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Using Inclusion-Exclusion, prove that the probability of occurrence of exactly $m$ events out of $\{A_1,\ldots,A_n\} = \sum_{i=0}^{n-m} (-1)^i \binom{m+i}{m} P_{m+i} $ where $P_k = \sum_{1 \leq i_1<\cdots \le i_k \leq n} P(A_{i_1} \cap\cdots \cap A_{i_k})$

Attempt: I was trying to understand this term on the RHS : $\binom{m+i}{m} P_{m+i}$. And then the inclusion-exclusion principle shall follow.

This is what I thought to be able to use inclusion-exclusion but I am not very sure if this is the right approach.

$m$ events can occur either from the first $m$ events, or from the first $m+1$ events, or from the first $m+2$ events, $\cdots$, or from the n events.

$$E_k = \text{the $m$ events occur from the first $m+k$ events}$$

Then $P(E_k) = \binom{m+k}{m} \sum_{1\leq i_1<\cdots i_m\leq m+k} P(A_{i_1} \cap \cdots \cap A_{i_m})$

But, then I could not move forward from here. I tried understanding the solution given here, but I would really appreciate something more intuitive.

Could someone please explain how do I move forward from here?

Michael Hardy
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MathMan
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  • Do you know what is $r$? – user Mar 14 '21 at 14:19
  • @user sorry. typo mistake. edited it – MathMan Mar 14 '21 at 15:05
  • There is still another one. – user Mar 14 '21 at 15:07
  • @user done now .. – MathMan Mar 14 '21 at 15:16
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    The $\bigcap$ symbol is used in things like $\bigcap_i A_i$ or $\displaystyle\bigcap_i A_i$ and the $\cap$ symbol in things like $A\cap B$ or $A_1\cap \cdots\cap A_n.$ I edited accordingly. I also changed the first expression below to the second: $$ \begin{align} & E_k = \text{the m events occur from the first m+k events} \ {} \ & E_k = \text{the $m$ events occur from the first $m+k$ events} \end{align} $$ I.e. both the isolated $m$ and $m+k$ look different. The first expression is considered incorrect usage according to standard conventions. – Michael Hardy Mar 14 '21 at 18:45

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I discovered a marvelous proof. However, since it's long, I haven't been able to type it. Posting clear images for the sake of anyone who might be searching for the same answer.enter image description hereenter image description hereenter image description here

MathMan
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