give a rigorous mathematical proof of:
\[A \cup (B \cap A^c)=A \cup B\]
proof:
Note that
\[A \cup (B \cap A^c)= (A \cup B) \cap (A\cup A^c)= (A \cup B) \cap S= A \cup B\] would not be considered a mathematically rigourous proof. Instead
assume \(x \in A\cup B\), then either \(x \in A\) or \(x \in B\). If \(x \in A\), then also \(x \in A \cup (B \cap A^c)\). If \(x \not\in A\), then \(x \in A^c\) and \(x \in B\), so \(x \in B\cap A^c\), and so again \(x \in A \cup (B \cap A^c)\). Therefore \[A \cup B \subset A \cup (B \cap A^c)\]
Now assume \(x \in A \cup (B \cap A^c)\), then either \(x \in A\), and therefore \(x \in A\cup B\), or \(x \not\in A\), but then \(x \in B\cap A^c\), so \(x \in B\) and so \(x \in A\cup B\). Therefore
\[A \cup (B \cap A^c) \subset A \cup B\]
and the theorem is proven.
Give a mathematically correct proof of Boole’s inequality
We use mathematical induction.
because \(P(A\cap B) \ge0\) always.
\[P(\bigcup_{i=1}^k A_i) \le \sum_{i=1}^{k} P(A_i)\] Then
\[ \begin{aligned} &P(\bigcup_{i=1}^n A_i) = \\ &P(\bigcup_{i=1}^{n-1} A_i \cup A_n) \le \\ &P(\bigcup_{i=1}^{n-1} A_i) + P(A_n) = \text{ (Induction hypothesis with k=2<n)} \le \\ &\sum_{i=1}^{n-1} P(A_i) + P(A_n) = \text{ (Induction hypothesis with k=n-1<n)}\\ &\sum_{i=1}^{n} P(A_i) \end{aligned} \]
A box contains 20 blue balls and 10 green balls. Define appropriate sample spaces and events and use them to find
Let’s define the sample space \(S=\{b1,..,b20, g1,..,g10\}\) and the event \(A=\{b1,..,b20\}\). Now each of these 35 balls is equally likely to be chosen, so
\[P(\text{select a blue ball}) = P(A) = \frac{n(A)}{n(S)} =\frac{20}{30}\]
Now the sample space is \(S=\{(b1,b2),..,(g9,g10)\}\) and \(A=\{(b1,g1),..,(b20,g10)\}\), and again all of the outcomes are equally likely. Now \(n(S)=30*29\) and \(n(A)=2*20*10\). Here the “2” is because both (b1,g1) and (g1,b1) result in a blue and a green ball. And so
\[P(A)=\frac{2*20*10}{30*29}=\frac{40}{87}=0.4598\]