1.

Let . Since and , it follows that and are disjoint.

Since , it follows that for each .

Suppose for some . By the previous claim, it follows that

Lastly, let be monotone decreasing. Noting that is monotone increasing,

2.

Since by additivity, it follows that .

If is contained in , then

As an immediate consequence of the previous two claims, it follows that .

Since , it follows that .

Lastly, we point out that by taking in the countable additivity property (Axiom 3), we obtain finite additivity: for any disjoint sets and .

3.

a)

Note that

Similarly,

b)

is in is in for each for each , we can find such that is in .

Remark. A shorthand for is .

c)

is in is in for some we can find such that is in for each .

Remark. A shorthand for is .

4.

Note that

Similarly,

5.

The sample space for the repeated coin flip experiment is : the set of all functions from the natural numbers to . Let be one if the -th toss is heads and zero otherwise. Then, the probability of stopping at the -th toss is

6.

Let be a probability measure on . By additivity, . Suppose is uniform. Then, for each and hence (we interpret ), a contradiction.

7.

Define as in the hint. By our findings in Questions 1 and 2,

8.

Since

it follows that

9.

First, note that . In particular, . Lastly, let be disjoint. Then,

10.

Without loss of generality, we can assume that the player picks door 1 and Monty reveals there is no prize behind door 2. Then, the player is left between choosing door or . It follows that

In particular,

Since the player should pick to maximize , the player should switch from door 1 to door 3.

11.

First, note that that

Using the independence of and ,

12.

Let (respectively, ) be the event that the side of the seen (respectively, unseen) card is green. Since , Then,

13.

a)

The sample space for this question is identical to that of Question 5.

b)

We stop at the third toss if and only if the first three flips are or . If is the probability of heads, then the probability of this is .

14.

Let and be events.

Suppose . Then, and hence .

Suppose . Then, and hence by our most recent findings, and are independent. By our findings in Question 11, it follows that and are independent.

Suppose now that is independent of itself. Then, and hence either or .

15.

Let be an indicator random variable that is one if and only if the -th child has blue eyes. Let . Let be the probability of having blue eyes and .

a)

Note that

Moreover, and . Therefore,

b)

Note that

16.

Let and be events with . follows by multiplying by on both sides of the definition of conditional probability. Moreover, if and are independent,

17.

Assuming and are positive, the result follows from combining

and

18.

If are a partition of the sample space, then . Moreover, for any event ,

If , then we can divide both sides by to get . Combining this with a previous equality, we get . Suppose now . Then,

It follows that for at least one .

19.

We use , , and to denote the event that the user uses Mac, Windows, and Linux, respectively. We use to denote the event that the user has the virus.

20.

a)

b)

c)

21.

TODO (Computer Experiment)

22.

TODO (Computer Experiment)

23.

TODO (Computer Experiment)