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lecture_05
add p.57
[ (1 + x)^n = \sum\limits_{k = 0}^{n} { n \choose k} x^k ]
[ P( A \cup B) = P(A) + P(B) - P(A \cap B) = P(A) + P(B) - P(AB) ]
[ P(A \cup B \cup C) = P(A) + P(B \cup C) - P( A ( B \cup C)) \ = P(A) + P(B) + P(C) - P(BC) - P(AB \cup AC)\ = P(A) + P(B) + P(C) - P(BC) - P(AB) - P(AC) + P(ABC) ]
[ P \left( \bigcup\limits_{i = 1}^{n} A_i \right) = \sum\limits_{i = 1}^{n}{ P( A_i) - \sum\limits_{i \notequal j}{ P( A_i A_g)}
- \sum\limits_{i, j,k distinct}{ P( A_i A_j A_k)
- \sum{ P(A_i A_g A_k A_l)}
- \cdots + (-1)^{n-1} \sum{ P( A_1 A_2 \cdots A_n)}
- pf. Say each [ A_i] has only a finite # of elements. Say an element [ B] in [ m] eof the events [A_1, A_2, \dots, A_n ]. How often is it counted?
[ H = m - {m \choose 2} + {m \choose 3} - {m \choose 4} + \cdots + (-1)^{m+1}{m \choose m}]
[ 1 - H = 1 - m + {m \choose 2} - {m \choose 3}+ \cdots + (-1)^m{m \choose m}]
[ = \sum\limits_{k = 0}{ m}{m \choose k}(-1)^k = (1 + (-1))^m = 0^m = 0]
[ \sum\limits_{ k= 0}^n{ {n \choose k} k^2} = 2^{k-2} n(n + 1)]
- can be proven with calculus or with combintorial logic.
- Equation (1)
- [ {n + m \choose r} = {n \choose 0}{m \choose r} + {n \choose 1}{m \choose r - 1} + {n \choose 2}{m \choose r - 2} + \cdots + {n \choose r}{m \choose 0}]
- [ {n + m \choose r}]
- Equation (2)
- [ (1+ x)^{n + m} = \sum\limits_{k = 0}^{n + m}{ {n + m \choose k} x^k}]
- [ (1+ x)^{n + m} = (1 + x)^n(1 + x)^m = \sum{ {n \choose s} x^s \cdot \sum{ {m \choose t} x^t} = \sum\limits_{s = 0}^{n}{ \sum\limits_{t = 0}^{m}{ {n \choose s}{m \choose t}x^{s + t}}}]
- get
- [ \sum\limits_{ s + t = r}{ {n \choose s}{m \choose t}} = {n \choose 0}{m \choose r} + {n \choose 1}{m \choose r - 1} + \cdots + {n \choose r}{m \choose 0}]
[ n2^{n-1} = \sum\limits_{k = 1}^{n}{ {n \choose k} k} ]
What [ B ] probability that at least one person gets his own hat?
Hats: [ 1,2,3, \dots , n]
Persons: [ 1, 2, 3, \dots, n]
[ n =] number of persons
Let [ E_i ] be the vent that person [ i ] gets own hat. want [ P( E_1 \cup E_2 \cup \cdots \cup E_n) = \sum{ P(A_i)} - \sum{ P( A_i A_j) } \sum{ P( A_i A_j A_k) } = \sum\limits_{i=1}^{n}{ \frac{ (n-1)!}{ n!}} - \sum{ \frac{ (n-2}!}{ n!}} + \sum{ \frac{ (n-3)!}{ n!}}]