Acknowledgements: Thanks to Ben S. for correcting some mistakes.

## 1.

Let $$i < j$$. Since $$B_i \subset A_i$$ and $$B_j \cap A_i = \emptyset$$, it follows that $$B_i$$ and $$B_j$$ are disjoint.

Since $$A_1 \subset A_2 \subset \cdots$$, it follows that $$A_n = \cup_{i = 1}^n A_i$$ for each $$n$$.

Suppose $$\cup_{i = 1}^n B_i = A_n$$ for some $$n$$. By the previous claim, it follows that

$$$\cup_{i = 1}^{n + 1} B_i = A_n \cup B_{n+1} = \left( \cup_{i = 1}^n A_i \right) \cup \left( A_{n + 1} \setminus \left( \cup_{i = 1}^n A_i \right) \right) = \cup_{i = 1}^{n + 1} A_i.$$$

Lastly, let $$A_1 \supset A_2 \supset \cdots$$ be monotone decreasing. Noting that $$A_1^c \subset A_2^c \subset \cdots$$ is monotone increasing,

$$$\mathbb{P}(\cap_n A_n) = 1 - \mathbb{P}(\cup A_n^c) = 1 - \lim_n \mathbb{P}(A_n^c) = \lim_n 1 - \mathbb{P}(A_n^c) = \lim_n \mathbb{P}(A_n).$$$

## 2.

Since $$\mathbb{P}(\emptyset \cup \emptyset) = 2 \mathbb{P}(\emptyset)$$ by additivity, it follows that $$\mathbb{P}(\emptyset) = 0$$.

If $$A$$ is contained in $$B$$, then

$$$\mathbb{P}(B) = \mathbb{P}(A \cup B) = \mathbb{P}(A \cup \left( B \setminus A \right)) = \mathbb{P}(A) + \mathbb{P}(B \setminus A) \geq \mathbb{P}(A)$$$

As an immediate consequence of the previous two claims, it follows that $$\mathbb{P}(A) \leq \mathbb{P}(\Omega) = 1$$.

Since $$\mathbb{P}(A) + \mathbb{P}(A^c) = \mathbb{P}(A \cup A^c) = \mathbb{P}(\Omega) = 1$$, it follows that $$\mathbb{P}(A) = 1 - \mathbb{P}(A^c)$$.

Lastly, we point out that by taking $$A_2 = A_3 = \cdots = \emptyset$$ in the countable additivity property (Axiom 3), we obtain finite additivity: $$\mathbb{P}(A_1 \cup A_2) = \mathbb{P}(A_1) + \mathbb{P}(A_2)$$ for any disjoint sets $$A_1$$ and $$A_2$$.

## 3.

### a)

Note that

$$$B_n = \cup_{i = n}^\infty A_i \supset \cup_{i = n + 1}^\infty A_i = B_{n + 1}.$$$

Similarly,

$$$C_n = \cap_{i = n}^\infty A_i \subset \cap_{i = n + 1}^\infty A_i = C_{n + 1}.$$$

### b)

$$\omega$$ is in $$\cap_n B_n$$ $$\iff$$ $$\omega$$ is in $$B_n$$ for each $$n$$ $$\iff$$ for each $$n$$, we can find $$i \geq n$$ such that $$\omega$$ is in $$A_i$$.

Remark. A shorthand for $$\cap_n B_n$$ is $$\limsup_n A_n$$.

### c)

$$\omega$$ is in $$\cup_n C_n$$ $$\iff$$ $$\omega$$ is in $$C_n$$ for some $$n$$ $$\iff$$ we can find $$n$$ such that $$\omega$$ is in $$A_i$$ for each $$i \geq n$$.

Remark. A shorthand for $$\cup_n C_n$$ is $$\liminf_n A_n$$.

## 4.

Note that

\begin{align} \omega \in \left(\cup_i A_i\right)^c & \iff \omega \notin \cup_i A_i \\ & \iff \omega \notin A_i \text{ for each } i \\ & \iff \omega \in A_i^c \text{ for each } i \\ & \iff \omega \in \cap_i A_i^c. \end{align}

Similarly,

\begin{align} \omega \in \left( \cap_i A_i \right)^c & \iff \omega \notin \cap_i A_i \\ & \iff \omega \notin A_i \text{ for some } i \\ & \iff \omega \in A_i^c \text{ for some } i \\ & \iff \omega \in \cup_i A_i^c. \end{align}

## 5.

The sample space for the repeated coin flip experiment is $$\{H,T\}^{\mathbb{N}}$$: the set of all functions from the natural numbers to $$\{H,T\}$$. Let $$X_n$$ be one if the $$n$$-th toss is heads and zero otherwise. Then, the probability of stopping at the $$k$$-th toss is

$\begin{multline*} \mathbb{P}(X_1 + \cdots + X_{k - 1} = 1) \times \mathbb{P}(X_k = 1) = \binom{k-1}{1} p \left( 1 - p \right)^{k - 2} \times p \\ = \left(k - 1\right) p^2 \left( 1 - p \right)^{k-2}. \end{multline*}$

The above simplifies to $$(k - 1) 2^{-k}$$ in the case of a fair coin.

## 6.

Let $$\mathbb{P}$$ be a probability measure on $$\mathbb{N}$$. By additivity, $$1 = \mathbb{P}(\mathbb{N}) = \sum_n \mathbb{P}(\{n\})$$. Suppose $$\mathbb{P}$$ is uniform. Then, $$\mathbb{P}(\{n\}) = c$$ for each $$n$$ and hence $$\mathbb{P}(\mathbb{N}) = c \cdot \infty$$ (we interpret $$0 \cdot \infty = 0$$), a contradiction.

## 7.

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

$$$\mathbb{P}(\cup_n A_n) = \mathbb{P}(\cup_n B_n) = \sum_n \mathbb{P}(B_n) \leq \sum_n \mathbb{P}(A_n).$$$

## 8.

Since

$$$\mathbb{P}(\cup_i A_i^c) \leq \sum \mathbb{P}(A_i^c) = 0,$$$

it follows that

$$$\mathbb{P}(\cap_i A_i) = 1 - \mathbb{P}( \left( \cap_i A_i \right)^c ) = 1 - \mathbb{P}(\cup_i A_i^c) \geq 1 - 0 = 1.$$$

## 9.

First, note that $$\mathbb{P}(A \mid B) = \mathbb{P}(A \cap B) / \mathbb{P}(B) \geq 0$$. In particular, $$\mathbb{P}(\Omega \mid B) = 1$$. Lastly, let $$A_1, A_2, \ldots$$ be disjoint. Then,

$$$\mathbb{P}(\cup_n A_n \mid B) = \frac{\mathbb{P}(\cup_n \left( A_n \cap B \right))}{\mathbb{P}(B)} = \sum_n \frac{\mathbb{P}(A_n \cap B)}{\mathbb{P}(B)} = \sum_n \mathbb{P}(A_n \mid B).$$$

## 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 $$i = 1$$ or $$i = 3$$. It follows that

$$$p_i \equiv \mathbb{P}(\omega_1=i\mid\omega_2=2) = \frac{\mathbb{P}(\omega_2=2\mid\omega_1=i)\mathbb{P}(\omega_1=i)}{\mathbb{P}(\omega_2=2)} = \frac{\mathbb{P}(\omega_2=2\mid\omega_1=i)}{3\mathbb{P}(\omega_2=2)}.$$$

In particular,

$$$\mathbb{P}(\omega_2=2\mid\omega_1=i) = \begin{cases} 1/2 & \text{if }i=1\\ 1 & \text{if }i=3. \end{cases}$$$

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

## 11.

First, note that that

$$$\mathbb{P}(A^c \cap B^c) = \mathbb{P}(A^c) - \mathbb{P}(B) + \mathbb{P}(A \cap B).$$$

Using the independence of $$A$$ and $$B$$,

\begin{align} \mathbb{P}(A^c\cap B^c) & = \mathbb{P}(A^c) - \mathbb{P}(B) + \mathbb{P}(A) \mathbb{P}(B) \\ & = \mathbb{P}(A^c) - \left( 1 - \mathbb{P}(A) \right) \mathbb{P}(B) \\ & = \mathbb{P}(A^c) - \mathbb{P}(A^c) \mathbb{P}(B) \\ & = \mathbb{P}(A^c) \left( 1 - \mathbb{P}(B) \right) \\ & = \mathbb{P}(A^c) \mathbb{P}(B^c). \end{align}

## 12.

Let $$G_0$$ (respectively, $$G_1$$) be the event that the side of the seen (respectively, unseen) card is green. Since $$\mathbb{P}(G_0) = 1/3 + 1/3 \cdot 1/2 = 1/2$$, Then,

$$$\mathbb{P}(G_1 \mid G_0) = \frac{\mathbb{P}(G_0 \cap G_1)}{\mathbb{P}(G_0)} = \frac{1/3}{1/2} = 2/3.$$$

## 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 $$HHT$$ or $$TTH$$. If $$p$$ is the probability of heads, then the probability of this is $$p^2 (1 - p) + (1 - p)^2 p = p (1 - p)$$. In the case of a fair coin, this simplifies to $$1 / 4$$.

## 14.

Let $$A$$ and $$B$$ be events.

Suppose $$\mathbb{P}(A) = 0$$. Then, $$\mathbb{P}(A \cap B) \leq \mathbb{P}(A) = 0$$ and hence $$\mathbb{P}(A \cap B) = 0 = \mathbb{P}(A) \mathbb{P}(B)$$.

Suppose $$\mathbb{P}(A) = 1$$. Then, $$\mathbb{P}(A^c) = 0$$ and hence by our most recent findings, $$A^c$$ and $$B^c$$ are independent. By our findings in Question 11, it follows that $$A$$ and $$B$$ are independent.

Suppose now that $$A$$ is independent of itself. Then, $$\mathbb{P}(A) = \mathbb{P}(A\cap A) = \mathbb{P}(A)\mathbb{P}(A)$$ and hence either $$\mathbb{P}(A) = 0$$ or $$\mathbb{P}(A) = 1$$.

## 15.

Let $$B_k$$ be an indicator random variable that is one if and only if the $$k$$-th child has blue eyes. Let $$B = B_1 + B_2 + B_3$$. Let $$p = 1/4$$ be the probability of having blue eyes and $$q = 1 - p$$.

### a)

Note that

$$$\mathbb{P}(B \geq 2 \mid B \geq1) = \frac{\mathbb{P}(B \geq 2)}{\mathbb{P}(B \geq 1)} = \frac{1 - \mathbb{P}(B \leq 1)}{1 - \mathbb{P}(B = 0)}.$$$

Moreover, $$\mathbb{P}(B = 0) = q^3$$ and $$\mathbb{P}(B = 1) = 3pq^2$$. Therefore,

$$$\mathbb{P}(B \geq 2 \mid B \geq1) = \frac{1 - q^3 - 3pq^2}{1 - q^3} = \frac{10}{37}.$$$

### b)

Note that

$\begin{multline} \mathbb{P}(B\geq2\mid B_1 = 1) = \frac{\mathbb{P}(B_1=1,B_2+B_3\geq1)}{\mathbb{P}(B_1=1)} = \mathbb{P}(B_2+B_3\geq1) \\ = 1 - \mathbb{P}(B_2 + B_3 = 0) = 1 - q^2 = \frac{7}{16}. \end{multline}$

## 16.

Let $$A$$ and $$B$$ be events with $$\mathbb{P}(B)>0$$. $$\mathbb{P}(A \cap B) = \mathbb{P}(A \mid B)\mathbb{P}(B)$$ follows by multiplying by $$\mathbb{P}(B)$$ on both sides of the definition of conditional probability. Moreover, if $$A$$ and $$B$$ are independent,

$$$\mathbb{P}(A\mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A)\mathbb{P}(B)}{\mathbb{P}(B)} = \mathbb{P}(A).$$$

## 17.

Assuming $$\mathbb{P}(BC)$$ and $$\mathbb{P}(C)$$ are positive, the result follows from combining

$$$\mathbb{P}(ABC) = \frac{\mathbb{P}(ABC)}{\mathbb{P}(BC)} \mathbb{P}(BC) = \mathbb{P}(A \mid BC) \mathbb{P}(BC)$$$

and

$$$\mathbb{P}(BC) = \frac{\mathbb{P}(BC)}{\mathbb{P}(C)}\mathbb{P}(C) = \mathbb{P}(B \mid C) \mathbb{P}(C).$$$

## 18.

If $$A_1, \ldots, A_k$$ are a partition of the sample space, then $$1 = \mathbb{P}(\cup_i A_i) = \sum_i \mathbb{P}(A_i)$$. Moreover, for any event $$B$$,

$$$\mathbb{P}(B) = \mathbb{P}(\left( \cup_i A_i \right) \cap B) = \mathbb{P}(\cup_i \left(A_i \cap B \right)) = \sum_i \mathbb{P}(A_i \cap B).$$$

If $$\mathbb{P}(B) > 0$$, then we can divide both sides by $$\mathbb{P}(B)$$ to get $$1 = \sum_i \mathbb{P}(A_i \mid B)$$. Combining this with a previous equality, we get $$\sum_i\mathbb{P}(A_i) = \sum_i\mathbb{P}(A_i\mid B)$$. Suppose now $$\mathbb{P}(A_1\mid B) < \mathbb{P}(A_1)$$. Then,

$$$\sum_{i \neq 1}\mathbb{P}(A_i) < \sum_{i \neq 1}\mathbb{P}(A_i \mid B).$$$

It follows that $$\mathbb{P}(A_i \mid B) > \mathbb{P}(A_i)$$ for at least one $$i$$.

## 19.

We use $$M$$, $$W$$, and $$L$$ to denote the event that the user uses Mac, Windows, and Linux, respectively. We use $$V$$ to denote the event that the user has the virus.

$\begin{multline} \mathbb{P}(W\mid V) = \frac{\mathbb{P}(V\mid W)\mathbb{P}(W)}{\mathbb{P}(V)} = \frac{\mathbb{P}(V\mid W)\mathbb{P}(W)}{\sum_{X\in\{M,W,L\}}\mathbb{P}(V\mid X)\mathbb{P}(X)}\\ = \frac{82\times50}{65\times30+82\times50+50\times20} = \frac{82}{141} \approx 0.58. \end{multline}$

## 20.

### a)

$$$\mathbb{P}(C_i \mid H) = \frac{p_i \mathbb{P}(C_i)}{\mathbb{P}(H)} = \frac{p_i \mathbb{P}(C_i)}{\sum_j p_j \mathbb{P}(C_j)} = \frac{p_i}{\sum_j p_j}.$$$

### b)

$$$\mathbb{P}(H_2 \mid H_1) = \frac{\mathbb{P}(H_1 \cap H_2)}{\mathbb{P}(H_1)} = \frac{\sum_i p_i^2 \mathbb{P}(C_i)}{\sum_i p_i \mathbb{P}(C_i)} = \frac{\sum_i p_i^2}{\sum_i p_i}.$$$

### c)

$$$\mathbb{P}(C_i \mid B_4) = \frac{\mathbb{P}(C_i \cap B_4)}{\mathbb{P}(B_4)} = \frac{\mathbb{P}(B_4 \mid C_i) \mathbb{P}(C_i)}{\sum_j \mathbb{P}(B_4 \mid C_j) \mathbb{P}(C_j)} = \frac{\left( 1 - p_i \right)^3 p_i}{\sum_j \left(1 - p_j \right)^3 p_j}.$$$

## 21.

TODO (Computer Experiment)

## 22.

TODO (Computer Experiment)

## 23.

TODO (Computer Experiment)