.
35)+(0
.
30)(0
.
62)+(0
.
24)(0
.
58)
= 0
.
331
Similar computation gives (b) 0.383 (c) 0.286
(d) (0
.
46)(0
.
35) + (0
.
30)(0
.
62) + (0
.
24)(0
.
58) = 0
.
4862, which means that 48.62 percent of voters
participated.
26.
Let
C, M, W
be event of being colorblind, man, woman, respectively.
We want to calculate:
P
(
M

C
) =
P
(
MC
)
P
(
C
)
=
P
(
M
)
P
(
C

M
)
P
(
M
)
P
(
C

M
)+
P
(
W
)
P
(
C

W
)
Case1:
(0
.
5)(0
.
05)
(0
.
5)(0
.
05)+(0
.
5)(0
.
0025)
= 0
.
95
Case2:
(2
/
3)(0
.
05)
(2
/
3)(0
.
05)+(1
/
3)(0
.
0025)
= 0
.
98
32.
Let
C
j
j
= 1
,
2
,
3
,
4 be the event that family has 1,2,3,4 children respectively. Let
E
be the event that
child is eldest child. Using Bayes’s formula:
(a)
P
(
C
1

E
) =
P
(
C
1
E
)
P
(
E
)
=
P
(
C
1
)
P
(
E

C
1
)
P
(
C
1
)
P
(
E

C
1
)+
P
(
C
2
)
P
(
E

C
2
)+
P
(
C
3
)
P
(
E

C
3
)+
P
(
C
4
)
P
(
E

C
4
)
=
1(0
.
1)
1(0
.
1)+(1
/
2)0
.
25+(1
/
3)0
.
35+(1
/
4)0
.
3
= 0
.
24
(b) Similarly, we get 0.18
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 Winter '08
 Liu
 Probability, Trigraph, SEPTA Regional Rail, AirTrain Newark, p1 p4 p2, p4 p2 p5, Ei Ek+1

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