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Homework # 1

2.3. (a.) The proof consists of two parts: Arguments for

∞

n=1 An

c⊂ 

∞

n=1 Ac

n

and for

∞

n=1 An

c⊃∞

n=1 Ac

n.

Indeed,

ω ∈ ∞

n=1

An

c=⇒  ω  ∈

∞n=1

An =⇒ ω  ∈ An   ∀n ≥  1 =⇒ ω  ∈  Ac

n  ∀n ≥  1 =⇒ ω  ∈

∞n=1

Ac

n

So we conclude that ∞n=1 Anc

⊂ ∞n=1 Acn.

The proof of 

∞

n=1 An

c⊃∞

n=1 Ac

n   is similar.

(b). Replace  An  by Acn  in (a).

∞n=1

Ac

n

c=

∞n=1

(Ac

n)c =∞

n=1

An

Therefore∞n=1

Ac

n = ∞

n=1

Ac

ncc

= ∞

n=1

Anc

2.15 Write

B1  =  A1   and   Bi =  Ai \ ∪i−1

k=1Aki = 2, 3, · · ·

Then {Bi}  is pairwise disjoint and

ni=1

Ai =n

i=1

Bi   and∞i=1

Ai  =∞i=1

Bi

Thus,

P 

  ∞i=1

Ai

= P 

  ∞i=1

Bi

=

∞i=1

P (Bi) ≤∞i=1

P (Ai)

where the last step follows from the relation  Bi ⊂  Ai.

Similar argument leads to

P 

  n

i=1Ai

≤n

i=1P (Ai)

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3.11. Let  Bi  denote the event that the  ith ball is blue.

P (B2) =  P (B1)P (B2|B1) + P (Bc

1)P (B2|Bc

1)

=  b

b + r

b + d

b + r + d +

  r

b + r

b

b + r + d =

  b

b + r

P (B1|B2) =  P (B1)P (B2|B1)

P (B1)P (B2|B1) + P (Bc1)P (B2|Bc

1)

=b

b+r

b+d

b+r+d

b

b+r

b+d

b+r+d +   r

b+r

b

b+r+d

=  b + d

b + r + d

3.12. Let  n ≥ 2 be fixed but arbitrary. Define

Ak  =′′

k  blues are taken in the first  n − 1 rounds′′

, k = 0, 1, · · · , n − 1.

We have that

P (Bn+1|Ak) = P (Bn|Ak)P (Bn+1|Ak ∩ Bn) + P (Bc

n|Ak)P (Bn+1|Ak ∩ Bc

n)

=  b + kd

b + r + (n − 1)d

b + (k + 1)d

b + r + nd  +

 r + (n − 1 − k)d

b + r + (n − 1)d

b + kd

b + r + nd

=  b + kd

b + r + (n − 1)d = P (Bn|Ak)   k = 0, 1, · · · , n − 1

Hence,

P (Bn+1) =n−1k=0

P (Ak)P (Bn+1|Ak) =n−1k=0

P (Ak)P (Bn|Ak) = P (Bn)   n = 2, 3 · · ·

So the conclusion follows from induction.

3.13. Write  A =  B2 ∩ · · · ∩ Bn+1

P (B1|A) =  P (B1)P (A|B1)

P (B1)P (A|B1) + P (Bc1)P (A|Bc

1)

We have

P (B1) =  b

b + r

  and   P (Bc

1) =  r

b + r

P (A|B1) =  b + d

b + r + d · · ·

  b + nd

b + r + nd  and   P (A|Bc

1) =  b

b + r + d · · ·

 b + (n − 1)d

b + r + nd

Put everything together,

P (B1|A) =  b + nd

b + r + nd

Clearly,limn→∞

P (B1|A) = 1

This shows if you get a long run of blue since the second round, then you have enoughreason to believe that the first round was blue.

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