Chapter 1
1. S = {(R, R), (R, G) , (R, B), (G, R), (G, G), (G, B),
(B, R), (B, G), (B, B)}.
The probability of each point in S is 1/9.
2. S
= {(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)}.
3. S
= {(e
1
, e
2
,...,e
n
), n ≥ 2} where e
i
∈ (heads,
tails}. In addition, e
n
= e
n−1
= heads and for
i
= 1,...,n − 2ife
i
= heads, then e
i+1
= tails.
P
{4 tosses} = P{(t, t, h, h)} + P{(h, t, h, h)}
=
2
1
2
4
=
1
8
.
4. (a) F
(E ∪ G)
c
= FE
c
G
c
.
(b) EFG
c
.
(c) E
∪ F ∪ G.
(d) EF
∪ EG ∪ FG.
(e) EFG.
(f)
(E ∪ F ∪G)
c
= E
c
F
c
G
c
.
(g)
(EF)
c
(EG)
c
(FG)
c
.
(h)
(EFG)
c
.
5.
3
4
. If he wins, he only wins $1, while if he loses, he
loses $3.
6. If E
(F ∪G) occurs, then E occurs and either F or G
occur; therefore, either EF or EG occurs and so
E
(F ∪G) ⊂ EF ∪ EG.
Similarly, if EF
∪ EG occurs, then either EF or EG
occur. Thus, E occurs and either F or G occurs; and
so E(F
∪ G) occurs. Hence,
EF
∪ EG ⊂ E(F ∪ G),
which together with the reverse inequality proves
the result.
7. If
(E ∪ F)
c
occurs, then E ∪ F does not occur, and
so E does not occur (and so E
c
does); F does not
occur (and so F
c
does) and thus E
c
and F
c
both
occur. Hence,
(E ∪ F)
c
⊂ E
c
F
c
.
If E
c
F
c
occurs, then E
c
occurs (and so E does not),
and F
c
occurs (and so F does not). Hence, neither
E or F occur and thus
(E ∪ F)
c
does. Thus,
E
c
F
c
⊂ (E ∪F)
c
and the result follows.
8. 1
≥ P(E ∪ F)=P(E)+P(F) − P( EF).
9. F
= E ∪ FE
c
, implying since E and FE
c
are disjoint
that P
(F)=P(E)+P(FE)
c
.
10. Either by induction or use
n
∪
1
E
i
= E
1
∪ E
c
1
E
2
∪ E
c
1
E
c
2
E
3
∪···∪E
c
1
···E
c
n
−1
E
n,
and as each of the terms on the right side are
mutually exclusive:
P
(∪
i
E
i
)=P(E
1
)+P(E
c
1
E
2
)+P(E
c
1
E
c
2
E
3
)+···
+
P(E
c
1
···E
c
n
−1
E
n
)
≤
P(E
1
)+P(E
2
)+···+ P(E
n
). (why?)
11. P{sum is i} =
i
−1
36
, i
= 2,...,7
13
−i
36
, i
= 8,...,12.
12. Either use hint or condition on initial outcome as:
P
{E before F}
=
P{E before F | initial outcome is E}P(E)
+
P{E before F | initial outcome is F}P(F)
+
P{E before F | initial outcome neither E
or F}[1 − P(E) − P(F)]
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