贝叶斯定理或贝叶斯定律以托马斯·贝叶斯牧师的名字命名。它基于可能与该事件相关的条件的先验知识来描述事件的概率。也可以考虑使用条件概率示例。例如：有3个袋子，每个袋子中都装有一些白色大理石和黑色大理石。如果随机绘制白色大理石。很有可能会发现这种白色大理石来自第一个袋子。在这种情况下，我们使用贝叶斯定理。它用于基于其他条件(也称为条件概率)计算特定事件发生的概率的情况。因此，在详细介绍之前，让我们简要讨论一下“总概率定理” 。
总概率定理
令E 1 ，E 2 ，…………..E n是与随机实验相关的互斥和穷举性事件，令E为与某些E i一起发生的事件。然后证明
P(E) = n∑i=1P(E/Ei) . P(Ej)
证明：
Let S be the sample space. Then,
S = E1 ∪ E2 ∪ E3 ∪………………… ∪ En and Ei ∩ Ej = ∅ for i ≠ j.
Therefore, E = E∩S = E ∩ (E1 ∪ E2 ∪ E3 ∪………………… ∪ En)
= (E ∩ E1) ∪ (E ∩ E2) ∪ ……∪ (E ∩ En)
=> P(E) = P{(E ∩ E1) ∪ (E ∩ E2)∪……∪(E ∩ En)}
= P(E ∩ E1) + P(E ∩ E2) + …… + P(E ∩ En)
= {Therefore, (E ∩ E1), (E ∩ E2),………….,(E ∩ En)} are pairwise disjoint}
= P(E/E1) . P(E1) + P(E/E2) . P(E2) +……………………+ P(E/En) . P(En) [by multiplication theorem]
= n∑i=1P(E/Ei) . P(Ei)
例子
示例1：一个人从事了工作。在有雨和无雨的情况下准时完成工作的概率分别为0.44和0.95。如果下雨的概率为0.45，那么确定该作业将按时完成的概率吗?
解决方案：
Let E1 be the event that the mining job will be completed on time and E2 be the event that it rains. We have,
P(A) = 0.45,
P(no rain) = P(B) = 1 − P(A) = 1 − 0.45 = 0.55
By multiplication law of probability,
P(E1) = 0.44
P(E2) = 0.95
Since, events A and B form partitions of the sample space S, by total probability theorem, we have
P(E) = P(A) P(E1) + P(B) P(E2)
= 0.45 × 0.44 + 0.55 × 0.95
= 0.198 + 0.5225 = 0.7205
So, the probability that the job will be completed on time is 0.684.
范例2：三个骨灰盒包含3个白色和2个黑色的小球； 2个白色和3个黑色的球； 1个黑球和4个白球。选择每种的可能性均等。一个球是随机选择的相等概率。抽出白球的几率是多少?
解决方案：
Let E1, E2, and E3 be the events of choosing the first, second, and third urn respectively. Then,
P(E1) = P(E2) = P(E3) =1/3
Let E be the event that a white ball is drawn. Then,
P(E/E1) = 3/5, P(E/E2) = 2/5, P(E/E3) = 4/5
By theorem of total probability, we have
P(E) = P(E/E1) . P(E1) + P(E/E2) . P(E2) + P(E/E3) . P(E3)
= (3/5 * 1/3) + (2/5 * 1/3) + (4/5 * 1/3)
= 9/15 = 3/5
贝叶斯定理
令E 1 ，E 2 ，E 3 ，………，E n是与随机实验相关的互斥和穷举事件，而E是与某些E i一起发生的事件。然后，
P(E|Ei) = P(E|Ei) . P(Ei)/n∑i=1 P(E|Ei) . P(Ei)
Where i = 1, 2, 3, 4…….., n
证明：
By the theorem of total probability, we have
P(E) = n∑i=1 P(E|Ei) . P(Ei) …………………………….(i)
Therefore, P(E|Ei) = P(E ∩ Ei)/P(E) [by multiplication theorem]
= P(E|Ei) . P(Ei)/P(E) [Therefore, P(E/Ei) = P(E ∩ Ei)/P(Ei)]
= P(E|Ei) . P(Ei)/n∑i=1 P(E|Ei) . P(Ei) [Using (i)]
Hence, P(E|Ei) = P(E|Ei) . P(Ei)/n∑i=1 P(E|Ei) . P(Ei)
贝叶斯定理公式
让我们将E和Ei视为两个事件，因此贝叶斯定理的公式为：
P(Ei|E) = P(E ∩ Ei)/P(E)
Here,
P(Ei|E) is in conditional probability when event Ei occurs before event E.
P(E ∩ Ei) is the probability of event E and event Ei.
P(E) as the Probability of E.
贝叶斯定理推导
As we know Bayes Theorem can be derived from events and random variables separately with the help of conditional probability and density. As per conditional probability, we assume that there are two events T and Q associated with the same rab = ndom experiment. Then, the probability of occurrence of T under the condition that Q has already occurred and P(Q)≠0 is called conditional probability, denoted by P(T/Q). So we ultimately define it as
P(T|Q) = P(T ⋂ Q)/P(Q), where P(Q) ≠ 0
P(Q|T) = P(Q ⋂ T)/P(T), where P(T) ≠ 0
Likewise, If the conditional distribution of J given G is in a continuous distribution, then its probability of density function is known as the conditional density function. Bayes theorem can derive these two continuous random variables namely F and G as given below:
ƒ J|G = g(j) = ƒ J,G(j,g)/ƒ G(g)
ƒ J|G = j(g) = ƒ J,G(j,g)/ƒ J(j)
ƒ J|G = g(j) = ƒ G|J=j(g) ƒJ(j) ƒG(g)
贝叶斯定理的例子
示例1：一包52张卡片中的一张卡片丢失。从该包的其余卡片中抽出两张卡片，发现它们都是心脏。发现丢失的卡片是一颗心的可能性?
解决方案：
Let E1, E2, E3, and E4 be the events of losing a card of hearts, clubs, spades, and diamonds respectively.
Then P(E1) = P(E2) = P(E3) = P(E4) = 13/52 = 1/4.
Let E be the event of drawing 2 hearts from the remaining 51 cards. Then,
P(E|E1) = probability of drawing 2 hearts, given that a card of hearts is missing
= 12C2/51C2 = (12 * 11)/2! * 2!/(51 * 50) = 22/425
P(E|E2) = probability of drawing 2 clubs ,given that a card of clubs is missing
= 13C2/51C2 = (13 * 12)/2! * 2!/(51 * 50) = 26/425
P(E|E3) = probability of drawing 2 spades ,given that a card of hearts is missing
= 13C2/51C2 = 26/425
P(E|E4) = probability of drawing 2 diamonds ,given that a card of diamonds is missing
= 13C2/51C2 = 26/425
Therefore,
P(E1|E) = probability of the lost card is being a heart, given the 2 hearts are drawn from the remaining 51 cards
= P(E1) . P(E|E1)/P(E1) . P(E|E1) + P(E2) . P(E|E2) + P(E3) . P(E|E3) + P(E4) . P(E|E4)
= (1/4 * 22/425) / {(1/4 * 22/425) + (1/4 * 26/425) + (1/4 * 26/425) + (1/4 * 26/425)}
= 22/100 = 0.22
Hence, The required probability is 0.22.
示例2：假设300名男性中的15名男性和1000名女性中的25名女性是好演说者。演说家是随机选择的。查找选择男性的可能性。假设男人和女人的人数相等?
解决方案：
Let there be 1000 men and 1000 women.
Let E1 and E2 be the events of choosing a man and a woman respectively. Then,
P(E1) = 1000/2000 = 1/2 , and P(E2) = 1000/2000 = 1/2
Let E be the event of choosing an orato. Then,
P(E|E1) = 50/1000 = 1/20, and P(E|E2) = 25/1000 = 1/40
Probability of selecting a male person ,given that the person selected is a good orator
P(E1/E) = P(E|E1) * P(E1)/ P(E|E1) * P(E1) + P(E|E2) * P(E2)
= (1/2 * 1/20) /{(1/2 * 1/20) + (1/2 * 1/40)}
= 2/3
Hence the required probability is 2/3.
例3：已知一个人讲谎言四分之一。他丢了一个骰子，并报告说已经六点了。找出实际上是六的概率?
解决方案：
In a throw of a die, let
E1 = event of getting a six,
E2 = event of not getting a six and
E = event that the man reports that it is a six.
Then, P(E1) = 1/6, and P(E2) = (1 – 1/6) = 5/6
P(E|E1) = probability that the man reports that six occurs when six has actually occurred
= probability that the man speaks the truth
= 3/4
P(E|E2) = probability that the man reports that six occurs when six has not actually occurred
= probability that the man does not speak the truth
= (1 – 3/4) = 1/4
Probability of getting a six ,given that the man reports it to be six
P(E1|E) = P(E|E1) * P(E1)/P(E|E1) * P(E1) + P(E|E2) * P(E2) [by Bayes’ theorem]
= (3/4 * 1/6)/{(3/4 * 1/6) + (1/4 * 5/6)}
= (1/8 * 3) = 3/8
Hence the probability required is 3/8.