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機率的運算法則

$A \cup \overline{A} = S$ , $P(A)+P(\overline{A})=P(S)=1$ .
Law of Complement, $P(A)=1-P(\overline{A})$
Addition Law, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
假如 $A\cap B=\phi$ , 則 $P(A\cap B)=0$ , $P(A\cup B)=P(A)+P(B)$ .

例: Refer the previous monkeys example are selected from four by lottery. What is the probability that the selected monkeys are either of the same type or the same age?

$\begin{array}{rl} A=&[same type]=\{e_1,e_6\} \\ B=&[same age]=\{e_2,e_3,e_6\} \\ \end{array}$

so, $\displaystyle P(A)=\frac{2}{6}$ and $\displaystyle P(B)=\frac{3}{6}$ . And we are to calculate $P(A \cup B)$ . To employ the addition law, we also need to calculate $P(A \cup B)$ . And $A\cup B=\{e_6\}$ , so $\displaystyle P(A \cup B)=\frac{1}{6}$ . Therefore,

$\begin{array}{rcl} P(A \cup B)&=&P(A)+P(B)-P(A\cup B) \\ &=&\displaystyle\frac{2}{6}+\frac{3}{6}-\frac{1}{6}=\frac{2}{3} \\ \end{array}$

which is confirmed by the observation that $A\cup B =\{e_1,e_2,e_3,e_6\}$ indeed has four outcomes.