Chapter 2--sections 2.5 - 2
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CHAPTER 2
2.5 Counting Rules Useful in Probability
Let A = “the event of getting an even number”
P ( A) =
3 Number of outcomes favorable to A = 6 Total number of equally likely outcomes
A quality inspector examines two manufactured items selected from a production line. Item 1 can be defective or nondefective, as can item 2. How many possible outcomes exists? Construct a 2x2 table for this case. Theorem. If the first task of an experiment can result in n1 possible outcomes and, for each such outcome, the second task can result in n2 possible outcomes, then there are n1 n2 possible outcomes for the two tasks together.
2.5 Counting Rules Useful in Probability
1. A firm is deciding where to build two new plants, one in the east and one in the west. Four eastern cities and two western cities are possibilities. Construct a tree diagram for this problem. If all choices are equally likely, find the probability that the 2nd western city is selected. Answer= ½ 2. Five motors (numbered 1 through 5) are available for use, and motor 2 is defective. Motor 1 and 2 come from supplier I, and motors 3,4 and 5 come from supplier II. Suppose two motors are randomly selected for use on a particular day. Let A denote the event that the defective motor is selected and B the event that at least one motor comes from supplier I. Find P(A) and P(B). Answer P(A)=0.4 and P(B)=0.7
2.5 Counting Rules Useful in Probability First Motor selected
1
Second Motor selected 2 3 4 5
2
1 3 4 5
3
4
1 2 4 5 1 2 3 5
5
1 2 3 4
2.5 Counting Rules Useful in Probability
From three pilots (A,B,C) a crew of two is to be selected to form a pilot-copilot team. Show how many possibilities exist?
Obs. This is an example of a permutation.
Theorem. The number of ordered arrangements, or permutations, of r objects selected from n distinct objects (r ≤ n) is given by
n! P = n(n − 1)...(n − r + 1) = (n − r )! n r
2.5 Counting Rules Useful in Probability From among 10 employees, three are to be selected for travel to three out-of-town plants, A,B,C, with one employee traveling to each plant. Since the plants are in different cities, the order of assisting the employees to the plants is an important consideration. The first person selected might, for instance, go to plant A, the second to plant B. In how many ways can the assignments be made?
Answer= 720
An assembly operation in a manufacturing plant involves four steps, which can be performed in any order. If the manufacturer wishes to experimentally compare the assembly times for each possible ordering of the steps, how many orderings will the experiment involve?
Answer =24
2.5 Counting Rules Useful in Probability
A manager is asked to rank four divisions of a company with respect to their ability to adapt new technology. If the divisions are labeled A,B,C and D what is the probability that B gets ranked highest? What is the probability that B ranks first and D ranks second?
Answer: ¼ 1/12
2.5 Counting Rules Useful in Probability
Theorem. The number of distinct unordered subsets, or combinations, of size r that can be selected from n distinct objects (r ≤ n) is given by
n n! = r r!(n − r )!
Since any particular set of r objects can be ordered among themselves in Prr = r! ways, it follows that
n r!= Prn r or n 1 n n! = Pr = r!(n − r )! r r!
2.5 Counting Rules Useful in Probability
Suppose that three employees are to be selected from ten to visit a new plant. In how many ways can the selection be made?
Answer =120
If two of the above employees are female and eight are male, what is the probability that exactly one female gets selected among the three?
Answer = 7/15
2.5 Counting Rules Useful in Probability
Five applicants for a job are ranked according to ability, with applicant number 1 being the best, number 2 the second best and so on. These rankings are unknown to an employer, who simply hires two applicants at random. What is the probability that this employer hires exactly one of the two best applicants?
Answer = 0.6
Work in Class
For a certain style of new automobile, the colors blue, white, black and green are in equal demand. Three successive orders are placed for automobiles of this style. Find the probability that:
A) one blue, one white and one green are ordered B) two blues are ordered C) at least one black is ordered D) exactly two of the orders are of the same color
Solution
A) P(B, W,G) = (number of ways of ordering a B,W,G) / (total number of ways a sample of 3 can be chosen) = P33 / (4 * 4 *4) = 6/64=0.09375 B) P(BB) = P( order 2B out of 3 cars) * P( number of ways of getting the third car) / P(total number of ways a sample of 3 cars can be chosen) = 3 *3 / 64 = 0.140625
C) P( at least one is black) = P(1 black)+ P( 2black)+ P( 3 black) = 1 – P( no black) = 1 – (get any of the 3 colors)/(get a sample of 3) = 1- 3*3*3/4*4*4 = 37/64 = =0.5718125
D) P( exactly 2 are the same color) = 4 * 3*3/64=0.5625
2.5 Counting Rules Useful in Probability
Theorem. The number of ways of partitioning n distinct objects into k groups containing n1, n2 … nk objects respectively, is
n! n1! n2 !...nk ! where k
n = n ∑ i= 1
i
2.5 Counting Rules Useful in Probability
Suppose that ten employees are to be divided among three jobs, with three employees going to job I, four going to job II and three going to job III. In how many ways can the job assignment be made?
Answer = 10! / 3!*4!*3! = 4,200
In the above problem, suppose the only three employees of a certain ethnic group all get assigned to job I. What is the probability of this happening under a random assignment of employees to jobs?
Answer = (how many ways we can assign the other 7 employees)/(how many ways to assign 10 employees to 3 jobs) = (7!/ (4!*3!)) / 4,200 = 35/4,200= 1/120
HOMEWORK
2.30 2.32
2.6 Conditional Probability and Independence
Percentage of net new workers in the labor force
Women Men ************************************************ White 42% 15% 57% Nonwhite 14% 7% 21% Immigrant 9% 13% 22%
How do the relative frequencies for the three racial/immigrant categories compares between women and men?
2.6 Conditional Probability and Independence
Definition: If A and B are any two events, then the conditional probability of A given B, denoted P(A|B), is
P ( A | B) =
P ( AB ) P ( B)
provided that P(B) >0.
Example: From five motors, of which one is defective, two motors are to be selected at random for use on a particular day. Find the probability that the second motor selected is nondefective, given that the first was nondefectice. Answer = (12/20) / (16/20) = ¾=0.75
2.6 Conditional Probability and Independence
Probabilities are usually very sensitive to the conditioning information. Sometimes, however, a probability does not change when a conditioning information is supplied.
If the extra information derived from knowing that an event B has occurred does not change the probability of A – if P(A|B)=P(A) – then the events A and B are said to be independent.
Using the previous definition we obtain P(AB)=P(A)P(B).
Definition: Two events are said to be independent if P(A|B)=P(A) or P(B|A)=P(B). This is equivalent to stating that P(AB)=P(A)P(B).
2.6 Conditional Probability and Independence
Suppose a foreman must select one worker for a special job from a pool of four available workers, numbered 1,2,3 and 4. He selects the worker by mixing the four names and randomly selecting one. Let A denote the event that worker 1 or 2 is selected, B the event that worker 1 or 3 is selected and C the event that worker 1 is selected. Are A and B independent? Are A and C independent?
Answer: Calculate the probabilities P(A), P(B), P(C) and P(AB), P(AC) respectively.
2.7 Rules of probability
Definition. The complement Ā of an event A is the set of all outcomes in a sample space S that are not in A. Thus Ā and A are mutually exclusive and their union is S. Therefore, Ā U A = S. P(Ā U A ) = P(Ā) + P(A) =P(S)=1 P(Ā) = 1 – P(A)
A quality-control inspector has ten assembly lines from which to choose products for testing. Each morning of a five-day week, she randomly selects one of the lines to work on for the day. Find the probability that a line is chosen more than once a week.
Answer: P (line is chosen more than once) = 0.7
2.7 Additive and Multiplicative Rule
Theorem. If A and B are any two events, then P(A U B) = P(A) + P(B) – P(AB).
If A and B are mutually exclusive, then P(A U B) = P(A) + P(B).
What is P(A U B U C) =?
Theorem. If A and B are any two events, then
P(A ∩ B) = P(A) P(B|A) = P(B)P(A|B). If A and B are independent, then P(AB) = P(A) P(B).
2.7 Additive and Multiplicative Rule
Records indicate that for the parts coming out of a hydraulic repair shop at an airplane rework facility, 20% will have a shaft defect, 10% will have a bushing defect and 75% will be defectfree. For an item chosen at random from this output, find the probability of the following:
A: The item has at least one type of defect. B: The item has only a shaft defect. C: The item has only a bushing defect.
2.7 Additive and Multiplicative Rule
Three different orders are to be mailed to three suppliers. However, an absent – minded secretary gets the order mixed up and just sends them randomly to suppliers. If a match refers to the fact that a supplier gets the correct order, find the probability of
A: no matches. B: exactly one match.
2.7 Bayes’ Rule
Let B1 , B2 … Bk are partitions of a sample space S and the following conditions exists: Bi Bj = Ø for any pair I and j B1 U B2 U… U Bk = S.
Bayes’ Rule.
P ( B j | A) =
P (B j )P ( A | B j ) k
∑ P (B ) P ( A | B ) i =1
i
i
2.7 Bayes’ Rule
A company buys tires from two suppliers, 1 and 2. Supplier 1 has a record of delivering tires containing 10% defectives, whereas supplier 2 has a defective rate of only 5%. Suppose 40% of the current supply came from supplier 1. If a tire is taken from this supply and observed to be defective, find the probability that it came from supplier 1.
Law of Total Probability
Allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space.
One way to partition S is to break into sets F and Fc, for any event F. Therefore the simplest form of the law of total probability:
P(E)=P(E∩F) + P(E ∩ Fc) = P(E|F) P(F) + P(E | Fc) P(Fc)
Example. A machine produces parts that are either good (90%) , slightly defective (2%) , or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which was able to detect any part that is obviously defective and discard it.
We assume that a one-year warranty is given for the parts that are shipped to customers. Suppose that a good part fails within the first year with probability 0.01, while a slightly defective part fails within the first year with probability 0.10. What is the probability that a customer receives a part that fails within the first year and therefore is entitled to a warranty replacement?
Solution Let G be the event that randomly chosen shipped part is good Let SD be the event that randomly chosen shipped part is slightly defective Let OD be the event that randomly chosen shipped part is obviously defective P(G)=90/92, P(SD)=2/92 Let E be the event that a randomly selected customer’s part fails in the first year P(E|G)=0.01 and P(E|SD)=0.1 P(E) = P(E|G) P(G) + P(E|SD) P(SD) = (0.01) 90/92 + (0.1)* 2/92 = 0.012
HOMEWORK
2.42 2.47
2.5 Counting Rules Useful in Probability
Let A = “the event of getting an even number”
P ( A) =
3 Number of outcomes favorable to A = 6 Total number of equally likely outcomes
A quality inspector examines two manufactured items selected from a production line. Item 1 can be defective or nondefective, as can item 2. How many possible outcomes exists? Construct a 2x2 table for this case. Theorem. If the first task of an experiment can result in n1 possible outcomes and, for each such outcome, the second task can result in n2 possible outcomes, then there are n1 n2 possible outcomes for the two tasks together.
2.5 Counting Rules Useful in Probability
1. A firm is deciding where to build two new plants, one in the east and one in the west. Four eastern cities and two western cities are possibilities. Construct a tree diagram for this problem. If all choices are equally likely, find the probability that the 2nd western city is selected. Answer= ½ 2. Five motors (numbered 1 through 5) are available for use, and motor 2 is defective. Motor 1 and 2 come from supplier I, and motors 3,4 and 5 come from supplier II. Suppose two motors are randomly selected for use on a particular day. Let A denote the event that the defective motor is selected and B the event that at least one motor comes from supplier I. Find P(A) and P(B). Answer P(A)=0.4 and P(B)=0.7
2.5 Counting Rules Useful in Probability First Motor selected
1
Second Motor selected 2 3 4 5
2
1 3 4 5
3
4
1 2 4 5 1 2 3 5
5
1 2 3 4
2.5 Counting Rules Useful in Probability
From three pilots (A,B,C) a crew of two is to be selected to form a pilot-copilot team. Show how many possibilities exist?
Obs. This is an example of a permutation.
Theorem. The number of ordered arrangements, or permutations, of r objects selected from n distinct objects (r ≤ n) is given by
n! P = n(n − 1)...(n − r + 1) = (n − r )! n r
2.5 Counting Rules Useful in Probability From among 10 employees, three are to be selected for travel to three out-of-town plants, A,B,C, with one employee traveling to each plant. Since the plants are in different cities, the order of assisting the employees to the plants is an important consideration. The first person selected might, for instance, go to plant A, the second to plant B. In how many ways can the assignments be made?
Answer= 720
An assembly operation in a manufacturing plant involves four steps, which can be performed in any order. If the manufacturer wishes to experimentally compare the assembly times for each possible ordering of the steps, how many orderings will the experiment involve?
Answer =24
2.5 Counting Rules Useful in Probability
A manager is asked to rank four divisions of a company with respect to their ability to adapt new technology. If the divisions are labeled A,B,C and D what is the probability that B gets ranked highest? What is the probability that B ranks first and D ranks second?
Answer: ¼ 1/12
2.5 Counting Rules Useful in Probability
Theorem. The number of distinct unordered subsets, or combinations, of size r that can be selected from n distinct objects (r ≤ n) is given by
n n! = r r!(n − r )!
Since any particular set of r objects can be ordered among themselves in Prr = r! ways, it follows that
n r!= Prn r or n 1 n n! = Pr = r!(n − r )! r r!
2.5 Counting Rules Useful in Probability
Suppose that three employees are to be selected from ten to visit a new plant. In how many ways can the selection be made?
Answer =120
If two of the above employees are female and eight are male, what is the probability that exactly one female gets selected among the three?
Answer = 7/15
2.5 Counting Rules Useful in Probability
Five applicants for a job are ranked according to ability, with applicant number 1 being the best, number 2 the second best and so on. These rankings are unknown to an employer, who simply hires two applicants at random. What is the probability that this employer hires exactly one of the two best applicants?
Answer = 0.6
Work in Class
For a certain style of new automobile, the colors blue, white, black and green are in equal demand. Three successive orders are placed for automobiles of this style. Find the probability that:
A) one blue, one white and one green are ordered B) two blues are ordered C) at least one black is ordered D) exactly two of the orders are of the same color
Solution
A) P(B, W,G) = (number of ways of ordering a B,W,G) / (total number of ways a sample of 3 can be chosen) = P33 / (4 * 4 *4) = 6/64=0.09375 B) P(BB) = P( order 2B out of 3 cars) * P( number of ways of getting the third car) / P(total number of ways a sample of 3 cars can be chosen) = 3 *3 / 64 = 0.140625
C) P( at least one is black) = P(1 black)+ P( 2black)+ P( 3 black) = 1 – P( no black) = 1 – (get any of the 3 colors)/(get a sample of 3) = 1- 3*3*3/4*4*4 = 37/64 = =0.5718125
D) P( exactly 2 are the same color) = 4 * 3*3/64=0.5625
2.5 Counting Rules Useful in Probability
Theorem. The number of ways of partitioning n distinct objects into k groups containing n1, n2 … nk objects respectively, is
n! n1! n2 !...nk ! where k
n = n ∑ i= 1
i
2.5 Counting Rules Useful in Probability
Suppose that ten employees are to be divided among three jobs, with three employees going to job I, four going to job II and three going to job III. In how many ways can the job assignment be made?
Answer = 10! / 3!*4!*3! = 4,200
In the above problem, suppose the only three employees of a certain ethnic group all get assigned to job I. What is the probability of this happening under a random assignment of employees to jobs?
Answer = (how many ways we can assign the other 7 employees)/(how many ways to assign 10 employees to 3 jobs) = (7!/ (4!*3!)) / 4,200 = 35/4,200= 1/120
HOMEWORK
2.30 2.32
2.6 Conditional Probability and Independence
Percentage of net new workers in the labor force
Women Men ************************************************ White 42% 15% 57% Nonwhite 14% 7% 21% Immigrant 9% 13% 22%
How do the relative frequencies for the three racial/immigrant categories compares between women and men?
2.6 Conditional Probability and Independence
Definition: If A and B are any two events, then the conditional probability of A given B, denoted P(A|B), is
P ( A | B) =
P ( AB ) P ( B)
provided that P(B) >0.
Example: From five motors, of which one is defective, two motors are to be selected at random for use on a particular day. Find the probability that the second motor selected is nondefective, given that the first was nondefectice. Answer = (12/20) / (16/20) = ¾=0.75
2.6 Conditional Probability and Independence
Probabilities are usually very sensitive to the conditioning information. Sometimes, however, a probability does not change when a conditioning information is supplied.
If the extra information derived from knowing that an event B has occurred does not change the probability of A – if P(A|B)=P(A) – then the events A and B are said to be independent.
Using the previous definition we obtain P(AB)=P(A)P(B).
Definition: Two events are said to be independent if P(A|B)=P(A) or P(B|A)=P(B). This is equivalent to stating that P(AB)=P(A)P(B).
2.6 Conditional Probability and Independence
Suppose a foreman must select one worker for a special job from a pool of four available workers, numbered 1,2,3 and 4. He selects the worker by mixing the four names and randomly selecting one. Let A denote the event that worker 1 or 2 is selected, B the event that worker 1 or 3 is selected and C the event that worker 1 is selected. Are A and B independent? Are A and C independent?
Answer: Calculate the probabilities P(A), P(B), P(C) and P(AB), P(AC) respectively.
2.7 Rules of probability
Definition. The complement Ā of an event A is the set of all outcomes in a sample space S that are not in A. Thus Ā and A are mutually exclusive and their union is S. Therefore, Ā U A = S. P(Ā U A ) = P(Ā) + P(A) =P(S)=1 P(Ā) = 1 – P(A)
A quality-control inspector has ten assembly lines from which to choose products for testing. Each morning of a five-day week, she randomly selects one of the lines to work on for the day. Find the probability that a line is chosen more than once a week.
Answer: P (line is chosen more than once) = 0.7
2.7 Additive and Multiplicative Rule
Theorem. If A and B are any two events, then P(A U B) = P(A) + P(B) – P(AB).
If A and B are mutually exclusive, then P(A U B) = P(A) + P(B).
What is P(A U B U C) =?
Theorem. If A and B are any two events, then
P(A ∩ B) = P(A) P(B|A) = P(B)P(A|B). If A and B are independent, then P(AB) = P(A) P(B).
2.7 Additive and Multiplicative Rule
Records indicate that for the parts coming out of a hydraulic repair shop at an airplane rework facility, 20% will have a shaft defect, 10% will have a bushing defect and 75% will be defectfree. For an item chosen at random from this output, find the probability of the following:
A: The item has at least one type of defect. B: The item has only a shaft defect. C: The item has only a bushing defect.
2.7 Additive and Multiplicative Rule
Three different orders are to be mailed to three suppliers. However, an absent – minded secretary gets the order mixed up and just sends them randomly to suppliers. If a match refers to the fact that a supplier gets the correct order, find the probability of
A: no matches. B: exactly one match.
2.7 Bayes’ Rule
Let B1 , B2 … Bk are partitions of a sample space S and the following conditions exists: Bi Bj = Ø for any pair I and j B1 U B2 U… U Bk = S.
Bayes’ Rule.
P ( B j | A) =
P (B j )P ( A | B j ) k
∑ P (B ) P ( A | B ) i =1
i
i
2.7 Bayes’ Rule
A company buys tires from two suppliers, 1 and 2. Supplier 1 has a record of delivering tires containing 10% defectives, whereas supplier 2 has a defective rate of only 5%. Suppose 40% of the current supply came from supplier 1. If a tire is taken from this supply and observed to be defective, find the probability that it came from supplier 1.
Law of Total Probability
Allows one to compute the probability of an event E by conditioning on cases, according to a partition of the sample space.
One way to partition S is to break into sets F and Fc, for any event F. Therefore the simplest form of the law of total probability:
P(E)=P(E∩F) + P(E ∩ Fc) = P(E|F) P(F) + P(E | Fc) P(Fc)
Example. A machine produces parts that are either good (90%) , slightly defective (2%) , or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which was able to detect any part that is obviously defective and discard it.
We assume that a one-year warranty is given for the parts that are shipped to customers. Suppose that a good part fails within the first year with probability 0.01, while a slightly defective part fails within the first year with probability 0.10. What is the probability that a customer receives a part that fails within the first year and therefore is entitled to a warranty replacement?
Solution Let G be the event that randomly chosen shipped part is good Let SD be the event that randomly chosen shipped part is slightly defective Let OD be the event that randomly chosen shipped part is obviously defective P(G)=90/92, P(SD)=2/92 Let E be the event that a randomly selected customer’s part fails in the first year P(E|G)=0.01 and P(E|SD)=0.1 P(E) = P(E|G) P(G) + P(E|SD) P(SD) = (0.01) 90/92 + (0.1)* 2/92 = 0.012
HOMEWORK
2.42 2.47
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