1. 1.7 Bayes Theorem Independent

2. Theorem 1.5 (Bayes’Rule) If the events B 1,B 2,…, B n constitute a partition of the sample space S,where then for any event A in S such that ----Bayes’Rule Proof:

3. probability that it was made by machine B 3 ? Example 1.17 With reference to Example 1.16,if a product were chosen randomly and found to be defective,what is the Solution:

4. 1. 条件概率 全概率公式 贝叶斯公式 小 结小 结 乘法定理

6. Now consider an experiment in which 2 cards are drawn in succession from an ordinary deck, with replacement.The events are defined as A: the second card is a spade,B: the first card is an ace. Since the first card is replaced, our sample space for both the first and second draws consists of 52 Independent That is,P(A|B)=P(A).When this is true, B are said to be independent. the events A and Consider Solution: cards, containing 4 aces and 13 spades. Hence

7. It show that B has occurred does not change the Definition 1.14 Two events A and B are independent if and only if P(A|B)=P(A) or P(B|A)=P(B) A and B are dependent.Otherwise, probability of A occurring. We have we must have So

8. Therefore B is independent of A,i.e.knowledge that A has occurred does not change the probability of B occurring.Hence,independence is a symmetric relation on the set of all events in a sample space. We have we must have So

9. otherwise they are called dependent events. Theorem 1.5 We say that A and B are independent events if and the fire engine will be available. Example 1.18 A small town has one fire engine and one ambulance available for emergencies.The probability that the fire engine is available when needed is 0.98, probability that the ambulance is available when called and the is 0.92.In the event of an injury resulting from a burning building,find the probability that both the ambulance Solution:

10. Example 1.19 An electrical system consists of four components.The system works if components A and B work and either of the components C or D work. The reliability of each component is also shown in Figure 1.5. Find the probability that(a) the entire system works, And (b) the component C does not work,given that the entire system works.Assume that four components work independently. 0.9 A 0.9 B 0.8 C 0.8 D Fig. 1.5 Solution:

11. We can generalize this notion to n events as follows. Definition 1.15 The n events A 1, A 2,…A n are mutually independent if for every subset

12. without replacement, from an ordinary deck of playing cards. Three cards are drawn in succession, Example 1.20 Find the probability that the event occurs, where A 1 is the event that the first card is a red ace, A 2 is the event that the second card is a 10 or a jack, and A 3 is the event that the third card is greater than 3 but less than 7. Solution:

13. Example 1.21 A coin is biased so that a head is twice as likely to occur as a tail.If the coin is tossed 3 times,what is the probability of getting 2 tails and 1 head ? Solution:

14. Notes: (ii) Do not confuse independence with disjointness. if A and B are disjoint and P(A)>0 andIn fact, P(B)>0 then they are dependent because the occurrence of one precludes the occurrence of the other. (i) Any event A with P(A)=0 or 1 is independent of every other event B.

15. 两事件相互独立 两事件互斥 例如 由此可见两事件相互独立，但两事件不互斥. 两事件相互独立与两事件互斥的关系. 请同学们思考 二者之间没 有必然联系

16. 由此可见两事件互斥但不独立.

17. 小 结

18. Proof: Thomas Bayes(1702---1761)

19. Solution:Using Bayes’ rule to write and then substituting the probabilities calculated in example 1.16, we have In view of the fact that a defective product was selected, this result suggests that it probably was not made by machine B 3.

20. Solution:Let A and B represent the respective events that the fire engine and the ambulance are available. Then and the fire engine will be available. A small town has one fire engine and one ambulance available for emergencies.The probability that the fire engine is available when needed is 0.98, probability that the ambulance is available when called and the is 0.92.In the event of an injury resulting from a burning building,find the probability that both the ambulance Example 1.18

21. b. To calculate the conditional probability in this case, 0.9 A 0.9 B 0.8 C 0.8 D Solution: a. Clearly the probability that the entire system works can be calculated as the following notice that

22. b. To calculate the conditional probability in this case, notice that

23. Solution: First we define the events A 1 : the first card is a red ace, A 2 : the second card is a 10 or jack, A 3 : the third card is greater than 3 but less than 7. Now

24. Solution:The sample space Assigning probabilities of w and 2w for getting a tail and a head,respectively. Then we have 3w=1 or w=1/3. HenceP(H)=2/3 and P(T) =1/3. Let A=the event of getting 2 tails and 1 head in the 3 tosses of the coin.Then, A={TTH, THT, HTT}. Similarly,