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:

Integral Region

Integral Region

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 \(P(A \cap B^\prime) = 0.1\).

Venn Diagram for P(A) = 0.4, P(B^\prime) = 0.3, and P(A \cap B^\prime) = 0.1

Venn Diagram for \(P(A) = 0.4\), \(P(B^\prime) = 0.3\), 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:

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

Solution:

Venn Diagram with P((A \cup B) \cap C^\prime) shaded.

Venn Diagram with \(P((A \cup B) \cap C^\prime)\) shaded.

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:

Venn Diagram with P(A \cup (B \cap C)) shaded.

Venn Diagram with \(P(A \cup (B \cap C))\) shaded.

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

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