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Probability

IndependenceIf knowing that Event A has occurred gives you information about Event B, then Events A & B are not independent.

Ex. Football players and knee problemsEx. Outcome on 2 fair coins

Mutually Exclusive EventsAlso called disjoint. They have no outcomes in common.Knowing that Event A has occurred does give you information about B. If A occurred, then B could not have occurred. Thus, disjoint events are not independent.The P(A and B) when A & B are disjoint is 0.

Adapted from Barron’s p. 394 #7

Suppose that for a certain Caribbean island in any 3 year period the probability of a major hurricane is .25, the probability of water damage is .44, and the probability of both a hurricane and water damage is .22.A Venn diagram helps to organize information.


Suppose that for a certain Caribbean island in any 3 year period the probability of a major hurricane is .25, the probability of water damage is .44, and the probability of both a hurricane and water damage is .22.Are the events hurricane and water damage independent?


Suppose that for a certain Caribbean island in any 3 year period the probability of a major hurricane is .25, the probability of water damage is .44, and the probability of both a hurricane and water damage is .22.Are the events hurricane and water damage independent? No, because P(hurricane) x P(water damage) doesn’t equal P(hurricane & water damage)


Suppose that for a certain Caribbean island in any 3 year period the probability of a major hurricane is .25, the probability of water damage is .44, and the probability of both a hurricane and water damage is .22.What is the probability of water damage given that there is a hurricane?


Suppose that for a certain Caribbean island in any 3 year period the probability of a major hurricane is .25, the probability of water damage is .44, and the probability of both a hurricane and water damage is .22.What is the probability of water damage given that there is a hurricane? Answer: .22/.25 = .88


Given the probabilities P(A) = .3 and P(B) = .2, what is the P(A union B) if A and B are mutually exclusive?


Given the probabilities P(A) = .3 and P(B) = .2, what is the P(A union B) if A and B are mutually exclusive?

Answer: .3 + .2 = .5


Given the probabilities P(A) = .3 and P(B) = .2, what is the P(A union B) if A and B are independent?


Given the probabilities P(A) = .3 and P(B) = .2, what is the P(A union B) if A and B are independent?

Answer: .3 + .2 - .06 =.44


Given the probabilities P(A) = .3 and P(B) = .2, what is the P(A union B) if B is a subset of A?

Answer: .3

Tree DiagramA plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is .005, and the rest from a company whose defect rate is 0.010. What proportion of the circulators can be expected to be defective?

Tree DiagramA plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is .005, and the rest from a company whose defect rate is 0.010. What proportion of the circulators can be expected to be defective?

Answer: (.6)(.005) + (.4)(.010) = .007

Tree Diagram-Barron’s P. 368 #6

A plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is .005, and the rest from a company whose defect rate is 0.010. If a circulator is defective,what is the probability that it came from the first company?

Tree DiagramA plumbing contractor obtains 60% of her boiler circulators from a company whose defect rate is .005, and the rest from a company whose defect rate is 0.010. If a circulator is defective,what is the probability that it came from the first company?Answer: (.6)(.005)/.007

Probability with Normal Distributions

Barron’s p. 373 #31The mean Law School Aptitude Test (LSAT) score for applicants to a particular law is 650 with a standard deviation of 45. Suppose that only applicants with scores above 700 are considered. What percentage of the applicants considered have scores below 740? Assume the scores are normally distributed.

Probability with Normal Distributions

The mean Law School Aptitude Test (LSAT) score for applicants to a particular law is 650 with a standard deviation of 45. Suppose that only applicants with scores above 700 are considered. What percentage of the applicants considered have scores below 740? Assume the scores are normally distributed.

Answer: P(X>700) = .1332, P(X is between 700 and 740) = .1105P(X<740 given that X>700) = .1105/.1332 = .8297

Binomial ProbabilityRemember BINSB-BinaryI-Independent trialsN-Set number of trialsS-Probability of success is the same for each trial

AP Free Response 2004 #3

At an archaeological site that was an ancient swamp, the bones from 20 brontosaur skeletons have been unearthed. The bones do not show any sign of disease or malformation. It is thought that these animals wandered into a deep area of the swamp and became trapped in the swamp bottom. The 20 left femur bones (thigh bones) were located and 4 of these left femurs are to be randomly selected without replacement for DNA testing to determine gender.A) Let X be the number out of the 4 selected left femurs that are from males. Based on how these bones were sampled, explain why the probability distribution of X is not binomial.


A) Let X be the number out of the 4 selected left femurs that are from males. Based on how these bones were sampled, explain why the probability distribution of X is not binomial.Answer: X is not binomial since the trials are not independent and the conditional probabilities of selecting a male change at each trial depending on the previous outcome(s), due to the sampling without replacement.


At an archaeological site that was an ancient swamp, the bones from 20 brontosaur skeletons have been unearthed. The bones do not show any sign of disease or malformation. It is thought that these animals wandered into a deep area of the swamp and became trapped in the swamp bottom. The 20 left femur bones (thigh bones) were located and 4 of these left femurs are to be randomly selected without replacement for DNA testing to determine gender.B) Suppose that the group of 20 brontosaurs whose remains were found in the swamp had been made up of 10 males and 10 females. What is the probability that all 4 in the sample to be tested are male?


B) Suppose that the group of 20 brontosaurs whose remains were found in the swamp had been made up of 10 males and 10 females. What is the probability that all 4 in the sample to be tested are male?Answer: (10/20)(9/19)(8/18)(7/17) = .043


At an archaeological site that was an ancient swamp, the bones from 20 brontosaur skeletons have been unearthed. The bones do not show any sign of disease or malformation. It is thought that these animals wandered into a deep area of the swamp and became trapped in the swamp bottom. The 20 left femur bones (thigh bones) were located and 4 of these left femurs are to be randomly selected without replacement for DNA testing to determine gender.C) The DNA testing revealed that all 4 femurs tested were from males. Based on this result and your answer from part (b), do you think that males and females were equally represented in the group of 20 brontosaurs stuck in the swamp? Explain.


C) The DNA testing revealed that all 4 femurs tested were from males. Based on this result and your answer from part (b), do you think that males and females were equally represented in the group of 20 brontosaurs stuck in the swamp? Explain.

Answer: No. If males and females were equally represented, the probability of observing four males is small (0.043).


At an archaeological site that was an ancient swamp, the bones from 20 brontosaur skeletons have been unearthed. The bones do not show any sign of disease or malformation. It is thought that these animals wandered into a deep area of the swamp and became trapped in the swamp bottom. The 20 left femur bones (thigh bones) were located and 4 of these left femurs are to be randomly selected without replacement for DNA testing to determine gender.D) Is it reasonable to generalize your conclusion in part (c) pertaining to the group of 20 brontosaurs to the population of all brontosaurs? Explain why or why not.


D) Is it reasonable to generalize your conclusion in part (c) pertaining to the group of 20 brontosaurs to the population of all brontosaurs? Explain why or why not.Answer: No, we can’t generalize to the population of all brontosaurs because it is not reasonable to regard this sample as a random sample from the population of all brontosaurs; there is reason to suspect that this sampling method might cause bias.

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