Trying to prove this using induction hypothesis

Question

I want to solve this using Induction hypothesis and not the inclusion-exclusion identity. I have an idea but finding hard time to approach the answers for each part.

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I started by assuming that P(Ui=1 to n Ai)= P1 - P2 + P3 .....+- Pn; and proved it for n+1. where, P1= sum of probabilities of Ai P2 = sum of probabilities for Ai intersection Aj .. and so on.

But how to follow to get the answers for each part.

Thanks for your help.

Answer 1

Since the sets $E_k$ are changing for different $n$, let $E_{k,h}$ be the set of all sample points that are contained in exactly $h$ of the events $A_1, A_2, \ldots, A_k$. Notice that $E_{k,h} = \emptyset$ if $k<h$.

The statement to prove can actually be written the following form, and the proof from the following right hand side to the above right hand side holds, as every pair of $E_{n,i}$ and $E_{n,j}$ ($i\ne j$) are disjoint.

$$\bigcup^n_{i=1}A_i = \bigcup^n_{i=1}E_{n,i}$$

Consider base case $n=1$. Trivially $A_1=E_{1,1}$.

Assume the statement is true for case $n=k$ for some $k\in\mathbb N$.

Then for case $n=k+1$, $$\begin{align} \bigcup^{k+1}_{i=1}A_i =& \bigcup^{k}_{i=1}A_i\cup A_{k+1}\\ =& \left(\bigcup^{k}_{i=1}A_i\cap \overline{A_{k+1}}\right) \cup \left(\bigcup^{k}_{i=1}A_i\cap A_{k+1}\right) \cup \left(\overline{\bigcup^{k}_{i=1}A_i}\cap A_{k+1}\right)\\ =& \left(\bigcup^{k}_{i=1}E_{k,i}\cap \overline{A_{k+1}}\right) \cup \left(\bigcup^{k}_{i=1}E_{k,i}\cap A_{k+1}\right) \cup \left(E_{k,0}\cap A_{k+1}\right)\\\ =& \left(\bigcup^{k}_{i=1}E_{k,i}\cap \overline{A_{k+1}}\right) \cup \left(\bigcup^{k}_{i=0}E_{k,i}\cap A_{k+1}\right)\\ =& \left(\bigcup^{k+1}_{i=1}E_{k,i}\cap \overline{A_{k+1}}\right) \cup \left(\bigcup^{k+1}_{i=1}E_{k,i-1}\cap A_{k+1}\right)\\ =& \bigcup^{k+1}_{i=1}\left[\left(E_{k,i}\cap \overline{A_{k+1}}\right) \cup \left(E_{k,i-1}\cap A_{k+1}\right)\right]\\ =& \bigcup^{k+1}_{i=1}E_{k+1,i}\\ \end{align}$$ The last step holds because if a sample point is in $E_{k+1,i}$, it is either in exactly $i$ of $A_1, A_2, \ldots, A_k$ and not in $A_{k+1}$, or it is in exactly $i-1$ of $A_1, A_2, \ldots, A_k$ and in $A_{k+1}$. The converse is also true.