Peter McFetridge, 25480138

Jan 22, 2016

Math 302

1. Section 1.2 Exercise 6

For any $i \in \Omega = \{1,2,3,4,5,6\}$ there is a distribution function $m(i) = C \cdot i$ where $C$ the probability of side $i$ facing up, $C$ is proportional to the number of dots on side $i$.

or

Let $E$ be the event that an even number is thrown. So, $E = \{2,4,6\}$

2. Section 1.2 Exercise 7

Let A and B be events such that $P(A \cap B) = 1/4$, $P(\tilde{A})=1/3$, and $P(B)= 1/2$. What is $P (A \cup B)$?

$$P(A) = 1 - P(\tilde{A})$ then $P(A) = 1 - 1/3 = 2/3$$

And $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Gives us

3. Section 1.2 Exercise 10

In order for a bill to get to the president it must pass the House of Representatives and the Senate.

Let $S =$ a bill passing the Senate

Let $H =$ a bill passing the House of Representatives

The probability that a bill is passed throught the Senate is $$P(S) = .8$$ and the probability that a bill is passed through the house is $$P(H) = .6$$ and the probability that a bill is passed through at least one of the two is $$P(S \cup H) = .9$$

Since $$P(S \cup H) = P(S) + P(H) - P(S \cap H)$$

Then $$P(S \cap H) = P(S) + P(H) - P(S \cup H)$$

So, $$P(S \cap H) = .8 + .6 - .9 = \mathbf{.5}$$

4. Section 1.2 Exercise 11

What odds should a person give in favor of the following events?

(a) A card chosen at random from a 52 card deck is an ace.

$E$ = ace is drawn There are 4 aces in a 52 card deck $$P(E) = \frac{4}{52} = \mathbf{\frac{1}{13}}$$

(b) Two heads will turn up when a coin is tossed twice.

$E$ = coin toss comes up heads $$P(E) = \frac{1}{2}$$ We want the probability that we will receive 2 heads or $E \cdot E$

Probability that two heads will turn up = $$E \cdot E = \left({\frac{1}{2}}\right)^2 = \mathbf{\frac{1}{4}}$$

This can also so seen by looking at the total number of outcomes. In this case there are 4 total outcome. $$\Omega = \{HH, \; HT, \; TH, \; TT\}$$

Only 1 of the 4 outcomes satisfies what we are looking for and therefore $$\mathbf{\frac{1}{4}}$$

(c) Two sixes will turn up when two dice are rolled.

Omega is the sample space of the results when a die is roll with an even distribution function. $$\Omega = \{1,2,3,4,5,6\}$$ Event $E$ = 6 facing up when die is rolled. We are looking for the probability that both die are 6 $$P(E) \cdot P(E) = P(E^2) = \left(\frac{1}{6}\right)^2 = \mathbf{\frac{1}{36}}$$

5. Section 1.2 Exercise 18

(a) For events $A_1,...,A_n$, prove that

Let $A_1,...,A_n$ are events in the set $\Omega$

Since $P \left( \bigcup_{i=1}^{n} A_i\right)$ is the event that at least one event in $A_i$ occurs and is bounded by $\Omega = 1$ by Axiom 2. Regardless if the events are distinct or not distinct.

By Axiom 1 $0 \leq P(A_i) \leq 1$ then it is easy to see that $\sum_{i=1}^{n} P(A_i) \geq 1$ unless event $A = \emptyset$.

(b) For events A and B, prove that

Let $A$ and $B$ be events in $\Omega$

We can rearrange the problem $$1 \geq P(A) + P(B) - P(A \cap B)$$ Which is equivalent to $$1 \geq P(A \cup B)$$ Since $P(A \cup B) = \Omega$ we have $$1 \geq \Omega$$ Which is true by Axiom 2

6. Section 1.2 Exercise 18

If $A, B, C$ are any three events, show that $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$$

We can expand

7. Section 1.2 Exercise 20

Explain why it is not possible to define a uniform distribution function (see Definition 1.3) on a countably infinite sample space. Hint: Assume m(ω) = a for all ω, where 0 ≤ a ≤ 1. Does m(ω) have all the properties of a distribution function?

For $\Omega = \{1,2,3,.....\}$

We can show that $m(n) = \frac{1}{2^n}$

Which is true for the geometric sequence, $$1 + r +r^2 +r^3 + \cdot \cdot \cdot = \frac{1}{1-r}$$ or $$r+r^2+r^3+r^4+ \cdot \cdot \cdot = \frac{r}{1-r}$$

for $-1 < r < 1$

When $r = .25$ $$P(E) = \frac{.25}{.75} = \frac{1}{3}$$

Which shows that $$P(E) = \frac{1}{3}$$ and $$P(\tilde{E}) = \frac{2}{3}$$

And that contradicts the properties of a uniform distribution.

8. Section 1.2 Exercise 26

Two cards are drawn from a deck of 52 cards. What is the probability that the second card is higher in rank than the first card. $$1 = P(Higher) + P(Lower) + P(Same) \\$$ $$0 = 1 - P(Higher) - P(Lower) - P(Same) \\$$ $$\text{Since} \; P(Higher) \; and \; P(Lower) \; \text{are the same we can re-write} \\$$ $$2P(Higher) = 1 - P(Same) \\$$ $$\text{Since one card has been removed} P(Same) = \frac{3}{51} \\$$ $$2P(Higher) = 1 - \frac{3}{51} \\$$ $$P(Higher) = \frac{48}{102} = \frac{24}{51} \\$$