Chapter 4
4 - 2
Solutions:
1.
Number of experimental Outcomes
=
(3) (2) (4)
=
24
2.
6
3
6!
33
654321
321 321
20
F
H
G
I
K
J
==
⋅⋅⋅⋅⋅
⋅⋅
=
!! (
)
(
)
ABC
ACE
BCD
BEF
ABD
ACF
BCE
CDE
ABE
ADE
BCF
CDF
ABF
ADF
BDE
CEF
ACD
AEF
BDF
DEF
3.
P
3
6
6!
63
654 1
2
0
=
−
()
!
BDF
BFD
DBF
DFB
FBD
FDB
4.
a.
H
T
H
T
H
T
H
T
H
T
H
T
H
T
(H,H,H)
(H,H,T)
(H,T,H)
(H,T,T)
(T,H,H)
(T,H,T)
(T,T,H)
(T,T,T)
1st Toss
2nd Toss
3rd Toss
b.
Let: H be head and T be tail
(H,H,H) (T,H,H)
(H,H,T) (T,H,T)
(H,T,H) (T,T,H)
(H,T,T) (T,T,T)
c.
The outcomes are equally likely, so the probability of each outcomes is 1/8.
5.
P(E
i
) = 1 / 5
for i = 1, 2, 3, 4, 5
P(E
i
)
≥
0
for i = 1, 2, 3, 4, 5
P(E
1
) + P(E
2
) + P(E
3
) + P(E
4
) + P(E
5
) = 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 + 1 / 5 = 1
The classical method was used.