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2019 Paper 3 Q12
D: 1500.0 B: 1485.6
probability independent events combinatorics subsets counting power set intersection of sets inclusion

The set \(S\) is the set of all integers from 1 to \(n\). The set \(T\) is the set of all distinct subsets of \(S\), including the empty set \(\emptyset\) and \(S\) itself. Show that \(T\) contains exactly \(2^n\) sets. The sets \(A_1, A_2, \ldots, A_m\), which are not necessarily distinct, are chosen randomly and independently from \(T\), and for each \(k\) \((1 \leq k \leq m)\), the set \(A_k\) is equally likely to be any of the sets in \(T\).

  1. Write down the value of \(P(1 \in A_1)\).
  2. By considering each integer separately, show that \(P(A_1 \cap A_2 = \emptyset) = \left(\frac{3}{4}\right)^n\). Find \(P(A_1 \cap A_2 \cap A_3 = \emptyset)\) and \(P(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset)\).
  3. Find \(P(A_1 \subseteq A_2)\), \(P(A_1 \subseteq A_2 \subseteq A_3)\) and \(P(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m)\).

Solution: For every element in \(S\) we can choose whether or not it appears in a subset of \(S\), therefore there are \(2^n\) choices so \(2^n\) distinct subsets.

  1. \(\mathbb{P}(1 \in A_1) = \frac12\) (since \(1\) is in exactly half the subsets)
  2. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12\right) \\ &&&= \left (\frac34 \right)^n \end{align*}
  3. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 \cap A_3 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2 \cap A_3) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2 \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2))\mathbb{P}(i \in \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12 \cdot \frac12\right) \\ &&&= \left (\frac78 \right)^n \end{align*} Similarly, \(\displaystyle \mathbb{P}(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset) = \left ( \frac{2^m-1}{2^m} \right)^n\)
  4. \(\,\) \begin{align*} && \mathbb{P}(A_1 \subseteq A_2) &= \mathbb{P}(A_1 \cap A_2^c = \emptyset) \\ &&&= \left (\frac34 \right)^n \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq A_3) &= \prod_{i=1}^n \mathbb{P}(\text{once }i\text{ appears it keeps appearing}) \\ &&&= \prod_{i=1}^n \frac{\#\{(0,0,0), (0,0,1), (0,1,1), (1,1,1) \}}{2^3} \\ &&&= \prod_{i=1}^n \frac{4}{8} \\ &&&= \frac{1}{2^n} \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m) &= \prod_{i=1}^n \frac{m+1}{2^m} \\ &&&= \left ( \frac{m+1}{2^m} \right)^n \end{align*}

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