To set up a mathematical model for the experiment described, assume that the balls in the urn are distinguishable; in particular, assume that they are numbered 1 to 6. Let the white balls bear numbers 1 to 4, and let the red balls be numbered 5 and 6.

Let us first consider that the balls are drawn without replacement. The sample description space \(S\) of the experiment is then given by (3.1) of Chapter 1; more compactly we write

\[S=\left\{\left(z_{1}, z_{2}\right): \text { for } i=1,2, z_{i}=1, \ldots, 6, \text { but } z_{1} \neq z_{2}\right\}. \tag{2.1}\] 

In words, one may read (2.1) as follows: \(S\) is the set of all 2-tuples \(\left(z_{1}, z_{2}\right)\) whose components are any numbers, 1 to 6, subject to the restriction that no two components of a 2-tuple are equal. The \(j\) th component \(z_{j}\) of a description represents the number of the ball drawn on the \(j\) th draw. Now let \(A\) be the event that both balls drawn are white, let \(B\) be the event that both balls drawn are red, and let \(C\) be the event that at least one of the balls drawn is white. The problem at hand can then be stated as one of finding (i) \(P[A]\) , (ii) \(P[A \cup B]\) , (iii) \(P[C]\) . It should be noted that \(C=B^{c}\) , so that \(P[C]=1-P[B]\) . Further, \(A\) and \(B\) are mutually exclusive, so that \(P[A \cup B]=P[A]+P[B]\) . Now \begin{align} A &= \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4), \\ &\quad\quad\quad (3,1),(3,2),(3,4),(4,1),(4,2),(4,3) \} \\ &= \{(z_1, z_2) \mid z_1, z_2 \in {1,2,3,4}, z_1 \neq z_2 \}, \tag{2.2} \end{align} whereas \(B=\{(5,6),(6,5)\}\) . Let us assume that all descriptions in \(S\) are equally likely. Then \[P[A]=\frac{N[A]}{N[S]}=\frac{4 \cdot 3}{6 \cdot 5}=0.4, \quad P[B]=\frac{2 \cdot 1}{6 \cdot 5}=0.066. \tag{2.3}\] 

The answers to the questions posed in example \(2 \mathrm{~A}\) are given, in the case of sampling without replacement, by (i) \(P[A]=0.4\) , (ii) \(P[A \cup B]=0.466\) , (iii) \(P[C]=0.933\) . These probabilities have been obtained under the assumption that the balls in the urn may be regarded as numbered (distinguishable) and that all descriptions in the sample description space \(S\) given in (2.1) are equally likely. In the case of sampling with replacement, a similar analysis may be carried out; one obtains the answers \[\begin{array}{cc} P[A]=\frac{4 \cdot 4}{6 \cdot 6}=0.444, & P[B]=\frac{2 \cdot 2}{6 \cdot 6}=0.11, \tag{2.4} \\[2mm] P[A \cup B]=0.555, & P[C]=0.888. \end{array}\] 

It is interesting to compare the values obtained by the foregoing model with values obtained by two other possible models. One might adopt as a sample description space \(S=\{(W, W),(W, R),(R, W),(R, R)\}\) . This space corresponds to recording the outcome of each draw as \(W\) or \(R\) , depending on whether the outcome of the draw is white or red. If one were to assume that all descriptions in \(S\) were equally likely, then \(P[A]=\frac{1}{4}\) , \(P[A \cup B]=\frac{1}{2}, P[C]=\frac{3}{4}\) . Note that the answers given by this model do not depend on whether the sampling is done with or without replacement. One arrives at a similar conclusion if one lets \(S=\{0,1,2\}\) , in which 0 signifies that no white balls were drawn, 1 signifies that exactly 1 white ball was drawn, and 2 signifies that exactly two white balls were drawn.

Under the assumption that all descriptions in \(S\) are equally likely, one would conclude that \(P[A]=\frac{1}{3}, P[A \cup B]=\frac{2}{3}, P[C]=\frac{2}{3}\) .

Let \(A\) denote the event that the first ball drawn is white, \(B\) denote the event that the second ball drawn is white, and \(C\) denote the event that both balls drawn are white. It should be noted that \(C=A B\) . Let the balls in the urn be numbered \(1\) to \(M\) , the white balls bearing numbers \(1\) to \(M_{W}\) , and the red balls bearing numbers \(M_{W}+1\) to \(M\) .

We consider first the case of sampling with replacement . The sample description space \(S\) of the experiment consists of ordered 2-tuples \(\left(z_{1}, z_{2}\right)\) , in which \(z_{1}\) is the number of the ball drawn on the first draw and \(z_{2}\) is the number of the ball drawn on the second draw. Clearly, \(N[S]=M^{2}\) . To compute \(N[A]\) , we use the fact that a description is in \(A\) if and only if its first component is a number 1 to \(M_{W}\) (meaning a white ball was drawn on the first draw) and its second component is a number 1 to \(M\) (due to the sampling with replacement the color of the ball drawn on the second draw is not affected by the fact that the first ball drawn was white). Thus there are \(M_{W}\) possibilities for the first component, and for each of these \(M\) possibilities for the second component of a description in \(A\) . Consequently, by (1.1) , the size of \(A\) is \(M_{W} M\) . Similarly, \(N[B]=M M_{W}\) , since there are \(M\) possibilities for the first component and \(M_{W}\) possibilities for the second component of a description in \(B\) . The reader may verify by a similar argument that the event \(A B\) , (a white ball is drawn on both draws), has size \(N[A B]=M_{W} M_{W}\) . Thus in the case of sampling with replacement one obtains the result, if all descriptions are equally likely, that \[P[A]=P[B]=\frac{M_{W}}{M}, \quad P[A B]=\left(\frac{M_{W}}{M}\right)^{2}. \tag{2.5}\] 

We next consider the case of sampling without replacement . The sample description space of the experiment again consists of ordered 2-tuples \(\left(z_{1}, z_{2}\right)\) , in which \(z_{j}\) (for \(j=1,2\) ) denotes the number of the ball drawn on the \(j\) th draw. As in the case of sampling with replacement, each \(z_{j}\) is a number 1 to \(M\) . However, in sampling without replacement a description \(\left(z_{1}, z_{2}\right)\) must satisfy the requirement that its components are not the same. Clearly, \(N[S]=(M) _{2}=M(M-1)\) . Next, \(N[A]=M_{W}(M-1)\) , since there are \(M_{W}\) possibilities for the first component of a description in \(A\) and \(M-1\) possibilities for the second component of a description in \(A\) ; the urn from which the second ball is drawn contains only \((M-1)\) balls. To compute \(N[B]\) , we first concentrate our attention on the second component of a description in \(B\) . Since \(B\) is the event that the ball drawn on the second draw is white, there are \(M_{W}\) possibilities for the second component of a description in \(B\) . To each of these possibilities, there are only \(M-1\) possibilities for the first component, since the ball which is to be drawn on the second draw is known to us and cannot be drawn on the first draw. Thus \(N[B]=(M-1) M_{W}\) by (1.1) . The reader may verify that the event \(A B\) has size \(N[A B]=M_{W}\left(M_{W}-1\right)\) . Consequently, in sampling without replacement one obtains the result, if all descriptions are equally likely, that \[P[A]=P[B]=\frac{M_{W}}{M}, \quad P[A B]=\frac{M_{W}\left(M_{W}-1\right)}{M(M-1)}. \tag{2.6}\] 

Another way of computing \(P[B]\) , which the reader may find more convincing on first acquaintance with the theory of probability, is as follows. Let \(B_{1}\) denote the event that the first ball drawn is white and the second ball drawn is white. Let \(B_{2}\) denote the event that the first ball drawn is red and the second ball drawn is white. Clearly, \(N\left[B_{1}\right]=\) \(M_{W}\left(M_{W}-1\right), N\left[B_{2}\right]=\left(M-M_{W}\right) M_{W}\) . Since \(P[B]=P\left[B_{1}\right]+P\left[B_{2}\right]\) , we have \[P[B]=\frac{M_{W}\left(M_{W}-1\right)}{M(M-1)}+\frac{\left(M-M_{W}\right) M_{W}}{M(M-1)}=\frac{M_{W}}{M}.\] 

To illustrate the use of (2.5) and (2.6) , let us consider an urn containing \(M=6\) balls, of which \(M_{W}=4\) are white. Then \(P[A]=P[B]=\frac{2}{3}\) and \(P[A B]=\frac{4}{9}\) in sampling with replacement, whereas \(P[A]=P[B]=\frac{2}{3}\) and \(P[A B]=\frac{2}{5}\) in sampling without replacement.

The reader may find (2.6) startling. It is natural, in the case of sampling with replacement, in which \(P[A]=P[B]\) , that the probability of drawing a white ball is the same on the second draw as it is on the first draw, since the composition of the urn is the same in both draws. However, it seems very unnatural, if not unbelievable, that in sampling without replacement \(P[A]=P[B]\) . The following remarks may clarify the meaning of (2.6) .

Suppose that one desired to regard the event that a white ball is drawn on the second draw as an event defined on the sample description space, denoted by \(S^{\prime}\) , which consists of all possible outcomes of the second draw. To begin with, one might write \(S^{\prime}=\{1,2, \ldots, M\}\) . However, how is a probability function to be defined on the subsets of \(S^{\prime}\) in the case in which the sample is drawn without replacement. If one knows nothing about the outcome of the first draw, perhaps one might regard all descriptions in \(S^{\prime}\) as being equally likely; then, \(P[B]=M_{W} / M\) . However, suppose one knows that a white ball was drawn on the first draw. Then the descriptions in \(S^{\prime}\) are no longer equally likely; rather, it seems plausible to assign probability 0 to the description corresponding to the (white) ball, which is not available on the second draw, and assume the remaining descriptions to be equally likely. One then computes that the probability of the event \(B\) (that a white ball will be drawn on the second draw), given that the event \(A\) (that a white ball was drawn on the first draw) has occurred, is equal to \(\left(M_{W}-1\right) /(M-1)\) . Thus \(\left(M_{W}-1\right) /(M-1)\) represents a conditional probability of the event \(B\) (and, in particular, the conditional probability of \(B\) , given that the event \(A\) has occurred), whereas \(M_{W} / M\) represents the unconditional probability of the event \(B\) . The distinction between unconditional and conditional probability is made precise in section 4 .

Let us call the first card looked at by the experimenter card 1; similarly, for \(k=2,3, \ldots, 8\) , let the \(k\) th card looked at by the experimenter be called card \(k\) . To describe the subject’s response during the course of the experiment, we write the subset \(\left\{z_{1}, z_{2}, z_{3}, z_{4}\right\}\) of the numbers \(\{1,2,3,4,5,6,7,8\}\) , which consists of the numbers of all the cards the subject said were red. The sample description space \(S\) then consists of all subsets of size 4 of the set \(\{1,2,3,4,5,6,7,8\}\) . Therefore, \(N[S]=\left(\begin{array}{l}8 \\ 4\end{array}\right)\) . The event \(A\) that the subject made exactly six correct guesses may be represented as the set of those subsets \(\left\{z_{1}, z_{2}, z_{3}, z_{4}\right\}\) , exactly three of whose members are equal to the numbers of cards that were, in fact, red. To compute the size of \(A\) , we notice that the three numbers in a description in \(A\) , corresponding to a correct guess, may be chosen in \(\left(\begin{array}{l}4 \\ 3\end{array}\right)\) ways, whereas the one number in a description in \(A\) , corresponding to an incorrect guess, may be chosen in \(\left(\begin{array}{l}4 \\ 1\end{array}\right)\) ways. Consequently, \(N[A]=\left(\begin{array}{l}4 \\ 3\end{array}\right)\left(\begin{array}{l}4 \\ 1\end{array}\right)\) , and \[P[A]=\frac{\left(\begin{array}{l} 4 \\ 3 \end{array}\right)\left(\begin{array}{l} 4 \\ 1 \end{array}\right)}{\left(\begin{array}{l} 8 \\ 4 \end{array}\right)}=\frac{8}{35}.\]