De Morgan’s law

$$ \overline{\bigcup^{n}{j=1}}A_j = \bigcap^{n}{j=1}\overline{A_j}, \overline{\bigcap^{n}{j=1}}A_j = \bigcup^{n}{j=1}\overline{A_j} $$

Simplified ver.

$$ \overline{A\cup B} = \bar{A} \bar{B}, \overline{AB} = \bar{A}\cup \bar{B} $$

Commutative law

Associative law

$$ A\cup (B\cup C) = (A \cup B)\cup C, A\cap (B\cap C) = (A\cap B)\cap C $$

Distributive law

$$ A(B\cup C) = (AB)\cup (AC), (AB)\cup C = (A\cup C)(B\cup C) $$

Subtraction formula

$$ P(A- B) = P(A)-P(AB) $$

specially, if $B\subset A$, then $P(A- B) = P(A)-P(B)$

Addition formula

$$ P(A\cup B) = P(A) +P(B)-P(AB) $$

$$ P(A\cup B \cup C) = P(A)+P(B)+P(C)-P(AB) -P(AC)-P(BC)+P(ABC) $$

<aside> ✔️ e.g.1 $P(AB)=P(\bar{A}\bar{B}), P(A)=r, \text{then } P(B) = ?$

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<aside> ✔️

e.g.2 $P(A)=P(B)=P(C) =0.4,$ the probability that at least two of A, B, C happens is 0.3, and the probability that all them happens is 0.05. solve the probability that none of them happens.

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<aside> ✔️ e.g.3 On n cards are numbers 1~n, after random arrangement, solve the probability that at least one card has its order is the same of its number.

$A_i=$ {the $i$ th card has a number $i$}

$P(A)=P(A_1 \cup A_2 \cup ...\cup A_n) = \sum P(A_i) -\sum P(A_i A_j)+...+(-1)^{n-1}P(A_1 A_2...A_n)$

since each card is equivalent to each other, then the formula equals $C_n^1P(A_1)-C_n^2P(A_1A_2)+...+C_n^nP(A_1...A_n)$

in which $P(A_1)= \frac{1}{n}, P(A_1A_2)=\frac{1}{n(n-1)}, ... ,P(A_1...A_k) = \frac{(n-k)!}{n!}$

therefore, the formula equals $\sum(-1)^{k-1}C^k_n\frac{(n-k)!}{n!}= \sum(-1)^{k-1}\frac{n!}{(n-k)!k!}\frac{(n-k)!}{n!} = \sum \frac{(-1)^{k-1}}{k!}\overset{\text{Taylor's formula}}{=}\frac{1}{e}$

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Conditional probability

Multiplication formula