Prelim 1 (10 8)!8! = 10 9 = 45. (b) How many ways can it be done if the first three questions are required? 2 = 21

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Math 1105 Fall 010 J.T. Gene Hwang, Instructor Robyn Miller, T.A. Prelim 1 Instructions This is a closed book exam. Graphing calculators are not permitted. Solutions should be written in your blue book in an orderly, legible manner. Showing all work is strongly recommended; incorrect answers without work will receive no credit. Each part of a problem is worth 4 points. 1. There are 10 questions on an exam, of which students must answer exactly 8. Keep in mind that submitted exams contain no evidence of the order in which problems were done, so order does not matter for parts (a) and (b). However for part (c), as the set of exam questions is fixed, the order of presentation is all that can be varied. (a) How many different ways could this be done? ( ) 10 10! 8 (10 8)!8! 10 9 45 (b) How many ways can it be done if the first three questions are required? ( ) 7 7! 5 (7 5)!5! 7 6 1 (c) When the instructor was creating the exam, he knew 3 of the 10 questions would have to appear in a fixed sequence at the beginning of the exam. Subject to this constraint, how many ways could he arrange the questions on the exam? 7! 5040. Find the number of distinct permutations of the letters in the following words. (a) FIRST 5! 5! 10 1!1!1!1!1! (b) LABEL 5!!1!1!1! 10 60 1

(c) UNUSUAL 7! 3!1!1!1!1! 5040 840 6 (d) DEEDED 6! 3!3! 70 36 0 3. Find the probability of getting the following sums when two fair dice are rolled independently. (a) 1 0 (b) 1 36 (c) 3 36 1 18 (d) 4 3 36 1 1 4. In a bag there are 5 blue, red and 4 green marbles, all identically shaped. We draw two marbles from the bag without replacement. (a) Given that the first draw is a blue marble, what is the probability that the second is green? Once a blue marble has been drawn, 4 blue, red and 4 green marbles remain. So P (nd draw green 1st draw was blue) 4 4 4 4 10 5 (b) What is the probability that the second draw is a green marble? Is the event of getting a blue marble in the first draw independent of the event of getting a green marble in the second? To receive credit you must supply a precise mathematical argument. Define four events: G 1 {1st draw is green}, R 1 {1st draw is red}, B 1 {1st draw is blue}, G {nd draw is green}. The probability that the second draw is green is the sum of the probabilities of the event intersections G 1 G, B 1 G and R 1 G. Recall that P (A B) P (A B)P (B), so: P (G ) P (G 1 G ) P (B 1 G ) P (R 1 G ) P (G G 1 )P (G 1 ) P (G B 1 )P (B 1 ) P (G R 1 )P (R 1 ) ( ) ( ) ( ) ( ) ( ) ( ) 3 4 5 10 11 5 11 5 11 1 110 0 110 8 110 40 110 4 11

(c) What is the probability that both are green? See above. P (G G 1 ) P (G G 1 )P (G 1 ) ( ) ( 3 4 ) 1 10 11 110 6 (d) What is the probability that at least one is green? Using your answers to (b) and (c): P ({at least one is green}) P (G 1 G ) P (G 1 ) P (G ) P (G 1 G ) 4 11 4 11 6 34 (e) What is the probability that both are the same color? Define two new events:b {nd draw is blue} and R {nd draw is red}. Again, use your answers and methods of answering previous parts of the problem: P ({both same color}) P (G 1 G ) P (B 1 B ) P (R 1 R ) P (G G 1 )P (G 1 ) P (B B 1 )P (B 1 ) P (R R 1 )P (R 1 ) ( ) ( ) ( ) ( ) ( ) ( ) 3 4 4 5 1 10 11 10 11 10 11 1 0 110 17 5. Let A be the set {0, 1,, 3, 4, 5, 6, 7, 8, 9}. (a) How many subsets does A have? A 10 104 (b) How many subsets contain 5? There are ( A 1) 9 subsets of A {1,, 3, 4, 6, 7, 8, 9} A. Appending a 5 to each subset B A (including the B ) produces every subset B A with 5 B. Thus there are 9 subsets of A containing 5. (c) Randomly pick a subset of A. What is the probability that this subset contains 5? 9 10 1 3

(d) What is the probability that the subset contains 5 and 6? Either show that the event that a subset B A contains 6 and the event that B contains 5 are independent, then justifiably claim that P ({5 B} {6 B}) P ({5 B})P ({6 B}) ( ( 1 1 ) ) or use the same reasoning as in (b) to show that the number of subsets B of A containing both 5 and 6 is ( A ) 8 and thus P ({5 B} {6 B}) 8 10 1 4 (e) What is the conditional probability that the chosen set contains 6, given that it contains 5? Let B be a randomly chosen subset of A. P (6 B 5 B) P ({6 B} {5 B}) P ({5 B}) ( 1 4) ( 1 ) 1 (f) Is the event of picking a subset of A containing 5 independent of the event of picking a subset of A containing 6? To receive credit you must supply a precise mathematical argument. These events are independent because for a randomly chosen subset B of A the following two equivalent conditions hold: i. P (6 B 5 B) 1 P (6 B) ii. P ({5 B} {6 B}) 1 4 P (5 B)P (6 B) 6. A recent Maryland highway safety study found that in 77% of accidents the driver was wearing a seat belt. Accident reports indicate that 9% of those drivers escaped serious injury (defined as a hospitalization or death), but only 63% of the non-belted drivers were so fortunate. In your written solutions please use B and NB to denote, respectively, the sets of belted and non-belted drivers; use I and N I for injured and non-injured drivers. (a) What is the probability that a driver in an accident was seriously injured and did not wear a seat belt? P (I NB) P (I NB)P (NB) (1 P (NI NB)) (1 P (B)) ( ) ( ) 37 3 100 100 851 10000 0.0851 4

(b) Given that a person was seriously injured, what is the probability that he was not wearing a seat belt? P (I NB)P (NB) P (NB I) P (I NB)P (NB) P (I B)P (B) ( 851 ) ( ) ( ) 10000 8 77 ) 100 100 ( 851 10000 851 851 615 0.58 (c) Is the event of wearing a seat belt independent of the event of escaping serious injury? To receive credit you must supply a precise mathematical argument. No, the events B and NI are dependent. P (NI B) 0.9 while P (NI NB) 0.63. If the events were independent then P (NI B) P (NI NB) P (NI), but P (NI B) P (NI NB), so B and NI are dependent events. 5