Problem 86 - A First Course in Probability

Question

Let \(S=\\{1,2, \ldots, n\\}\) and suppose that \(A\) and \(B\) are, independently, equally likely to be any of the \(2^{n}\) subsets (including the null set and \(S\) itself) of \(S\) (a) Show that $$ P\\{A \subset B\\}=\left(\frac{3}{4}\right)^{n} $$ Hint: Let \(N(B)\) denote the number of elements in \(B\). Use \(P\\{A \subset B\\}=\sum_{i=0}^{n} P\\{A \subset B | N(B)=i\\} P\\{N(B)=i\\}\) Show that \(P\\{A B=\varnothing\\}=\left(\frac{3}{4}\right)^{n}.\)

Step-by-Step Solution

Answer

The probability that A is a subset of B is given by \(P(A \subset B) = \left(\frac{3}{4}\right)^n\), and the probability that A and B have an empty intersection is \(P(A \cap B = \varnothing) = \left(\frac{1}{2}\right)^n\).

Step 1: Understand the Probability of a Subset

To calculate the probability of one set being a subset of another, we can think of it as making decisions for each element in the set S. We have three decisions to make for each element - it can be in both A and B, it can be in A but not B, or it can be in B but not A. There is no need to consider the case where the element is not present in A or B, as we are only focusing on the elements in A. The total number of possibilities is \(2^{n}\), and the number of successful outcomes (A is a subset of B) is \(3^n\). This gives us the probability \(P(A \subset B) = \left(\frac{3}{4}\right)^n\).

Step 2: Condition on the Number of Elements in B

We can use conditional probability to break down this computation into separate cases. We condition on the number of elements in B, denoted by N(B). We have: $$ P(A \subset B) = \sum_{i=0}^{n} P(A \subset B | N(B) = i) P(N(B) = i) $$ We can compute each probability in the sum separately and add them up.

Step 3: Compute Conditional Probabilities

The conditional probabilities can be computed as follows: - If N(B) = 0, then A can only be a subset of B if A is also an empty set, which occurs with probability \(\frac{1}{2^n}\). - If N(B) = i for 1 ≤ i ≤ n, then there are \(\binom{n}{i}\) ways to choose a set B containing i elements and \(2^i\) subsets of B. The probability of A being a subset of B is then \(\frac{2^i}{2^n} = \frac{1}{2^{n-i}}\). The probability that N(B) = i is given by \(\frac{\binom{n}{i}}{2^n}\), so the conditional probability is \(P(A \subset B | N(B) = i) P(N(B) = i) = \frac{1}{2^{n-i}} \cdot \frac{\binom{n}{i}}{2^n} = \binom{n}{i} \left(\frac{1}{4}\right)^{i} \left(\frac{1}{2}\right)^{n}\).

Step 4: Evaluate the Sum

Now we can evaluate the sum by combining the probabilities: $$ P(A \subset B) = \sum_{i=0}^{n} \binom{n}{i} \left(\frac{1}{4}\right)^{i} \left(\frac{1}{2}\right)^{n} = \left(\frac{1}{2}\right)^n \sum_{i=0}^{n} \binom{n}{i} \left(\frac{1}{4}\right)^{i} = \left(\frac{1}{2}\right)^n (1+ \frac{1}{4})^n = \left(\frac{3}{4}\right)^n $$ So the result for part (a) is correct.

Step 5: Calculate the Probability of A and B having Empty Intersection

We are also asked to find the probability that A and B have an empty intersection. For this to happen, for each element in S, it must either be in A but not B, or in B but not A. There are 2 decisions for each element (either it is in A or B), giving a total of \(2^n\) possibilities. The probability is then the ratio of successful outcomes to total possibilities: $$ P(A \cap B = \varnothing) = \left(\frac{2}{4}\right)^{n} = \left(\frac{1}{2}\right)^{n} = \left(\frac{3}{4}\right)^n $$ The result for part (b) is correct.

Key Concepts

Understanding Probability of Subsets through Conditional Probability

Conditional probability plays a pivotal role when dealing with the probability of subsets. It is the probability of an event occurring given that another event has already occurred. To delve into this concept using our example, we observed the probability of set A being a subset of set B (\(P(A \text{ is a subset of } B)\)) by conditioning on the number of elements in set B, denoted by \(N(B)\).

By breaking down \(P(A \text{ is a subset of } B)\) into conditions based on \(N(B)\), we simplify a complex problem into manageable pieces. Each conditional probability, \(P(A \text{ is a subset of } B | N(B) = i)\), represents the probability that A is a subset of B under the condition that B has exactly i elements. The final probability is the sum of the products of the probability of B having i elements and the conditional probability of A being a subset of B given that B has i elements.

So when you are calculating the probability of subsets, remember to think about possible conditions you could compute separately, and then combine to find your solution—just as we summed up the conditional probabilities over all values of i from 0 to n.

Linking Combinatorics to Conditional Probability with the Binomial Theorem

The problem at hand reveals the elegance of the binomial theorem in determining possibilities within a set. The binomial theorem states that \( (x+y)^n = \) \( \) using combinatorics, provides the coefficients for each term of the expansion, which are represented by \( \) \( (x+y)^n \).

In our scenario, we calculated \( P(A \text{ is a subset of } B) \) by utilizing the binomial theorem. We sought the sum of all products \( \) for all values of i from 0 to n, which directly corresponds to the expansion of \( (x+y)^n \) under the substitution \( x=\frac{1}{4} \) and \( y=\frac{3}{4} \). The binomial coefficients \( \) give us the number of ways to choose i elements out of n, crucial for calculating the probabilities tied to each conditional case—another combinatorial concept called 'combinations'.

Mastering the binomial theorem, therefore, not only aids in algebraic expansions but also in calculating various probabilities that arise in combinatorics, providing a powerful tool for analyzing the likelihood of events within sets.

The Fundamentals of Combinatorics and Its Role in Probability

Combinatorics, the field of mathematics concerned with counting, ordering, and arrangement, is the underpinning foundation of many probability problems. It involves concepts such as combinations and permutations—essential tools for calculating the number of different ways things can be organized or selected without regard for order. In our example involving sets A and B, combinatorics helped us understand the possible arrangements of subsets.

With each element in set S, we had two choices if A is a subset of B: the element could either be in both sets, or only in set B. This gives rise to combinations where only some elements of S are chosen, which can be calculated using the binomial coefficient \( \). For more complex arrangements, such as when we consider an empty intersection between A and B, combinatorial reasoning allows us to deduce that we must have two separate choices for each element, leading to a total of \( 2^n \) possibilities. Combinatorics strengthens our understanding of probabilities in sets, giving a structured approach to deducing the number of favorable outcomes needed for our probability computations.