## Exercise 1

(a) Evaluate the following integral. Do not use a calculator or computer, except to check your work.

$\int_{0}^{\infty}x e^{-2x}dx$

Solution:

Here we have integration by parts. We set

$u = x, \quad dv = e^{-2x}dx$

Thus we have

$du = dx, \quad v = -\frac{1}{2}e^{-2x}$

Then we obtain

$\int_{0}^{\infty} x e^{-2x}dx = -\frac{1}{2}xe^{-2x} \Bigg |_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{2}e^{-2x}dx = \boxed{\frac{1}{4}} = \boxed{0.25}$

Note that we are being somewhat abusive with notation since we are dealing with an improper intergral.

(b) Evaluate the following integral. Do not use a calculator or computer, except to check your work.

$\int_{0}^{\infty}x e^{-x^2}dx$

Solution:

Here we use integration by substitution. We set

$u = x^2$

Thus we have

$du = 2xdx$

Then we obtain

$\int_{0}^{\infty} x e^{-x^2}dx = \int_{0}^{\infty} \frac{1}{2}e^{-x^2}(2xdx) = \frac{1}{2} \int_{0}^{\infty} e^{-u}du = -\frac{1}{2}e^{-u} \Bigg |_{0}^{\infty} = \boxed{\frac{1}{2}} = \boxed{0.50}$

## Exercise 2

Find the value $$c$$ such that

$\iint\limits_A cx^2y^3 dydx = 1$

where $$A = \{ (x,y) : 0 < x < 1, \ 0 < y < \sqrt{x} \}$$. Do not use a calculator or computer, except to check your work.

Solution:

First,

$\iint\limits_A cx^2y^3 dydx = 1 = \int_{0}^{1} \left(\int_{0}^{\sqrt{x}} cx^2y^3 dy \right) dx = \int_{0}^{1} \frac{c}{4}x^4 dx = \frac{c}{20}$

Then,

$\frac{c}{20} = 1 \implies c = \boxed{20}$

## Exercise 3

Suppose $$S = \{2, 3, 4, 5, \ldots \}$$ and

$P(k) = c \cdot \frac{2^k}{k!}, \quad k = 2, 3, 4, 5, \ldots$

Find the value of $$c$$ that makes this a valid probability distribution.

Solution:

First note that,

\begin{aligned} \sum_{\text{all } x} P(x) &= \sum_{k = 2}^{\infty} c \cdot \frac{2^k}{k!} \\ &= c \cdot \left( \sum_{k = 0}^{\infty}\frac{2^k}{k!} - \frac{2^0}{0!} - \frac{2^1}{1!} \right) \\ &= c \cdot (e^2 - 1 - 2) \\ &= c \cdot (e^2 - 3) \end{aligned}

Then since we need to have

$\sum_{\text{all } x} P(x) = 1$

we obtain

$c \cdot (e^2 - 3) = 1 \implies c = \boxed{\frac{1}{e^2 - 3}} \approx \boxed{0.22784}.$

## Exercise 4

Suppose $$S = \{2, 3, 4, 5, \ldots \}$$ and

$P(k) = \frac{6}{3^k}, \quad k = 2, 3, 4, 5, \ldots$

Find $$P(\text{outcome is greater than 3})$$.

Solution:

\begin{aligned} P(\text{outcome is greater than 3}) &= P(4) + P(5) + P(6) + \ldots \\ &= \sum_{k = 4}^{\infty}\frac{6}{3^k} = \frac{\text{first term}}{1 - \text{base}} = \frac{\frac{6}{3^4}}{1 - \frac{1}{3}} \\ &= \boxed{\frac{1}{9}} = \boxed{0.111\overline1} \end{aligned}

Or alternatively,

\begin{aligned} P(\text{outcome is greater than 3}) &= 1 - P(2) - P(3) \\ &= 1 - \frac{6}{3^2} - \frac{6}{3^3} \\ &= \boxed{\frac{1}{9}} = \boxed{0.111\overline1} \end{aligned}

## Exercise 5

Suppose $$P(A) = 0.4$$, $$P(B^\prime) = 0.3$$, and $$P(A \cap B^\prime) = 0.1$$.

$$B$$ $$B^\prime$$
$$A$$ 0.30 0.10 0.40
$$A^\prime$$ 0.40 0.20 0.60
0.70 0.30 1.00

(a) Find $$P(A \cup B)$$.

Solution:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.40 + 0.70 - 0.30 = \boxed{0.80}$

$P(A \cup B) = 1 - P(A^\prime \cap B^\prime) = 1 - 0.20 = \boxed{0.80}$

$P(A \cup B) = P(A \cap B) + P(A \cap B^\prime) + P(A^\prime \cap B) = 0.30 + 0.10 + 0.40 = \boxed{0.80}$

(b) Find $$P(B^\prime \mid A)$$.

Solution:

$P(B^\prime \mid A) = \frac{P(A \cap B^\prime)}{P(A)} = \frac{0.10}{0.40} = \boxed{\frac{1}{4}} = \boxed{0.25}$

(c) Find $$P(B \mid A^\prime)$$.

Solution:

$P(B \mid A^\prime) = \frac{P(A^\prime \cap B)}{P(A^\prime)} = \frac{0.40}{0.60} = \boxed{\frac{2}{3}} = \boxed{0.66\overline6}$

## Exercise 6

Suppose:

• $$P(A) = 0.6$$
• $$P(B) = 0.5$$
• $$P(C) = 0.4$$
• $$P(A \cap B) = 0.3$$
• $$P(A \cap C) = 0.2$$
• $$P(B \cap C) = 0.2$$
• $$P(A \cap B \cap C) = 0.1$$

(a) Find $$P((A \cup B) \cap C^\prime)$$.

Solution:

After finding the probabilities for each disjoint region and shading the appropriate venn diagram, we have

$P((A \cup B) \cap C^\prime) = \boxed{0.50}$

(b) Find $$P(A \cup (B \cap C))$$.

Solution:

After finding the probabilities for each disjoint region and shading the appropriate venn diagram, we have

$P(A \cup (B \cap C)) = \boxed{0.70}$