Chapter 4 labs
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Disjoint events
events cannot happen at the same time

Find compliment of A
1P(A)

Addition rule (key work "or")
P(A OR B) = P(A) + P(B)  P(A AND B)

Unususal event
less than 0.05

P(B/A) represents
The probability of event B occuring, given that event A already occured

Two events A and B are independent if the occurrence of onedoes not affect the probability of the
occurrence of the other. (Several events
are similarly independent if the occurrence of any does not affect the probabilities of the occurrence of the others.)
If A and B are not independent, they are said to be dependent.

Multiplication rule (key word "and")
 P(A and B) = P(A) • P(B/A)
 Must subtract 1

When calculating the probability of an event such as "both accidents involved intoxicated driver" you can reexpress the problem as:
What is the probability that the first accident involved an intoxicated driver and then the second accident also involved an intoxicated driver?
Event A= 1st accident involved an intoxicated driver
Event B= 2nd accident involved an intoxicated driver
P(both)=P(A)*P(BA)
Assume that the accidents are being selected without replacement.

Suppose 3 CDs are selected for testing. The entire batch is accepted only if all 3 CDs are ok.
The probability that the entire batch is accepted is equal to the product of the probabilities that each CD selected is ok.
P(enter batch is accepted)=P(1st is ok)P(2nd is ok)P(3rd CD is ok)
Remember that the CDs are selected WITHOUT replacement.

Part BP(both fail)=P(1st clock fails)*P(2nd clock fails)
Part CRemember that you will be awaken as long as 1 clock works.Sample SpaceA=1st clock works and 2nd clock worksB=1st clock works and 2nd clock does not workC=1st clock does not work and 2nd clock does workD=1st clock does not work and 2nd clock does not workThere are 4 outcomes in the sample spaceP(being awaken by at least 1 clock)=P(A)+ P(B) + P(C)ORP(being awaken by at least 1 clock)=1P(both clocks do not work)=1P(D)but you already calcuolated P(D) in part B

Series ConfigurationAs long as 1 surge protector works then the TV is protected.find P(at least 1 surge protector works)Parallel ConfigurationBoth surge protectors must work for the TV to be protected.find P(both surge protectors work)

Total number of events in the sample space for a family of n children.2^nwhere n is the number of children in the family.If there are 6 children in the family then the number of possible events in the sample space is:2^6 = 64Since the event of having 6 girls in a family of 6 children can only happen 1 way thenP(6 girls in a family of 6 children)=1/64Since the event of having 0 girls (all boys) in a family of 6 children can only happen 1 way thenP(no girls in a family of 6 children)=1/64

Permutations or Combinations:Ask yourself the following:If you test the yellow wire and then the blue wire and then the green wire.....Would that be any different than testing the green wire and then the blue wire and then the yellow wire?????If there's no difference, then the problem is a combination problem.