Answers to exercises
Exercise 1
 \(A\), \(C\), \(D\) and \(E\) are all events, because they each consist of a collection of possible outcomes. \(B\) is not an event for this random procedure, because if we only know the outcome of the roll of the dice, we do not know whether or not \(B\) has occurred.
 \(\begin{aligned}[t] A = \{&(1,1), (1,2), (2,1), (1,4), (4,1), (2,3), (3,2),\\ &(1,6), (6,1), (2,5), (5,2), (3,4), (4,3), (5,6), (6,5)\}.\end{aligned}\)
 \(D = 11\).
 \(A \cap E = \varnothing\), the empty set. None of the possible outcomes is in both \(A\) and \(E\): there is only one outcome in \(E\), and it is clearly not in \(A\) (since \(6+6 = 12\), which is not prime).
Exercise 2
For this random procedure, each possible outcome consists of an ordered sequence of 31 numbers, such as \((0.1, 1.2, 0.7, 3.4, \dots)\). The first number is the rainfall on 1 August, the second is the rainfall on 2 August, and so on. There are infinitely many possible outcomes, in principle. Cases like this are not at all unusual. Any time we take a sequence of measurements of a quantity of interest, we have a random procedure of this type.

 We know which dates are Mondays in August 2017. So, if we have observed an outcome, we can determine whether or not \(A\) has occurred. Thus \(A\) is an event.
 Although \(B\) seems related to rainfall, it is not an event for this random procedure. For any given outcome, we cannot say for sure whether \(B\) has occurred.
 The outcomes in \(C\) are those sequences for which the sum of the 31 numbers (the daily rainfall amounts) is greater than 20.0. Thus \(C\) is an event.
 The pairs of mutually exclusive events are: \(A\) and \(F\); \(E\) and \(F\).

 \(A'\) = "there is a rainfall amount greater than zero recorded on at least one day that is not a Monday".
 \(A \cap E\) = "either no rainfall is recorded for the whole month, or else the only day with a rainfall amount greater than zero is the last Monday of the month".
 \(A \cup C\) is a complicated event to put in words: the best way of describing it is to say that either \(A\) occurs, or \(C\) occurs, or both. An example of an outcome that is not in \(A \cup C\) is rainfall of 0.1 mm every day; for such an outcome there is nonzero rainfall on days other than Mondays (so \(A\) does not occur) and the total rainfall for the month is 3.1 mm (so \(C\) does not occur). Since neither \(A\) nor \(C\) occurs, the union \(A \cup C\) does not occur for this outcome.
Exercise 3
This problem requires you to recognise that, in a fourchild family, the event ``quot;at least one girl" is the complement of "all boys". Let \(A\) = "all four children are boys". Then \(A'\) = "there is at least one girl". Since \(\Pr(A) = 0.07\), it follows by property 4 that \(\Pr(A') = 1  0.07 = 0.93\).
Exercise 4
 Using the specified assumption of random mixing, we get \(\Pr(A) = \Pr(B) = \dfrac{1}{16}\).
 The child's wishes will be satisfied if at least one of the two purchases is a Pirate Captain; this event is \(A \cup B\). By the addition theorem (property 5), we have \begin{align*} \Pr(A \cup B) &= \Pr(A) + \Pr(B)  \Pr(A \cap B)\\ &= \dfrac{1}{16} + \dfrac{1}{16}  \dfrac{1}{256} = \dfrac{31}{256} \approx 0.121, \end{align*} using part (a) and the specified assumption that \(\Pr(A \cap B) = \dfrac{1}{256}\). \item

 \(B'\) = "a figure other than a Pirate Captain is purchased at the second shop"
 \(A' \cup B'\) = "either the first purchase is not a Pirate Captain, or the second purchase is not a Pirate Captain, or both"
 \(A \cap B'\) = "the first purchase is a Pirate Captain, and the second purchase is not a Pirate Captain"
 \(A' \cap B'\) = "neither purchase is a Pirate Captain"
 In part (b), we found that \(\Pr(A \cup B) = \dfrac{31}{256} \approx 0.121\). Note that \(A' \cap B' = (A \cup B)'\). This is an important general result: The complement of the union is the intersection of the complements. By an application of property 4 to this result, we find that \[ \Pr(A' \cap B') = 1  \Pr(A \cup B) = 1  \dfrac{31}{256} = \dfrac{225}{256} \approx 0.879. \] So not purchasing a Pirate Captain is the more probable event.
Exercise 5
This question reminds us that we are often using an idealised model in probability, in which random mixing is assumed. Whether the mixing is sufficiently random for this assumption to be reasonably satisfied is party a matter of physics (more specifically, the concrete physical processes involved in the mixing). But in practice it usually boils down to an assumption, informed by consideration of the procedure involved.
 A normal flip with many spins is likely to be random, if executed by a person with no particular skill. However, it is (apparently) possible to learn how to flip a coin to produce the desired result. In those cases the mixing is not random at all.
 For reasons associated with exercise 6, a flip that lands on the floor may show a bias towards heads or tails, depending on the specific coin. This does not mean the mixing is not random, but it demonstrates the subtleties of randomising devices.
 With a sufficiently vigorous and lengthy shake, the mixing is random, leading to no preference for any outcome, assuming a symmetric die.
 An overhand shuffle needs to be done for a very long time to produce randomness in the order of the pack, relative to the starting order. If the shuffle is done just a few times, traces of the original order will persist.
 A riffle shuffle produces a random order much more effectively than an overhand shuffle.
 It seems clear that this blast of air produces such chaotic and extensive mixing of the balls that, assuming the balls are of uniform size, shape and weight, all outcomes will be (close to) equally likely.
 Historically, random number generators have been of varying quality; it depends on the method used. Most computer games now use algorithms that will not produce detectable nonrandomness.
 A simple physical device like this may well have inadequate mixing.
Exercise 6
What you get depends on the specific coin. Were you surprised by your results?
Exercise 7
There are indeed five possibilities for the number of aces in the hand. But these different possibilities are not equally likely. A standard deck has 52 cards. There are thousands of equiprobable distinct hands of four cards that can be dealt: \(270\ 725\), to be precise. Only one of these possible hands contains all 4 aces, whereas there are many hands that contain no aces. So the chance of four aces is \(\dfrac{1}{270\ 725}\), while the probability of a hand with no aces is much greater.
Exercise 8
There is some simple reasoning to be applied here. Consider the ballot for the first half of the year, in which 29 February could be selected. In most years, no men born on 29 February could be recruited, even if that marble came up, since their birth year was not a leap year (for example, in 1965 the birth year was 1945). In years such as 1968, the men turning 20 were born in the leap year of 1948, and some would have been born on 29 February. For those men born on 29 February 1948, their chances of conscription were the same as other men born in the first half of 1948.
In fact, there were only two relevant leap years, 1948 and 1952, involving conscription in the years 1968 and 1972. The date 29 February came up in the selected marbles in 1972 and not in 1968.
Exercise 9
These answers are discussed further in the section Independence.
 Two lottery draws are usually regarded as mathematically independent because of their physical independence. In principle, there are issues to be considered here. Does the outcome depend at all on the initial position of the balls? If so, could this in any way be related to the position they finished after last week's draw?
 \(A\) and \(B\) are dependent, because they are mutually exclusive. Any mutually exclusive events are not independent.
 \(A\) and \(B\) are not likely to be independent, given the seasonal cycle of weather in most places, as discussed after the exercise.
 \(A\) and \(B\) are independent if the toss is truly random.
Exercise 10
To answer this exercise it is important to recognise that the independence of the experiments means that their outcomes are independent.
 \(Y\) can take the values \(2, 3, 4, 5, 6, 7\); \(U\) can take the values \(2, 3, 4, \dots, 12\); \(V\) can take the values \(1, 2, 3, 4, 5, 6\).
 By careful consideration of the outcomes, and the use of independence, we find that \(\Pr(U > Y) = \dfrac{160}{216}\) and \(\Pr(U < Y) = \dfrac{35}{216}\). So it is much more likely that the sum \(U\) will be greater than \(X+1\), rather than the other way around. Note also that \(\Pr(U = Y) = \dfrac{21}{216}\).
 \(\Pr(V > Y) = \dfrac{90}{216}\) and \(\Pr(V < Y) = \dfrac{91}{216}\). (Very close!) Hence \(\Pr(V > Y) < \Pr(V < Y)\). Note also that \(\Pr(V = Y) = \dfrac{35}{216}\).
Exercise 11

 Assuming independence, we can calculate the probability of two nonsightings as \(0.3 \times 0.3 = 0.09\).
 \(0.3^7 \approx 0.0002\).
 If all components have to fail for the machine to fail, the probability of failure of the machine is obtained by multiplying the individual failure probabilities, so the probability equals \(0.1 \times 0.01 \times 0.01 \times 0.005 = 5 \times 10^{8}\). (Very small indeed.)
 The probability of at least one component failing is one minus the probability that none fails. Assuming independence, the chance that none fails is equal to \(0.9 \times 0.99 \times 0.99 \times 0.995 \approx 0.8777\). Hence the probability of failure of the machine is approximately \(1  0.8777 = 0.1223\). (Much higher.)
 Independence seems doubtful here; the six people could all be delayed by a common causative event, such as inclement weather. Assuming independence, the chance of the building being opened on time is \(1  (0.05 \times 0.1 \times 0.2 \times 0.25 \times 0.5 \times 0.9) \approx 0.9999\).
 This assessment entails the multiplication of probabilities, which assumes independence. It seems plausible that circumstances that precipitate a lossofcoolant accident might also predispose towards the failure of safety functions: for example, an earthquake or human sabotage.
Exercise 12
 The player loses at the first roll if 1, 4 or 6 is obtained. The probability of this occurring is 0.5.
 To win $6, the player needs to be paid $10. This only occurs if the first roll is 5 and the second roll is 5; the probability of this is \(\dfrac{1}{36}\).
 The probability of winning is obtained by adding up the probabilities of outcomes for which the payout is greater than $4. There are eight such outcomes: (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3) and (5,5), each with probability equal to \(\dfrac{1}{36}\). So the probability of winning equals \(\dfrac{8}{36} = \dfrac{2}{9}\).
Exercise 13
 \(\Pr(C'T_{}) \approx 0.9984\). So, if the test is negative, there is a very high probability that the woman does not have breast cancer.
 The positive predictive value is calculated as follows: \begin{align*} \Pr(CT_{+}) &= \dfrac{\Pr(C) \Pr(T_{+}C)}{\Pr(C) \Pr(T_{+}C) + \Pr(C') \Pr(T_{+}C')}\\ &= \dfrac{0.01 \times \Pr(T_{+}C)}{0.01 \times \Pr(T_{+}C) + 0.99 \times 0.07}. \end{align*} This is maximised when the sensitivity \(\Pr(T_{+}C)\) is 1. For \(\Pr(T_{+}C) = 1\), the positive predictive value is \(\Pr(CT_{+}) \approx 0.1261\), which is still quite small.
 Some algebra shows that, for \(\Pr(C) = 0.01\) and \(\Pr(T_{+}C)=0.85\):
 \(\Pr(CT_{+}) = 0.20\) requires \(\Pr(T_{}C') \approx 0.9657\)
 \(\Pr(CT_{+}) = 0.50\) requires \(\Pr(T_{}C') \approx 0.9914\)
 \(\Pr(CT_{+}) = 0.80\) requires \(\Pr(T_{}C') \approx 0.9979\)
 \(\Pr(CT_{+}) = 0.99\) requires \(\Pr(T_{}C') \approx 0.9999\).
 The positive predictive value is \(\Pr(CT_{+}) \approx 0.0012\) and the negative predictive value is \(\Pr(C'T_{}) \approx 0.99998\).
The specificity values required to achieve high positive predictive values are very close to 1.