Answers to exercises
Exercise 1
 We need \(\sum p_X(x) = 1\), but this holds for any value of \(\theta\). We also need \(p_X(x) \geq 0\). This holds if and only if \(1 \leq \theta \leq 1\).
 \(\Pr(X \geq 4) = 3 \times \bigl(\dfrac{1 + \theta}{6}\bigr) = \dfrac{1}{2}(1 + \theta)\).
 \(\Pr(3 \leq X \leq 4) = \Pr(X=3) + \Pr(X=4) = \dfrac{1\theta}{6} + \dfrac{1+\theta}{6} = \dfrac{1}{3}\).
 \(\Pr(\text{\(X\) is even}) = \dfrac{1}{6}(3 + \theta)\).
Exercise 2
 Assuming independence, \(\bigl(\dfrac{6}{7}\bigr)^3 \approx 0.6297\).
 \(1  \bigl(\dfrac{6}{7}\bigr)^3 \approx 0.3703\).
 0.3703.
 \(0.6297 \times 0.3703 \approx 0.2332\).
 \(0.6297^x \times 0.3703\).
 \(X\) has a geometric distribution with parameter \(p = 0.3703\). The probability function is given by \(p_X(x) = 0.3703 \times 0.6297^x\), for \(x = 0,1,2,\dots\).
 \(p_X(2) = 0.1468\), \(p_X(3) = 0.0925\), \(p_X(4) = 0.0582\), \(p_X(10) = 0.0036\).
Exercise 3
 There are nine possible outcomes in a single play, three of which are ties. So the probability of a tie is \(\dfrac{1}{3}\).
 The chance of Julia winning on any single play, including on the first play, is \(\dfrac{1}{3}\).
 \(X\) has a geometric distribution with parameter \(p = \dfrac{1}{3}\).
 \(p_X(x) = \dfrac{1}{3}\bigl(\dfrac{2}{3}\bigr)^x\), for \(x = 0,1,2,\dots\).
 \(p_X(5) = \dfrac{1}{3}\bigl(\dfrac{2}{3}\bigr)^5 \approx 0.0439\).
 There are at least two ways to approach this problem. One method is to calculate:
\[
\Pr(X \geq 5) = \sum_{x=5}^{\infty} p_X(x) = \sum_{x=5}^{\infty} \dfrac{1}{3}\Bigl(\dfrac{2}{3}\Bigr)^x = p_X(5) \sum_{n=0}^{\infty} \Bigl(\dfrac{2}{3}\Bigr)^n = \dfrac{p_X(5)}{1\dfrac{2}{3}} \approx 0.1317,
\]
using the formula for the sum of an infinite geometric series.
Alternatively, note that the event "\(X \geq 5\)" (the first success is after at least five failures) occurs precisely when the first five trials are failures. The probability of failure in a single trial is \(1  p = \dfrac{2}{3}\) and the trials are independent, so \(\Pr(X \geq 5) = \bigl(\dfrac{2}{3}\bigr)^5 \approx 0.1317\).
Exercise 4
 \(\mu_X = \sum x\, p_X(x) = 6\bigl(\dfrac{1\theta}{6}\bigr) + 15\bigl(\dfrac{1+\theta}{6}\bigr) = \dfrac{1}{2}(7 + 3\theta)\).
 From exercise 1, the largest value for \(\theta\) is 1. So the largest mean is \(\dfrac{1}{2}(7 + 3) = 5\).
Exercise 5
 In each case, the probabilities are all clearly nonnegative and the sum of the probabilities equals one, as required.
 You should be able to get fairly close to the actual means (see part c) visually.
 In each case, the mean is given by \(\mu_X = \sum x\, p_X(x)\).

a 
b 
c 
d 
e 
f 
\(\mu_X\) 
6.7 
6.8 
2.2 
7.0 
5.0 
6.6 
Exercise 6
 The chance of a single entry winning first prize is equal to \(\dfrac{1}{\binom{45}{6}} = \dfrac{1}{8\ 145\ 060}\).
 The random variable \(X\), the number of draws before a first prize is obtained, has a geometric distribution with parameter \(p = \dfrac{1}{8\ 145\ 060}\).
 The mean of \(X\) is given by \(\dfrac{1p}{p} = 8\ 145\ 059\).
 About 157 000 years.
Exercise 7
 \(\mu_X = 1.52\).
 \(\operatorname{var}(X) = 0.8216,\ \ \operatorname{sd}(X) \approx 0.9064\).
 The units are `number of languages'.
Exercise 8
 For each graph, we have \(p_X(x) \geq 0\) and \(\sum p_X(x) = 1\).
 You should be able to get close to the actual means (see part c) visually.
 Each of the four means is equal to 6.
 Guessing variances is much more difficult than guessing means. But for these four distributions, the order (largest to smallest) should be clear visually: \(\text{b} > \text{a} > \text{d} > \text{c}\).


a 
b 
c 
d 
\(\operatorname{var}(X)\) 
11.2 
16.2 
0.4 
4.0 
\(\operatorname{sd}(X)\) 
3.35 
4.02 
0.63 
2.0 