The inclusion-exclusion formula is included in the list of Problem Solving Strategies.
I showed you in Example 4.5 a use of Erdos's probabilistic method in combinatorics. If you liked that, you might like to read about other examples, for instance in this link.
I made a small slip at the end of the lecture.
$$
1 - \frac{1}{2!}+\frac{1}{3!}+\cdots =1-e^{-1}.
$$ So the probability of derangement tends to $e^{-1}=0.368$.
In Example 4.4 we considered an infinite sequence of tosses of biased coins and saw that if $\sum_{i=1}^\infty p_i<\infty$ then $P(\text{infinite number of heads})=0.$
Now let's prove that $\sum_{i=1}^\infty p_i=\infty \implies P(\text{finite number of heads})=0.$ This is trickier.
We are assuming the coin tosses are independent (which is terminology from Lecture 5, but you can guess intuitively what this means). Let $A_k$ be the event that the $k$th toss is a head. Let $E_n=\bigcap_{k=n}^\infty A_k^c=[\text{no heads after the $(n-1)$th toss}]$. Then for $m > n$
$P(E_n)\leq P(\bigcap_{k=n}^m A_k^c)=\prod_{k=n}^m P(A_k^c) =\prod_{k=n}^m (1-p_k)\leq \prod_{k=n}^m e^{-p_k} =e^{-\sum_{k=n}^m p_k}$.
The right hand side $\to 0$ as $m\to\infty$.
Hence $P(E_n)=0$, and then using Property (v) of $P$ (continuity), we have that
$P(\text{finite number of heads})=P(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k^c)=\lim_{n\to\infty}P(E_n)=0.$
These two results are examples of applications of the Borel-Cantelli lemmas, which appear in the Part II course Probability and Measure.
I did not talk through Example 4.8 (counting surjective functions), but it is another nice use of the inclusion-exclusion formula. See also Examples Sheet 1, #9.
I showed you in Example 4.5 a use of Erdos's probabilistic method in combinatorics. If you liked that, you might like to read about other examples, for instance in this link.
I made a small slip at the end of the lecture.
$$
1 - \frac{1}{2!}+\frac{1}{3!}+\cdots =1-e^{-1}.
$$ So the probability of derangement tends to $e^{-1}=0.368$.
In Example 4.4 we considered an infinite sequence of tosses of biased coins and saw that if $\sum_{i=1}^\infty p_i<\infty$ then $P(\text{infinite number of heads})=0.$
Now let's prove that $\sum_{i=1}^\infty p_i=\infty \implies P(\text{finite number of heads})=0.$ This is trickier.
We are assuming the coin tosses are independent (which is terminology from Lecture 5, but you can guess intuitively what this means). Let $A_k$ be the event that the $k$th toss is a head. Let $E_n=\bigcap_{k=n}^\infty A_k^c=[\text{no heads after the $(n-1)$th toss}]$. Then for $m > n$
$P(E_n)\leq P(\bigcap_{k=n}^m A_k^c)=\prod_{k=n}^m P(A_k^c) =\prod_{k=n}^m (1-p_k)\leq \prod_{k=n}^m e^{-p_k} =e^{-\sum_{k=n}^m p_k}$.
The right hand side $\to 0$ as $m\to\infty$.
Hence $P(E_n)=0$, and then using Property (v) of $P$ (continuity), we have that
$P(\text{finite number of heads})=P(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k^c)=\lim_{n\to\infty}P(E_n)=0.$
These two results are examples of applications of the Borel-Cantelli lemmas, which appear in the Part II course Probability and Measure.
I did not talk through Example 4.8 (counting surjective functions), but it is another nice use of the inclusion-exclusion formula. See also Examples Sheet 1, #9.
