But there are 3 6 different orderings so the number of combinations is 3 3 1

# But there are 3 6 different orderings so the number

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Example 2.11.Continuing from Example 2.10, assume instead of a board we are interested in thenumber of possiblecommittees, where each member has equal power. How many such committees are possible? 20 Application of Combinations to Probability Problems . Example 2.13.A lot of 100 articles contains 10 defective. If a sample of 5 articles is chose at random,what is the probability that it contains 3 defective? Example 2.14.A player serving at tennis is only allowed one fault. At a double fault, the sever losesa point and the other player gains a point. Given the following information: 21 What is the probability that the server loses a point,i.e.,what is the probability that both servicesresult in faults?Example 2.15.Referring to Example 2.6. Given that the length of a rod is too long, what is theprobability that the diameter is OK? Diameter Length Too Thin OK Too Thick Sum Too Short 10 3 5 18 OK 38 900 4 942 Too Long 2 25 13 40 Sum 50 928 22 1000 22 2.4.1Independent EventsWhen the given occurrence of one event does not influence the probability of a potential outcome ofanother event, then the two events are said to be independent.Two eventsAandBareindependentif the probability of each remains the same, whether or notthe other has occurred.AandBare independent ifP(A)6= 0 andP(B)6= 0, thenP(B|A) =P(B)⇐⇒P(A|B) =P(B)IfP(A) = 0 orP(B) = 0, then the two events are independent.More generally, the eventsA1, ...Anare independent if for eachAiand each collectionAj1, ..., Ajmofevents withP(Aj1∩ · · · ∩Ajm)6= 0P(Ai|Aj1, ..., Ajm) =P(Ai)As a consequence of independence, the rule of multiplication then saysP(AB) =P(A|B)P(B) =P(A)P(B)In general,Pk\i=1Ai!=kYi=1P(Ai)k1Example 2.16.Of the microprocessors manufactured by a certain process, 20% of them are defec-tive. Assume they function independently. Five microprocessors are chosen at random. What is theprobability that they will all work?  • • • 