## 1.

Let $% $. 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

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,

## 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

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.

Note that

Similarly,

### 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 $\liminf_n A_n$.

Note that

Similarly,

## 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

## 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,

Since

it follows that

## 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,

## 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

In particular,

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

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

## 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,

## 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 $p^2 (1 - p) + (1 - p)^2 p = p (1 - p)$.

## 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

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

Note that

## 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,

## 17.

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

and

## 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 $B$,

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 $% $. Then,

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.

## 21.

TODO (Computer Experiment)

## 22.

TODO (Computer Experiment)

## 23.

TODO (Computer Experiment)