The sample description space of the experiment described consists of 2-tuples \(\left(z_{1}, z_{2}\right)\) , in which \(z_{1}\) is the number of the urn chosen and \(z_{2}\) is the “name” of the ball chosen. The probability function \(P[\cdot]\) on the subsets of \(S\) is specified by means of the functions listed in (4.1) , with \(n=2\) , which the assumptions stated in the problem enable us to compute. In particular, let \(C_{1}\) be the event that urn I is chosen, and let \(C_{2}\) be the event that urn II is chosen. Then \(P\left[C_{1}\right]=P\left[C_{2}\right]=\frac{1}{2}\) . Next, let \(B\) be the event that a white ball is chosen. Then \(P\left[B \mid C_{1}\right]=\frac{5}{8}\) , and \(P\left[B \mid C_{2}\right]=\frac{3}{10}\) . The events \(C_{1}\) and \(C_{2}\) are the complements of each other. Consequently, by (4.5) of Chapter 2,

\[P[B]=P\left[B \mid C_{1}\right] P\left[C_{1}\right]+P\left[B \mid C_{2}\right] P\left[C_{2}\right]=\frac{37}{80} \tag{4.5}\] 

Let us write a 3-tuple \(\left(z_{1}, z_{2}, z_{3}\right)\) to describe the history of the mother and her two sons with regard to hemophilia. Let \(z_{1}\) equal \(s\) or \(f\) , depending on whether the mother is or is not a hemophilia carrier. Let \(z_{2}\) equal \(s\) or \(f\) , depending on whether the first son is or is not hemophilic. Let \(z_{3}\) equal \(s\) or \(f\) , depending on whether the second son will or will not have hemophilia. On this sample description space, we define the events \(A_{1}, A_{2}\) , and \(A_{3}: A_{1}\) is the event that the mother is a hemophilia carrier, \(A_{2}\) is the event that the first son has hemophilia, and \(A_{3}\) is the event that the second son will have hemophilia. To specify a probability function on the subsets of \(S\) , we specify all conditional probabilities of the form given in (4.1) :

\begin{align} & P\left[A_{1}\right]=2 m, \quad P\left[A_{1}^{c}\right]=1-2 m, \\ & P\left[A_{2} \mid A_{1}\right]=\frac{1}{2}, \quad P\left[A_{2}^{c} \mid A_{1}\right]=\frac{1}{2}, \\ & P\left[A_{2} \mid A_{1}^{c}\right]=m, \quad P\left[A_{2}^{c} \mid A_{1}^{c}\right]=1-m, \\ & P\left[A_{3} \mid A_{1}, A_{2}\right]=P\left[A_{3} \mid A_{1}, A_{2}^{c}\right]=\frac{1}{2}, \\ & P\left[A_{3}^{c} \mid A_{1}, A_{2}\right]=P\left[A_{3}^{c} \mid A_{1}, A_{2}^{c}\right]=\frac{1}{2}, \\ & P\left[A_{3} \mid A_{1}^{c}, A_{2}\right]=P\left[A_{3} \mid A_{1}^{c}, A_{2}^{c}\right]=m, \\ & P\left[A_{3}^{c} \mid A_{1}^{c}, A_{2}\right]=P\left[A_{3}^{c} \mid A_{1}^{c}, A_{2}^{c}\right]=1-m . \end{align} 

In making these assumptions (4.6) we have used the fact that the woman has no family history of hemophilia. A boy usually carries an \(X\) chromosome and a \(Y\) chromosome; he has hemophilia if and only if, instead of an \(X\) chromsome, he has an \(X^{\prime}\) chromosome which bears a gene causing hemophilia. Let \(m\) be the probability of mutation of an \(X\) chromosome into an \(X^{\prime}\) chromosome. Now the mother carries two \(X\) chromosomes. Event \(A_{1}\) can occur only if at least one of these \(X\) chromosomes is a mutant; this will happen with probability \(1-(1-m)^{2} \doteq 2 m\) , since \(m^{2}\) is much smaller than \(2 \mathrm{~m}\) . Assuming that the woman is a hemophilia carrier and exactly one of her chromosomes is \(X^{\prime}\) , it follows that her son will have probability \(\frac{1}{2}\) of inheriting the \(X^{\prime}\) chromosome.

We are seeking \(P\left[A_{3} \mid A_{2}\right]\) . Now

\[P\left[A_{3} \mid A_{2}\right]=\frac{P\left[A_{2} A_{3}\right]}{P\left[A_{2}\right]} \tag{4.7}\] 

To compute \(P\left[A_{2} A_{3}\right]\) , we use the formula

\begin{align} P\left[A_{2} A_{3}\right]= & P\left[A_{1} A_{2} A_{3}\right]+P\left[A_{1}^{c} A_{2} A_{3}\right] \tag{4.8} \\ = & P\left[A_{1}\right] P\left[A_{2} \mid A_{1}\right] P\left[A_{3} \mid A_{2}, A_{1}\right] \\ & +P\left[A_{1}^{c}\right] P\left[A_{2} \mid A_{1}^{c}\right] P\left[A_{3} \mid A_{2}, A_{1}^{c}\right] \\ = & 2 m\left(\frac{1}{2}\right)^{\frac{1}{2}}+(1-2 m) m m \\ = & \frac{1}{2} m \end{align} 

since we may consider \(1-2 m\) as approximately equal to 1 and \(m^{2}\) as approximately equal to 0. To compute \(P\left[A_{2}\right]\) , we use the formula

\begin{align} P\left[A_{2}\right] & =P\left[A_{2} \mid A_{1}\right] P\left[A_{1}\right]+P\left[A_{2} \mid A_{1}^{c}\right] P\left[A_{1}^{c}\right] \tag{4.9} \\ & =\frac{1}{2} 2 m+m(1-2 m) \\ & \doteq 2 m. \end{align} 

Consequently,

\[P\left[A_{3} \mid A_{2}\right]=\frac{\frac{1}{2} m}{2 m}=\frac{1}{4} \tag{4.10}\] 

Thus the conditional probability that the second son of a woman with no family history of hemophilia will have hemophilia, given that her first son has hemophilia, is approximately \(\frac{1}{4}\) !

To describe the results of the experiment that consists in selecting five tubes from the output of the machine and then selecting one tube from among the five previously selected, we write a 6-tuple \(\left(z_{1}, z_{2}, z_{3}\right.\) , \(z_{4}, z_{5}, z_{6}\) ); for \(k=1,2, \ldots, 5, z_{k}\) is equal to \(s\) or \(f\) , depending on whether the \(k\) th tube selected is defective or non-defective, whereas \(z_{6}\) is equal to \(s\) or \(f\) , depending on whether the tube selected from those previously selected is defective or non-defective. For \(j=0, \ldots, 5\) let \(C_{j}\) denote the event that \(j\) defective tubes were selected from the output of the machine.

Assuming that the selections were independent, \(P\left[C_{j}\right]=\left(\begin{array}{l}5 \\ j\end{array}\right)(0.2)^{j}(0.8)^{5-j}\) . Let \(B\) denote the event that the sixth tube selected from the box, is defective. We assume that \(P\left[B \mid C_{j}\right]=j / 5\) ; in words, each of the tubes in the box is equally likely to be chosen. By (4.11), it follows that

\[P[B]=\sum_{j=0}^{5} \frac{j}{5}\left(\begin{array}{l} 5 \tag{4.13} \\ j \end{array}\right)(0.2)^{j}(0.8)^{5-j}\] 

To evaluate the sum in (4.13), we write it as \[\sum_{j=1}^{5} \frac{j}{5}\left(\begin{array}{l} 5 \tag{4.14} \\ j \end{array}\right)(0.2)^{j}(0.8)^{5-j}=(0.2) \sum_{j=1}^{5}\left(\begin{array}{c} 4 \\ j-1 \end{array}\right)(0.2)^{j-1}(0.8)^{4-(j-1)}=0.2\] in which we have used the easily verifiable fact that \[\frac{j}{n}\left(\begin{array}{l} n \tag{4.15} \\ j \end{array}\right)=\left(\begin{array}{c} n-1 \\ j-1 \end{array}\right)\] and the fact that the last sum in (4.14) is equal to 1 by the binomial theorem. Combining (4.13) and (4.14), we have \(P[B]=0.2\) . In words, we have proved that selecting an item randomly from a sample which has been selected randomly from a larger population is statistically equivalent to selecting the item from the larger population. Note the fact that \(P[B]=0.2\) does not imply that the box containing five tubes will always contain one defective tube.

Let us next consider part (ii) of example 4D. To describe the results of the experiment that consists in selecting five tubes from the output of the machine and then selecting two tubes from among the five previously selected, we write a 7 -tuple \(\left(z_{1}, z_{2}, \ldots, z_{7}\right)\) , in which \(z_{6}\) and \(z_{7}\) denote the tubes drawn from the box containing the first five tubes selected. Let \(C_{0}, \ldots, C_{5}\) and \(B\) be defined as before. Let \(A\) be the event that the seventh tube is defective. We seek \(P[A \mid B]\) . Now, if two tubes, each of which has probability 0.2 of being defective, are drawn independently, the conditional probability that the second tube will be defective, given that the first tube is defective, is equal to the unconditional probability that the second tube will be defective, which is equal to 0.2. We now proceed to prove that \(P[A \mid B]=0.2\) . In so doing, we are proving a special case of the principle that a sample of size 2, drawn without replacement from a sample of any size whose members are selected independently from a given population, has statistically the same properties as a sample of size 2 whose members are selected independently from the population! More general statements of this principle are given in the theoretical exercises of section 4, Chapter 4 . We prove that \(P[A \mid B]=0.2\) under the assumption that \(P\left[A B \mid C_{j}\right]=(j)_{2} /(5)_{2}\) for \(j=0, \ldots, 5\) . Then, by (4.11), \begin{align} P[A B] & =\sum_{j=0}^{5} \frac{(j)_{2}}{(5)_{2}}\left(\begin{array}{l} 5 \\ j \end{array}\right)(0.2)^{j}(0.8)^{5-j} \\ & =(0.2)^{2} \sum_{j=2}^{5}\left(\begin{array}{c} 3 \\ j-2 \end{array}\right)(0.2)^{j-2}(0.8)^{3-(j-2)}=(0.2)^{2} \end{align} 

Consequently, \(P[A \mid B]=P[A B] / P[B]=(0.2)^{2} /(0.2)=0.2\) .

To describe the results of the experiment we write a 101-tuple \(\left(z_{1}, z_{2}, \ldots, z_{101}\right)\) . The components \(z_{2}, \ldots, z_{101}\) are \(H\) or \(T\) , depending on whether the outcome of the respective toss is heads or tails. What are the possible values that may be assumed by the first component \(z_{1}\) ? We assume that there is a set of \(N\) numbers, \(p_{1}, p_{2}, \ldots, p_{N}\) , each between 0 and 1, such that any coin in the urn has as its probability of falling heads some one of the numbers \(p_{1}, p_{2}, \ldots, p_{\mathrm{N}}\) . Having selected a coin from the urn, we let \(z_{1}\) denote the probability that the coin will fall heads; consequently, \(z_{1}\) is one of the numbers \(p_{1}, \ldots, p_{X}\) . Now, for \(j=1,2, \ldots, N\) let \(C_{j}\) be the event that the coin selected has probability \(p_{j}\) of falling heads, and let \(B\) be the event that the coin selected yielded 55 heads in 100 tosses. Let \(j_{0}\) be the number, 1 to \(N\) , such that \(p_{j_{0}}=\frac{1}{2}\) . We are now seeking \(P\left[C_{j_{0}} \mid B\right]\) , the conditional probability that the coin selected is a fair coin, given that it yielded 55 heads in 100 tosses. In order to use (4.16) to evaluate \(P\left[C_{j_{0}} \mid B\right]\) , we require a knowledge of \(P\left[C_{j}\right]\) and \(P\left[B \mid C_{j}\right]\) for \(j=1, \ldots, N\) . By the binomial law,

\[P\left[B \mid C_{j}\right]=\left(\begin{array}{c} 100 \tag{4.19} \\ 55 \end{array}\right)\left(p_{j}\right)^{55}\left(1-p_{j}\right)^{45}\] 

The probabilities \(P\left[C_{j}\right]\) cannot be computed but must be assumed. The probability \(P\left[C_{j}\right]\) represents the proportion of coins in the urn which has probability \(p_{j}\) of falling heads. It is clear that the value we obtain for \(P\left[C_{j_{0}} \mid B\right]\) depends directly on the values we assume for \(P\left[C_{1}\right], \ldots, P\left[C_{v}\right]\) . If the latter probabilities are unknown to us, then we must resign ourselves to not being able to compute \(P\left[C_{j_{8}} \mid B\right]\) . However, let us obtain a numerical answer for \(P\left[C_{j_{0}} \mid B\right]\) under the assumption that \(P\left[C_{1}\right]=\cdots=P\left[C_{N}\right]=\) \(1 / N\) , so that a coin selected from the urn is equally likely to have any one of the probabilities \(p_{1}, \ldots, p_{N}\) . We then obtain that \[P\left[C_{j_{0}} \mid B\right]=\frac{(1 / N)\left(\begin{array}{c} 100 \tag{4.20} \\ 55 \end{array}\right)\left(p_{j_{0}}\right)^{55}\left(1-p_{j_{0}}\right)^{45}}{(1 / N) \sum_{j=1}^{N}\left(\begin{array}{c} 100 \\ 55 \end{array}\right)\left(p_{j}\right)^{55}\left(1-p_{j}\right)^{45}}\] 

Let us next assume that \(N=9\) , and \(p_{j}=j / 10\) for \(j=1,2, \ldots, 9\) . Then \(j_{0}=5\) , and \begin{align} P\left[C_{5} \mid B\right] & =\frac{\left(\begin{array}{c} 100 \\ 55 \end{array}\right)(1 / 2)^{100}}{\sum_{j=1}^{7}\left(\begin{array}{c} 100 \\ 55 \end{array}\right)(j / 10)^{55}[(10-j) / 10]^{45}} \tag{4.21} \\ & =\frac{0.048475}{0.097664}=0.496 . \end{align} 

The probability \(P\left[C_{5}\right]=\frac{1}{9}\) is called the prior (or a priori) probability of the event \(C_{5}\) ; the conditional probability \(P\left[C_{5} \mid B\right]=0.496\) is called the posterior (or a posteriori) probability of the event \(C_{5}\) . The prior probability is an unconditional probability that is known to us before any observations are taken. The posterior probability is a conditional probability that is of interest to us only if it is known that the conditioning event has occurred.

To describe the results of our observations, we write an \(\left(n+n^{\prime}+1\right)\) -tuple \(\left(z_{1}, z_{2}, \ldots, z_{n+n^{\prime}+1}\right)\) in which the components \(z_{2}, \ldots\) , \(z_{n+1}\) describe the outcomes of the coin tosses which have been made and the components \(z_{n+2}, \ldots, z_{n+n^{\prime}+1}\) describe the outcomes of the subsequent coin tosses. The first component \(z_{1}\) describes the probability that the coin tossed has of falling heads; we assume that there are \(N\) known numbers, \(p_{1}, p_{2}, \ldots, p_{N}\) , which \(z_{1}\) can take as its value. We have italicized this assumption to indicate that it is considered controversial. For \(j=1,2, \ldots\) , \(N\) let \(C_{j}\) be the event that the coin tossed has probability \(p_{j}\) of falling heads. Let \(B\) be the event that the coin yields \(n\) heads in its first \(n\) tosses, and let \(A\) be the event that it yields \(n^{\prime}\) heads in its subsequent \(n^{\prime}\) tosses. We are seeking \(P[A \mid B]\) . Now

whereas

\begin{align} P[A B] & =\sum_{j=1}^{N} P\left[A B \mid C_{j}\right] P\left[C_{j}\right] \tag{4.22} \\ & =\sum_{j=1}^{N}\left(p_{j}\right)^{n+n^{\prime}} P\left[C_{j}\right] \end{align} 

\[P[B]=\sum_{j=1}^{N}\left(p_{j}\right)^{n} P\left[C_{j}\right] \tag{4.23}\] 

Let us now assume that \(p_{j}\) is equal to \(j / N\) and that \(P\left[C_{j}\right]=1 / N\) . Then

\[P[A \mid B]=\frac{(1 / N) \sum_{j=1}^{N}(j / N)^{n+n^{\prime}}}{(1 / N) \sum_{j=1}^{N}(j / N)^{n}} \tag{4.24}\] 

The sums in (4.24) may be approximately evaluated in the case that \(N\) is large by means of the integral calculus. The sums can be regarded as approximating sums of Riemann integrals, and we have

\begin{align} \frac{1}{N} \sum_{j=1}^{N}\left(\frac{j}{N}\right)^{n+n^{\prime}} & \doteq \int_{0}^{1} x^{n+n^{\prime}} d x=\frac{1}{n+n^{\prime}+1} \\ \frac{1}{N} \sum_{j=1}^{N}\left(\frac{j}{N}\right)^{n} & \doteq \int_{0}^{1} x^{n} d x=\frac{1}{n+1} . \tag{4.25} \end{align} 

Consequently, given that the first \(n\) tosses yielded a head, the conditional probability that \(n^{\prime}\) subsequent tosses of the coin will yield a head, under the assumption that the probability of the coin falling heads is equally likely to be any one of the numbers \(1 / N, 2 / N, \ldots, N / N\) , and \(N\) is large, is given by

\[P[A \mid B]=\frac{n+1}{n+n^{\prime}+1} \tag{4.26}\] 

Equation (4.26) is known as Laplace’s general rule of succession. If we take \(n^{\prime}=1\) , then

\[P[A \mid B]=\frac{n+1}{n+2}. \tag{4.27}\] 

Equation (4.27) is known as Laplace’s special rule of succession.

Equation (4.27) has been interpreted by some writers on probability theory to imply that if a theory has been verified in \(n\) consecutive trials then the probability of its being verified on the \((n+1)\) st trial is \((n+1)\) / \((n+2)\) . That the rule has a certain appeal at first acquaintance may be seen from the following example:

Consider a tourist in a foreign city who scarcely understands the language. With trepidation, he selects a restaurant in which to eat. After ten meals taken there he has felt no ill effects. Consequently, he goes quite confidently to the restaurant the eleventh time in the knowledge that, according to the rule of succession, the probability is \(\frac{11}{12}\) that he will not be poisoned by his next meal.

However, it is easy to exhibit applications of the rule that lead to absurd answers. A boy is 10 years old today. The rule says that, having lived ten years, he has probability \(\frac{11}{12}\) of living one more year. On the other hand, his 80 -year-old grandfather has probability \(81 / 82\) of living one more year! Yet, in fact, the boy has a greater probability of living one more year.

Laplace gave the following often-quoted application of the special rule of succession. “Assume”, he says, “that history goes back 5000 years, that is, \(1,826,213\) days. The sun rose each day and so you can bet \(1,826,214\) against 1 that the sun will rise again tomorrow”. However, before believing this assertion, ask yourself if you would believe the following consequence of the general rule of succession; the sun having risen on each of the last \(1,826,213\) days, the probability that it will rise on each of the next \(1,826,214\) days is \(\frac{1}{2}\) , which means that the probability is \(\frac{1}{2}\) that on at least one of the next \(1,826,214\) days the sun will not rise.