设试验E的样本空间为 S S S, A A A为 E E E的事件
{ B i ∣ i ∈ I } \set{B_i|i\in I} {Bi∣i∈I}是一个 S S S的划分, P ( B i ) > 0 , i ∈ I P(B_i)>0,i\in{I} P(Bi)>0,i∈I,则 P ( A ) = ∑ i = 1 n P ( A ∣ B i ) P ( B i ) P(A)=\sum_{i=1}^{n}P(A|B_i)P(B_i) P(A)=∑i=1nP(A∣Bi)P(Bi)
证明:
显然 A B i ⊂ B i AB_i\sub B_i ABi⊂Bi,又 B i B j = ∅ B_iB_j=\varnothing BiBj=∅,所以 ( A B i ) ( A B j ) = A ( B i B j ) = A ∅ = ∅ (AB_i)(AB_j)=A(B_iB_j)=A\emptyset=\emptyset (ABi)(ABj)=A(BiBj)=A∅=∅, ( i ≠ j ) (i\neq{j}) (i=j)
P ( A ∣ B i ) P ( B i ) = P ( A B i ) P(A|B_i)P(B_i)=P(AB_i) P(A∣Bi)P(Bi)=P(ABi)
证法1:
证法2:
某个含有20个球的箱子
含有0,1,2只次品的概率分别为0.8,0.1,0.1
记 B B B={抽中的4件产品都是正品}
那么发生事件 B B B的概率?
显然 A 0 , A 1 , A 2 A_0,A_1,A_2 A0,A1,A2构成试验E{观察箱子中的全部球的正品数,样本空间为{0,1,2}}的样本空间的一个划分
而事件 B B B进行的试验 F F F{观察取出的4个抽样品的正品数,样本空间为{0,1,2}}
容易根据古典概型公式计算(因为此时的样本空间已知)以下条件概率(这里不是用条件概率公式展开计算,这会绕回来)
P ( B ∣ A 0 ) = ( 20 4 ) ( 20 4 ) = 1 P(B|A_0)=\frac{\binom{20}{4}}{\binom{20}{4}}=1 P(B∣A0)=(420)(420)=1
P ( B ∣ A 1 ) = ( 19 4 ) ( 20 4 ) = ( 19 ∗ 18 ∗ 17 ∗ 16 ) ∗ ( 4 ∗ 3 ∗ 2 ∗ 1 ) ( 20 ∗ 19 ∗ 18 ∗ 17 ) ∗ ( 4 ∗ 3 ∗ 2 ∗ 1 ) = 4 5 P(B|A_1)=\frac{\binom{19}{4}}{\binom{20}{4}}=\frac{(19*18*17*16)*(4*3*2*1)}{(20*19*18*17)*(4*3*2*1)}=\frac{4}{5} P(B∣A1)=(420)(419)=(20∗19∗18∗17)∗(4∗3∗2∗1)(19∗18∗17∗16)∗(4∗3∗2∗1)=54
P ( B ∣ A 2 ) = ( 18 4 ) ( 20 4 ) = 12 19 P(B|A_2)=\frac{\binom{18}{4}}{\binom{20}{4}}=\frac{12}{19} P(B∣A2)=(420)(418)=1912
根据全概率公式 P ( B ) = ∑ i = 1 3 P ( B ∣ A i ) P ( A i ) = 0.943 P(B)=\sum\limits_{i=1}^{3}P(B|A_i)P(A_i)=0.943 P(B)=i=1∑3P(B∣Ai)P(Ai)=0.943
Note:这里试验 F F F的样本空间恰好和E的样本空间重复,但如果箱子中的次品高达4个以上,那么 F F F的样本空间 0 , 1 , 2 , 3 , 4 {0,1,2,3,4} 0,1,2,3,4是包含于 E E E的样本空间的
次品来源问题:设一批零件来自三个供应商
供应商 | 次品率 | 进货份额 |
---|---|---|
1 | 0.02 | 0.15 |
2 | 0.01 | 0.80 |
3 | 0.03 | 0.05 |
试验内容:从零件中抽取一件
求该样品是次品的概率:
从中取出一件,发现是次品,那么来自产商 i i i的概率是多少 ( i = 1 , 2 , 3 ) (i=1,2,3) (i=1,2,3)
机器与产品合格率问题
设机器正常时,生产的产品合格率为0.9,否则合格率为0.3
如果机器开机后,正常的概率是0.75(先验概率)
某天该机器第一件产品是合格的,机器正常的概率是多少?
分析:
根据假设可知:
后验概率是对先验概率的一种修正
后验概率和先验概率的解释分为两派
Chain rule (probability) - Wikipedia
In probability theory, the chain rule (also called the general product rule[1][2]) permits the calculation of any member of the joint distribution of a set of random variables using only conditional probabilities.
The rule is useful in the study of Bayesian networks, which describe a probability distribution in terms of conditional probabilities.
更一般的,如果反复套用上述公式,我们可以得到:
P ( ∏ i = 1 n A i ) = P ( ( ∏ i = 1 n − 1 A i ) A n ) = P ( A n ∣ ∏ i = 1 n − 1 A i ) P ( ∏ i = 1 n − 1 A i ) 设通项 P k = P ( ∏ i = 1 k A i ) = P ( ( ∏ i = 1 k − 1 A i ) A k ) = P ( A k ∣ ∏ i = 1 k − 1 A i ) P ( ∏ i = 1 k − 1 A i ) T ( k ) = ∏ i = 1 k A i k = n , n − 1 , n − 2 , ⋯ , 1 P k = P ( T ( k ) ) = P ( A k ∣ T ( k − 1 ) ) P ( T ( k − 1 ) ) ⋮ P(\prod_{i=1}^{n}A_i)=P((\prod_{i=1}^{n-1}A_i)A_n) =P(A_n|\prod_{i=1}^{n-1}A_i)P(\prod_{i=1}^{n-1}A_i) \\ 设通项P_k= P(\prod_{i=1}^{k}A_i)=P((\prod_{i=1}^{k-1}A_i)A_k) =P(A_k|\prod_{i=1}^{k-1}A_i)P(\prod_{i=1}^{k-1}A_i) \\T(k)=\prod_{i=1}^{k}A_i \\k=n,n-1,n-2,\cdots,1 \\P_k=P(T(k))=P(A_k|T(k-1))P(T(k-1)) \\\vdots P(i=1∏nAi)=P((i=1∏n−1Ai)An)=P(An∣i=1∏n−1Ai)P(i=1∏n−1Ai)设通项Pk=P(i=1∏kAi)=P((i=1∏k−1Ai)Ak)=P(Ak∣i=1∏k−1Ai)P(i=1∏k−1Ai)T(k)=i=1∏kAik=n,n−1,n−2,⋯,1Pk=P(T(k))=P(Ak∣T(k−1))P(T(k−1))⋮
特别的 : P 2 = P ( A 1 A 2 ) = P ( A 2 ∣ A 1 ) P ( A 1 ) P 3 = P ( A 1 A 2 A 3 ) = P ( A 3 ∣ A 1 A 2 ) P ( A 1 A 2 ) = P ( A 3 ∣ A 1 A 2 ) P ( A 2 ∣ A 1 ) P ( A 1 ) 类似的 : P 4 = P ( A 1 A 2 A 3 A 4 ) = P ( A 4 ∣ A 1 A 2 A 3 ) P ( A 1 A 2 A 3 ) = P ( A 4 ∣ A 1 A 2 A 3 ) P ( A 3 ∣ A 1 A 2 ) P ( A 2 ∣ A 1 ) P ( A 1 ) P n = ∏ i = 1 n P ( A n − i + 1 ∣ ∏ j = 0 n − i A j ) 严格的说 , ∏ j = 1 n A j 应该作 ⋂ j = 1 n A j , 表示积事件 定义 P ( A 0 ) = 1 特别的:P_2=P(A_1A_2)=P(A_2|A_1)P(A_1) \\P_3=P(A_1A_2A_3)=P(A_3|A_1A_2)P(A_1A_2) \\=P(A_3|A_1A_2)P(A_2|A_1)P(A_1) \\类似的: \\P_4=P(A_1A_2A_3A_4)=P(A_4|A_1A_2A_3)P(A_1A_2A_3) \\=P(A_4|A_1A_2A_3)P(A_3|A_1A_2)P(A_2|A_1)P(A_1) \\ P_n=\prod_{i=1}^{n}P(A_{n-i+1}|\prod_{j=0}^{n-i}A_j) \\严格的说,\prod_{j=1}^{n}A_j应该作\bigcap\limits_{j=1}^{n}A_j,表示积事件 \\定义P(A_0)=1 特别的:P2=P(A1A2)=P(A2∣A1)P(A1)P3=P(A1A2A3)=P(A3∣A1A2)P(A1A2)=P(A3∣A1A2)P(A2∣A1)P(A1)类似的:P4=P(A1A2A3A4)=P(A4∣A1A2A3)P(A1A2A3)=P(A4∣A1A2A3)P(A3∣A1A2)P(A2∣A1)P(A1)Pn=i=1∏nP(An−i+1∣j=0∏n−iAj)严格的说,j=1∏nAj应该作j=1⋂nAj,表示积事件定义P(A0)=1
其他写法
P n = ∏ i = 1 n P ( A n − i + 1 ∣ ⋂ j = 0 n − i A j ) 根据乘法交换律 ( 积事件调整书写顺序含义不变 ) P n = ∏ i = 1 n P ( A n − i + 1 ∣ ⋂ j = 0 n − i A j ) = ∏ i = 1 n P ( A i ∣ ⋂ j = 1 i − 1 A j ) 约定 ⋂ j = 1 0 A j 时省略该项 ( 作为必然事件 ) P_n=\prod_{i=1}^{n}P(A_{n-i+1}|\bigcap\limits_{j=0}^{n-i}A_j) \\根据乘法交换律(积事件调整书写顺序含义不变)\\ P_n=\prod_{i=1}^{n}P(A_{n-i+1}|\bigcap\limits_{j=0}^{n-i}A_j) =\prod_{i=1}^{n}P(A_{i}|\bigcap\limits_{j=1}^{i-1}A_j) \\约定\bigcap\limits_{j=1}^{0}A_j时省略该项(作为必然事件) Pn=i=1∏nP(An−i+1∣j=0⋂n−iAj)根据乘法交换律(积事件调整书写顺序含义不变)Pn=i=1∏nP(An−i+1∣j=0⋂n−iAj)=i=1∏nP(Ai∣j=1⋂i−1Aj)约定j=1⋂0Aj时省略该项(作为必然事件)
通常,公式右边的条件概率都是比较容易计算的
多次摸多颜色球问题
设有5红,3黑,2白
问,第三次才摸到白球的概率
即,前两次的摸球结果都不是白色的
为了方便讨论问题,记: A i = 第 i 次摸出白球 ; i = 1 , 2 , 3 A_i={第i次摸出白球};i=1,2,3 Ai=第i次摸出白球;i=1,2,3
P = P ( A 1 ‾ A 2 ‾ A 3 ) = P ( A 3 ∣ A 1 ‾ A 2 ‾ ) P ( A 2 ‾ ∣ A 1 ‾ ) P ( A 1 ‾ ) = 2 10 − 2 8 − 1 10 − 1 8 10 = 2 8 7 9 8 10 = 7 45 P=P(\overline{A_1}\ \overline{A_2}A_3) =P(A_3|\overline{A_1}\ \overline{A_2}) P(\overline{A_2}|\overline{A_1})P(\overline{A_1}) \\=\frac{2}{10-2}\frac{8-1}{10-1}\frac{8}{10} =\frac{2}{8}\frac{7}{9}\frac{8}{10} =\frac{7}{45} P=P(A1 A2A3)=P(A3∣A1 A2)P(A2∣A1)P(A1)=10−2210−18−1108=8297108=457
其中 P ( B n ∣ B n − 1 ⋯ B 1 ) P(B_n|B_{n-1}\cdots{B_1}) P(Bn∣Bn−1⋯B1)表示已经有 n − 1 n-1 n−1个求被摸出,现在再摸出一个球,发生事件 B n B_n Bn的概率
例如 P ( A 2 ‾ ∣ A 1 ‾ ) P(\overline{A_2}|\overline{A_1}) P(A2∣A1)表示已经摸出一个球(而且不是白球)的情况下,再摸出一个求,而且仍然不是白球的概率
实时上,稍微熟练点的高中生,就可以直接写出 p = 2 8 7 9 8 10 p=\frac{2}{8}\frac{7}{9}\frac{8}{10} p=8297108
Consider an indexed collection of random variables X 1 , … , X n {\displaystyle X_{1},\ldots ,X_{n}} X1,…,Xn taking possible values x 1 , … , x n x_{1},\dots ,x_{n} x1,…,xn respectively.
Then, to find the value of this member of the joint distribution, we can apply the definition of conditional probability to obtain:
P ( X n = x n , ⋯ , X 1 = x 1 ) = P ( X n = x n ∣ X n − 1 = x n − 1 , … , X 1 = x 1 ) ⋅ P ( X n − 1 = x n − 1 , … , X 1 = x 1 ) {\displaystyle \mathrm {P} \left(X_{n}=x_{n},\cdots ,X_{1}=x_{1}\right)=\mathrm {P} \left(X_{n}=x_{n}|X_{n-1}=x_{n-1},\ldots ,X_{1}=x_{1}\right)\cdot \mathrm {P} \left(X_{n-1}=x_{n-1},\ldots ,X_{1}=x_{1}\right)} P(Xn=xn,⋯,X1=x1)=P(Xn=xn∣Xn−1=xn−1,…,X1=x1)⋅P(Xn−1=xn−1,…,X1=x1)
Repeating this process with each final term and letting A k A_{k} Ak denote the event X k = x k {\displaystyle X_{k}=x_{k}} Xk=xk creates the product:
P ( ⋂ k = 1 n A k ) = ∏ k = 1 n P ( A k ∣ ⋂ j = 1 k − 1 A j ) = ∏ k = 1 n P ( X k = x k ∣ X 1 = x 1 , … X k − 1 = x k − 1 ) . {\displaystyle \mathrm {P} \left(\bigcap _{k=1}^{n}A_{k}\right)=\prod _{k=1}^{n}\mathrm {P} \left(A_{k}\,{\Bigg |}\,\bigcap _{j=1}^{k-1}A_{j}\right)=\prod _{k=1}^{n}\mathrm {P} \left(X_{k}=x_{k}\,|\,X_{1}=x_{1},\dots X_{k-1}=x_{k-1}\right).} P(k=1⋂nAk)=k=1∏nP(Ak j=1⋂k−1Aj)=k=1∏nP(Xk=xk∣X1=x1,…Xk−1=xk−1).