91Ó°ÊÓ

In order to calculate the probability that \(A\) is a subset of \(B\), we can use the conditional probability given in the hint. We have: $$ P\{A \subset B\}=\sum_{i=0}^{n} P\{A \subset B \mid N(B)=i\} P\{N(B)=i\} $$ First, let's calculate \(P\{N(B)=i\}\). For any \(i\), there are \(\binom{n}{i}\) ways to form a subset with \(i\) elements. Since the total number of subsets is \(2^n\), the probability of picking a subset with \(i\) elements is: $$ P\{N(B)=i\} = \frac{\binom{n}{i}}{2^n} $$ Next, let's calculate \(P\{A \subset B \mid N(B)=i\}\). Given \(B\) has \(i\) elements, there are \(2^i\) possible subsets of \(B\). Moreover, in this case, \(A\) can be any of these \(2^i\) subsets since \(B\) already has \(i\) elements. Therefore, the probability that \(A\) is a subset of \(B\) given \(N(B)=i\) is: $$ P\{A \subset B \mid N(B)=i\} = \frac{2^i}{2^n} $$ Now we can use these probabilities to calculate the overall probability of \(A\) being a subset of \(B\): $$ P\{A \subset B\} = \sum_{i=0}^{n} P\{A \subset B \mid N(B)=i\} P\{N(B)=i\} = \sum_{i=0}^{n} \frac{2^i}{2^n} \cdot\frac{\binom{n}{i}}{2^n} $$ Notice that this sum looks like a binomial expansion with the base \(\frac{1}{2}\) and \(\frac{2}{2}\). Therefore, we can rewrite it as: $$ P\{A \subset B\} = \sum_{i=0}^{n} \binom{n}{i} \left(\frac{1}{2}\right)^i \left(\frac{2}{2}\right)^{n-i} = \left(\frac{3}{4}\right)^n $$ which proves part (a).

To calculate the probability of \(A\) and \(B\) having an empty intersection, we can use the following relation: $$ P\{A B=\varnothing\} = P\{A \subset \overline{B}\} $$ In this case, we want to find the probability that \(A\) is a subset of the complement of \(B\). So, we can follow the same steps as in part (a), but with \(\overline{B}\) instead of \(B\). Let's first calculate \(P\{N(\overline{B})=i\}\). Since \(\overline{B}\) has \(n-i\) elements, \(P\{N(\overline{B})=i\}\) is the same as \(P\{N(B)=n-i\}\): $$ P\{N(\overline{B})=i\} = \frac{\binom{n}{n-i}}{2^n} $$ Now let's calculate the probability \(P\{A \subset \overline{B} \mid N(\overline{B})=i\}\). Given that \(\overline{B}\) has \(i\) elements, \(A\) can be any of the \(2^i\) subsets of \(\overline{B}\), and the probability is: $$ P\{A \subset \overline{B} \mid N(\overline{B})=i\} = \frac{2^{i}}{2^n} $$ Now, we can use the conditional probability formula and calculate the overall probability: $$ P\{A B=\varnothing\} = \sum_{i=0}^{n} P\{A \subset \overline{B} \mid N(\overline{B})=i\} P\{N(\overline{B})=i\} = \sum_{i=0}^{n} \frac{2^i}{2^n} \cdot \frac{\binom{n}{n-i}}{2^n} $$ By substituting \(j=n-i\) and noticing that this sum also looks like a binomial expansion with the base \(\frac{1}{2}\) and \(\frac{2}{2}\), we can rewrite it as: $$ P\{A B=\varnothing\} = \sum_{j=0}^{n} \binom{n}{j} \left(\frac{1}{2}\right)^j \left(\frac{2}{2}\right)^{n-j} = \left(\frac{3}{4}\right)^n $$ which proves part (b).