P (A B ) = P (B ) =.58

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1 Review Exam 1 This is a sample of problems that would be good practice for the exam. This is by no means a guarantee that the problems on the exam will look identical to those on the exam but it should give you an idea of the level of difficulty and the scope of the topics that will be covered. On your exam there will be space for you to complete the problems. 1. In an IT department 25% of all calls to the help desk are about and 42% of calls are about passwords. Let A be the event that the next call to the help desk is about and let B be the event that the next call is about a password problem. Compute P (B ), P (A B ) and P (A B ). State any assumptions you make. Solutions: First before I compute any of the quantities I am assuming that the intersection of A and B is the empty set. Namely, the next call to the help desk cannot be bot about and passwords. Thus P (A B) = 0. Now it is essentially given that P (B) =.42. This P (B ) = 1 P (B) = 1.42 =.58. Next we note that since A does not intersect B then all of A is not in B or A B = B since there are no elements that are in A that are not also in B. Consequently, P (A B ) = P (B ) =.58 To compute P (A B ) we note that A B = (A B) and thus, P (A B ) = 1 P (A B) = 1 [P (A) + P (B) P (A B)] Since P (A B) = 0 we then have: P (A B ) = 1 P (A) P (B) = = Suppose that a box contains 3 green, 4 blue, and 5 red balls. Mary first selects a ball and after Mary has chosen Jim chooses a ball. Let G 1, R 1 and B 1 be the events that Mary chooses a green, red or blue ball respectively. Similarly define G 2, B 2 and R 2 as the event that Jim chooses a green, blue or red ball respectively. (a) Let E be the event that Jim chooses a red ball but Marry does not. Express E in terms of the events defined above and then compute P (E). Solution: The event E = R 2 R 1. and P (R 2 R 1 ) = P (R 1 )P (R 2 R 1 ) = = Here I have used that if R 1 occurs then thee are still 5 red balls remaining out of the total 11 balls that remain after Mary has chosen, thus P (R 2 R 1 ) = 5/11.

2 (b) What is the probability that Jim chooses a red ball if it is known that Mary chose Green? Solution: The If Mary chooses Green then there are 11 balls left in the box and 5 of them are red, thus P (R 2 G 1 ) = 5/11. (c) What is the probability that Mary chose a red ball if it is known that Jim chose a red ball? Solution: Here we use Bayes Theorem where our mutually exclusive, exhaustive events are R 1, G 1 and B 1. We seek to compute P (R 1 R 2 ). Applying Bayes Theorem we have: P (R 1 R 2 ) = = P (R 1 )P (R 2 R 1 ) P (R 1 )P (R 2 R 1 ) + P (B 1 )P (R 2 B 1 ) + P (G 1 )P (R 2 G 1 ) = Prove that if P (A B) = P (A) then P (B A) = P (B). = Solution: Assume that P (A B) = A. By definition of conditional probability we have that: P (B A) = P (A B) P (A) Now by the multiplication rule, P (A B) = P (B)P (A B). Thus, P (B A) = P (B)P (A B) P (A) By our assumption P (A B) = P (A). Thus, This proves the desired result. P (B A) = P (B) P (A) P (A) = P (B)

3 4. Suppose that a clinical trial is being performed to test side effects of a drug meant to lower cholesterol. It is known that 50% of all people have normal (N) cholesterol, 20% have low (L) cholesterol and 30% have high (H) cholesterol. It is also known that the drug causes headaches in 30% of those with normal cholesterol, in 90% of those with low cholesterol and in 2% of those with high cholesteral. (a) What is the probability that a randomly selected person has both high cholesterol and gets a headache from using the drug? Solution: First define some events: H = { a randomly selected person has high cholesterol} N = { a randomly selected person has normal cholesterol} L = { a randomly selected person has low cholesterol} A = { a randomly selected person gets a headache from the drug} In this part of the question you are asked to compute P (H A). To do this we use the multiplication rule. P (H A) = P (H)P (A H) = (.3)(.02) =.006. (b) What is the probability that if randomly selected person is known to get headaches from the drug that they have high cholesterol? Solution: We are asked to compute P (H A). Note that H, L and A are mutually exclusive and exhaustive events. Consequently we can use Bayes Theorem which tells us that: P (H)P (A H) P (H A) = P (H)P (A H) + P (N)P (A N) + P (L)P (A L). Note that in the statement of the problem it is given that P (N) =.5, P (L) =.2 and P (H) =.3. IT is also given that P (A N) =.3, P (A L) =.9 andp (A H) =.02. Substituting these values in to Bayes formula gives: P (H A) = (.3)(.02).3(.02) +.5(.3) +.2(.9) =

4 5. A certain factory produces glass pipets for use in laboratories. Unfortunately these pipets are useless if they have any imperfections and they are very difficult to produce. Suppose that any pipet that comes off the assembly has no imperfections with probability.92. (a) If pipets are shipped in boxes of 24 what is the probability that the box has no imperfections? Solution: Let A i = { the i th pipet put into the box has no imperfection}. Since the probability that any pupet has no imperfections is independent of any other then we have that: P (A 1 A 2... A 24 ) = P (A 1 )P (A 2 )... P (A 24 ) = (.92)(.92)... (.92) = (.92) (b) If pipets are shipped in boxes of 24 what is the probability that a box has at least one imperfection? Solution: The event that at least one pipet has an imperfection is the complement of the event that non have imperfections. The probability that non have imperfections was computed above, thus we have that, P (at least one has an imperfection} = 1 (.92) (c) If pipets are shipped in boxes of 24 what is the probability that a box has exactly 4 with imperfections? Solution: The probability that the 1 st four put in the box are defective could be written as: P (A 1 A 2 A 3 A 4 A 5 A 6... A 24 ) = P (A 1)P (A 2)P (A 3)P (A 4)p(A 5 )P (A 6 )... P (A 24 ) = (.08) 4 (.92) However, the four that are defective are equally likely to be any four. And there are ( ) 24 4 different ways to select which of the four are defective. Thus the each of these events is just as likely to occur as the one with the first four being defective, this the total probability for this situation is: P (exactly four are defective) = ( ) 24 (.08) 4 (.92)

5 6. Anxiety disorders and symptoms can often be effectively treated with benzodiazepine medications. It is known that animals exposed to stress exhibit a decrease in benzodiazepine receptor binding in the frontal cortex. The paper Decreased Benzodiazepine Receptor Binding in Prefrontal Cortex in Combat-Related Post-traumatic Stress Disorder (Amer. J. of Psychiatry, 2000: ) described the first study of benzodiazepine receptor binding in individuals suffering from PTSD. The accompanying data on a receptor binding measure (adjusted) was read from a graph in the paper. PTSD : 7, 20, 25, 28, 31, 35, 37, 38, 39, 39, 42, 46 (a) Find the sample mean, sample median, sample standard deviation, fourth spread and 25% trimmed mean for this data set. The sample mean is given by: x = = The sample median is the average of the 6 th and 7 th largest values since there are 12 total observations. Since the observations are already ordered this means taking the average of 35 and 37. Thus: x = 36. The sample standard deviation can be computed by: x 2 i s = (P x i ) 2 n n 1 In this case n = 12 thus we have: ( )2 12 s = 11 Computing this gives: s = The fourth spread is computed by taking the difference of the lower and upper fourths. The upper fourth is the median of the 6 largest values. This it is the average of 39 and 39, thus the upper fourth is 39. Similarly the lower fourth is the median of the 6 smallest values which in this case is the average of 25 and 28 giving that the lower fourth is The fourth spread is the difference of these. Namely, f s = = 12.5 To compute the 25% trimmed mean we first note that 25% of 12 is 3, thus we disregard the largest 3 values and the smallest three values and average the remaining giving: x 25% trim =

6 (b) Draw a stem and leaf diagram for this data. Solution: The stem is the tens digit and the leaf is the ones digit (c) What is the general shape. The shape seems to be unimodal and perhaps slightly negatively skewed (which is also confirmed since the mean is less than the median). The peek is somewhere around 3 and there seems to be a gap between 7 and 20. (d) Decide if any of the data points are outliers or extreme outliers. Outliers are values that are more then 1.5(f s ) below the lower fourth or more then 1.5(f s ) above the upper fourth. 1.5(f s ) = (1.5)(12.5) = The lower fourth is 26.5 and = Since 7 is below this value it is an outlier = There are no values above this so there are no outliers at the upper end. Now we should check that 7 is or is not an extreme outlier as well. In order to be an extreme outlier it would need to be more then 3f s = 3(12.5) = 37.5 below the lower fourth, but = 11. Seven is certainly greater than this, so there are no extreme outliers. (e) Draw a box plot for this data

7 7. Suppose that 5 couples are seated in 10 chairs on one side of a rectangular table and that each person is equally likely to be sitting in any seat. (a) Find the probability that every couple sits next to each other. Solution: First note that there are 10! possibilities of how the 10 people could sit at the table. There are numerous ways to count the total possibilities of arrangements where all the couples are seated next to each other. One way to think about it is to first order the 5 couples in 5! ways and then within each couple order the couple or 2 5 ways. Thus a total of 2 5 5! ways. Another way to think about it is to seat the first person (10 possible choices). There is then only one person (the spouse of the first person) that can sit in the second seat. For the third seat there is now 8 choices, and so on giving, possible choices. The total probability for all the couples seated next to their spouse is: ! = (b) Find the probability that Joe and Sally (husband and wife) sit next to each other. Again there are several ways to think about this problem. One way is to think about Joe and Sally as one unit and there are 8 other units. So we must order the 9 units in 9! ways and then order Joe and Sally in their unit in 2 ways giving 9! 2 total possible ways in which Joe and Sally sit next to each other. Thus the probability that they sit next to each other is: 9!2 10! = 2 10 =.2 Another way of thinking about it is to define the event A = { Joe is seated in an end seat}. and N = { Joe and Sally are seated next to each other}. Using the law of total probability we have that: P (N) = P (A)P (N A) + P (A )P (N A ) Since Joe is equally likely to sit in any seat then P (A) = 2/10 and P (A ) = 8/10. If Joe is seated in an end seat then there is only one seat in which Sally would be sitting next to him. Since she is equally likely to be in any of the other seats then P (N A) = 1/9. Moreover, if Joe is not in an end seat there are two choices of seats in which Sally could be seated next to him, this P (N A ) = 2/9. The total probability is then: P (S) = (.2)(1/9) + (8/10)(2/9) = 18/90 =.2 A third way to think about it is that there are 10 9 Possible choices of the seats that Joe and Sally could be seated in. If they are to be seated together then there are 9 choices of the seats that they could be in ((one and two), (two and three), (three and four), (four and five), (five and six), (six and seven), (seven and eight), (eight and nine) or (nine and 10)). Then there are 2 ways that Joe and Sally can sit in those seats (either Sally on the Right and Joe on the left or Sally on the left and Joe on the right). Thus there the probability of ways in which they can sit next to each other is 9 2/(10 9) =.2.