Midterm 1 - Solution

Transcription

1 Midterm 1 - Solution Instructors: Dr. A. Grine and Dr. A. Ben Ghorbal Sections: 170, 171, 172, 173 Name (Printed) Section Student ID Your Signature Total Points Exercise 1 5 Exercise 2 8 Exercise 3 7 Total 20 Score

2 Instructions. Your exam should contain 5 Pages total and 3 Exercises. completeness. Read instructions for each problem Carefully. Please check your test for NO book, NO notes but you may use a calculator which does not graph and which is not programmable. Show all your work to get full credit. You must explain how you get your answers using techniques developed in this class so far. Answer with no supporting work, obtained by guessand-check, or via other methods will result in little or no credit, even correct. Place a box around Your Final Answer to each question. Answer the equation in the space provided on the question sheets. If you need more room (space), use the backs of the pages and indicate to the reader that you have done so. If you are not sure what a question means, raise your hand and ask me. Check your work! Exercise 1 (5 Marks) If S = {x 0 < x < 12}, M = {x 1 < x < 9} and N = {x 0 < x < 5}, find 1. M N 2. M N 3. M c N c Solution: Since we have Then, we get Solution: We have M = (1, 9) and N = (0, 5). M N = (0, 9) = {x 0 < x < 9}. M N = (0, 9). Solution: Since the universal set is S = (0, 12), then we have M = (0, 1] [9, 12) and N = (0, 5). 1

3 Exercise 2 (8 Marks) How many different linear arrangements are there of the letters A, B, C, D, E, F for which 1. A and B are next to each other; 2. A is before B; 3. A is before B and B is before C; 4. A is before B and C is before D; 2

4 5. A and B are next to each other and C and D are also next to each other; 6. E is not last in line? Exercise 3 (7 Marks) A president, treasurer, and secretary, all different, are to be chosen from a club consisting of 10 people, denoted by A, B, C, D, E, F, G, H, I and J, respectively. How many different choices of officers are possible if 1. there are no restrictions; 3

5 2. A and B will not serve together; 3. C and D will serve together or not at all; 4

6 4. E must be an officer; 5. F will serve only if he is president. Good Luck! 5

7 Solutions - Midterm 2 Instructors: Dr. A. Grine and Dr. A. Ben Ghorbal Sections: 170, 171, 172, 173 Total Marks Exercise 1 6 Exercise 2 4 Exercise 3 6 Exercise 4 4 Total 20 Score 1

8 Exercise 1 (6 Marks) Suppose that A, B and C are independent events in an experiment with P(A) = 0.3, P(B) = 0.5 and P(C) = 0.8. Express each of the following events in set and find its the probability 1. None of the three events occurs. Solution: The event is represented by the following set A c B c C c. Since the three events A, B and C are independent then the probability that none of the three events occurs is given by P (A c B c C c ) = P (A c ) P (B c ) P (C c ) = [1 P (A)] [1 P (B)] [1 P (C)] = (1 0.3) (1 0.5) (1 0.8) = =

9 2. At least one of the three events occurs. Solution: The event is represented by the following set By using the probability axioms we obtain A B C. P (A B C) = 1 P (A c B c C c ) = = 0.93 There is another way to find the result, in fact since we asked that at least one of the three events occurs so it means that we have one of the following situation A B c C c, P (A B c C c ) = = 0.03, OR A c B C c, P (A c B C c ) = = 0.07, OR A c B c C, P (A c B c C) = = 0.28, OR A B C c, P (A B C c ) = = 0.03, OR A B c C, P (A B c C) = = 0.12, OR A B C, P (A c B C) = = 0.28, OR A B C, P (A B C) = = 0.12, OR. Then the probability that at least one of the three events occurs is given by the following P (A B C) = P (A B c C c ) + P (A c B C c ) + P (A c B c C) + P (A B C c ) +P (A B c C) + P (A c B C) + P (A B C) = =

10 3. Exactly one of the three events occurs. Solution: The event is represented by the following set (A B c C c ) (A c B C c ) (A c B c C). Then the probability that exactly one of the three events occurs is P ((A B c C c ) (A c B C c ) (A c B c C)) = P (A B c C c ) + P (A c B C c ) + P (A c B c C) = =

11 4. Exactly two of the three events occurs. Solution: The event is represented by the following set (A B C c ) (A B c C) (A c B C). Then the probability that exactly one of the three events occurs is P ((A B C c ) (A B c C) (A c B C)) = P (A B C c ) + P (A B c C) + P (A c B C) = =

12 Exercise 2 (7 Marks) A supermarket in Riyadh buys its eggs from three different chicken ranches. They buy 1/3 of their eggs from Eggs R Us, 1/2 of their eggs from Al Mazra ah, and 1/6 of their eggs from Cheap Eggs. The supermarket determines that 1% = 0.01 of the eggs from Eggs R Us are cracked, 2% = 0.02 of the eggs from the Al Mazra ah are cracked, and 5% = 0.05 of the eggs from Cheap Eggs are cracked. What is the probability that an egg chosen at random is from Cheap Eggs, given that the egg is cracked? Solution: We have the following tree diagram We have P (cracked) = P (cracked Eggs R Us) P(Eggs R Us) + P (cracked Al Mazra ah) P (Al Mazra ah) +P (cracked Cheap Eggs) P (Cheap Eggs) = = and so, P (Cheap Eggs cracked) P (Cheap Eggs cracked) = P (cracked) P (cracked Cheap Eggs) P (Cheap Eggs) = P (cracked) = =

13 Exercise 3 (6 Marks) Let X be a random variable with probability density function c for 0 < x < 1 f X (x) = x 0 otherwise 1. Find the value of the constant c. Solution: Since f X is the probability density function of a given random variable then we need to choose c such that the sum of probability distribution function is equal to 1, i.e. + f X (x)dx = 1 The integral of the probability distribution function is and so we have + f X (x)dx = 1 0 c [ dx = 2 c ] 1 x = 2c x 0 2c = 1 = c = 1 2 = 0.5. Therefore, the continuous random variable X has the probability density function f X (x) = 1 2 x for 0 < x < 1 0 otherwise 7

14 2. Give the cumulative distribution function of X. Solution: By definition the cumulative distribution function is given by F X (x) = x f X (t)dt Then we distinguish the following three situations: 1 st case: x 0, so t x 0. Thus we have f X (t) = 0 and so F X (x) = 0 2 nd case: 0 < x < 1, then F X (x) = = x 0 f X (t)dt f X (t)dt + x 0 f X (t)dt = 0 + = x t dt [ ] x t = x 0 3 rd case: x 1, we have F X (x) = = x 0 f X (t)dt f X (t)dt + 1 f X (t)dt + x 0 1 f X (t)dt = 0 + = t dt + 0 Therefore, the continuous random variable X has the cumulative distribution function 0 for x 0 F X (x) = x for 0 < x < 1 1 for x 1 8

15 ( 1 3. Find P 2 < X < 3 ) (, P X = 1 ) ( and P X > 2 ) Solution: We have ( 1 P 2 < X < 3 ) 4 Since X is a continuous random variable ( P X = 1 ) = 0 8 ( = P X 3 ) ( P X 1 ) 4 2 ( ) ( ) 3 1 = F F = = 2 By using the same arguments as above we get ( P X > 2 ) ( = 1 P X 2 ) 5 5 ( ) 2 = 1 F 5 2 = = 5 9

16 4. Find E[X], Var(X) and σ X. Solution: variable we get Using the definition of the the expectation of a given continuous random E[X] = = + 0 = xf X (x)dx xf X (x)dx + x xdx = [ = 1 x = 1 3 ] x dx + 0 Now using the definition of the variance we get Var[X] = E [ X 2] (E[X]) 2 + ( 1 = x 2 f X (x)dx 3 = 0 = 0 + = 1 2 [ = x 2 f X (x)dx + x 2 x dx 9 x ] x dx xf X (x)dx + ) 2 x 2 f X (x)dx xf X (x)dx x 2 f X (x)dx 1 9 Therefore, = = 9 5 = σ Y = Var[X] = 4 45 =

17 Exercise 4 (4 Marks) Let X be a random variable defined as follows. An urn contains three balls numbered 1, 2 and 3. A fair coin is flipped; if the coin comes up heads, a ball is drawn from the urn and X is the number on the ball; if the coin comes up tails two balls are drawn without replacement from the urn and X is the sum of the numbers on the balls. 1. Compute the probability mass function. Solution: We first obtain the following table: s P({s}) X(s) H, 1 1/6 1 H, 2 1/6 2 H, 3 1/6 3 T, 1, 2 1/12 3 T, 1, 3 1/12 4 T, 2, 1 1/12 3 T, 2, 3 1/12 5 T, 3, 1 1/12 4 T, 3, 2 1/12 5 Thus the probability mass function is given the following table x f X (x) = P(X = x) = = =

18 2. Find the expectation E[X] and the variance Var(X). Solution: Using the definition of the expectation of a discrete random variable we get Now for the variance we have Therefore, E[X] = = 3 E[X 2 ] = = = 64 6 = 32 6 Var[X] = E [ X 2] (E [X]) 2 = = =

19 Instructors: Dr. A. Grine and Dr. A. Ben Ghorbal Sections: 170, 171, 172, 173 Solution Midterm 2 Total Points Exercise 1 5 Exercise 2 8 Exercise 3 7 Total 20 Score

20 Exercise 1 (??? Marks) The game Yahtzee (dice poker) is played by simultaneously rolling 5 dice and making a poker hand from the top faces of the dice. Find the probability that 1. All five dice have different values. Solution: If we assume that all 5 dice are distinguishable (as suggested) then there are N = 65 = 7776 possible Yahtzee hands. If all 5 dice are to have different values then there are 6 ways we can select the first number, 5 ways to select the second, 4 ways to select the third, 3 ways to select the fourth and 2 ways to select the fifth distinct number. Thus there are n = = 6! = 720 possible hands with all 5 numbers being distinct and so P (5distinct) = = There is exactly one pair in the five values. Solution: To get a single pair, we first select what number the pair will be. There ( are ) 5 6 ways to do this. Now we decide on which 2 dice the will pair appear. There are 2 ways to select the dice. Finally we need to select the numbers on the three remaining dice. There are 543 ways to do that. ( ) 5 n = = = The required probability is then P (one pair) = =

21 3. There are two different pairs of values. Solution: For two pairs, we follow ( ) the same logic as before. Now we need to select 6 2 numbers to be pairs. There are ways of doing that. Then we must place these 2 ( ) 5 two pairs. Let us place the numerically smallest pair first. There are ways to place ( ) 2 3 this pair. Then there are only 3 spots left so there are ways to place the numerically 2 higher pair. Finally there are 4 remaining numbers that can fill the last spot. ( ) ( ) ( ) n = 4 = = The probability of two pairs is then P (two pairs) = = There is a full house (3 of one value and 2 of another value). Solution: To get a full house there are 6 ways to choose the number ( ) to appear three 5 times and 5 ways to select the number to appear in the pair. There are ways to place 2 the pair and then the other number must fill up all remaining spots. Thus we get ( ) 5 n = 6 5 = = and hence P (full house) = =

22 Exercise 2 (8 Marks) Suppose that the probability of exposure to the flu virus during flu season is 0.4 = 40%. People can get a flu vaccine which prevents the vaccinated person getting the flu, if exposed, in 0.75 = 75% of cases. People who are not vaccinated get the flu 0.8 = 80% of the time if they are exposed to the virus. It is impossible for a person to get the flu if they are not exposed to the virus. Suppose that two friends, Scott and Kevin spend flu season in different places and are not in physical contact with the same people. Scott received the flu vaccine but Kevin did not. 1. What is the probability that exactly one of these two will get the flu? Solution: Let A 1 be the event that Scott is exposed to the flu virus, B 1 that Scott gets the flu and C 1 the event that Scott gets the vaccine. Similarly define A 2, B 2 and C 2 for Kevin; A 2 be the event that Kevin is exposed to the flu virus, B 2 that Kevin gets the flu and C 2 the event that Kevin gets the vaccine. The information in the question tells us that P (A i ) = 0.4 for i = 1, 2 P (B c i A i C i ) = 0.75 for i = 1, 2 P (B i A i C c i ) = 0.8 for i = 1, 2 Note that a person cannot get the flu unless exposed so P (B i A c i) = 0 for i = 1, 2. Another way of putting this is that B i A i. Since the two friends are not in contact with each other or the same people we can also assume that the events for Scott are independent of those for Kevin. We wish to find P ((B 1 B c 2) (B c 1 B 2 ) C 1 C c 2). Since the two events are mutually exclusive we can write P ((B 1 B c 2) (B c 1 B 2 ) C 1 C c 2) = P (B 1 B c 2 C 1 C c 2) + P (B c 1 B 2 C 1 C c 2). Furthermore due to the independence of the events we can write P (B 1 B c 2 C 1 C c 2) = P (B 1 C 1 ) P (B c 2 C c 2) P (B c 1 B 2 C 1 C c 2) = P (B c 1B 2 C 1 ) P (B 2 C c 2) 4

23 We can apply the law of total probability to find these probabilities P (B 1 C 1 ) = P (B 1 A 1 C 1 ) P (A 1 ) + P (B 1 A c 1 C 1 ) P (A c 1) = = 0.10 P (B2 C c 2) c = P (B2 A c 2 C2) c P (A 2 ) + P (B2 A c c 2 C2) c P (A c 2) = = 0.68 P (B1 C c 1 ) = 1 P (B 1 C 1 ) = 0.90 P (B 2 C2) c = 1 P (B2 C c 2) c = 0.32 Hence we have P (Exactly one gets the flu) = = If exactly one of them does get the flu, what is the probability it is Kevin? Solution: We have P (Kevin gets flu Exactly one gets the flu) P (Kevin gets flu AND exactly one gets flu) = P (Exactly one gets the flu) = P (Bc 1 B 2 C 1 C2) c = =

24 Exercise 3 (7 Marks) A supermarket in Riyadh buys its eggs from three different chicken ranches. They buy 1/3 of their eggs from Eggs R Us, 1/2 of their eggs from Al Mazra ah, and 1/6 of their eggs from Cheap Eggs. The supermarket determines that 1% = 0.01 of the eggs from Eggs R Us are cracked, 2% = 0.02 of the eggs from the Al Mazra ah are cracked, and 5% = 0.05 of the eggs from Cheap Eggs are cracked. What is the probability that an egg chosen at random is from Cheap Eggs, given that the egg is cracked? Solution: We have the following tree diagram We have P (cracked) = P (cracked Eggs R Us) P (Eggs R Us) + P (cracked Al Mazra ah) P (Al Mazra ah) +P (cracked Cheap Eggs) P (Cheap Eggs) = = and so, P (Cheap Eggs cracked) = P (Cheap Eggs cracked) P (cracked) = P (cracked Cheap Eggs) P (Cheap Eggs) P (cracked) = =

25 Exercise 4 (7 Marks) A discrete random variable, Y, has probability mass function (p.m.f.) f Y (y) = P (Y = y) = c (y 3) for y = 2, 1, 0, 1, Find the value of the constant c. Solution: Writing the probability mass function as a table we have y f Y (y) 25c 16c 9c 4c c We need to choose c such that the sum of the probability mass function is equal to 1. The sum of the probability mass function is and so we have 2 P (Y = y) = 55c y= 2 55c = 1 = c = Therefore, the discrete random variable, Y, has probability mass function (p.m.f.) f Y (y) = P (Y = y) = 1 (y 3) for y = 2, 1, 0, 1,

26 2. Give the cumulative distribution function of Y. Solution: c. First we will write the probability mass function with the appropriate value of y f Y (y) Thus we get the cumulative distribution function 0 for y < 2 25 for 2 y < for 1 y < 0 F (y) = for 0 y < for 1 y < for y 2 8

27 3. Find the mean and variance of Y. Solution: The mean (or expected value) of Y is E[Y ] = ( 2) ( 1) = = The variance is best calculated as Var(Y ) = E [ (Y E[Y ]) 2] = E [ Y 2] (E[Y ]) 2 where E [ Y 2] = ( 2) ( 1) = = =

28 Exercise 5 (7 Marks) Let X be a random variable with probability density function c for 0 < x < 1 f X (x) = x 0 otherwise 1. Give the cumulative distribution function of X. 10

29 ( 1 2. Find P 2 < X < 3 ) (, P X = 1 ) ( and P X > 2 )

30 3. Find E[Y ], Var(Y ) and σ Y. 12

31 بسم اهلل الرحمن الرحيم SEMESTER: FIRST YEAR: 1428/1429 COURSE: Math 301 DATE: 25/10/1428 DURATION: 2 HOURS Midterm 1 - SOLUTION Instructor: Dr. A. S. Ben Ghorbal SECTIONS: Name Section Student ID Your Signature TOTAL MARKS EXERCISE 1 05 EXERCISE 2 03 EXERCISE 3 04 EXERCISE 4 04 EXERCISE 5 04 TOTAL 20 SCORE Dr. A. S. BEN GHORBAL Page 1 of 8

32 بسم اهلل الرحمن الرحيم SEMESTER: FIRST YEAR: 1428/1429 COURSE: Math 301 DATE: 25/10/1428 DURATION: 2 HOURS Midterm 1 Instructor: Dr. A. Ben Ghorbal SECTIONS: Instructions: Your exam should contain 08 Pages total (including the first page) and 5 Exercises Please check your test for completeness. Read instructions for each problem carefully. NO book, NO notes but you may use a calculator which does not graph and which is not programmable. Show all your work to get full credit. You must explain how you get your answers using techniques developed in this class so far. Answer with no supporting work, obtained by guess-and-check, or via other methods will result in little or no credit, even correct. Place a box around Your Final Answer to each question. Answer the equation in the space provided on the question sheets. If you need more room (space), use the backs of the pages and indicate to the reader that you have done so. If you are not sure what a question means, raise your hand and ask me. Check your work! Good Luck! مع دعائنا لكم بالتوفيق Dr. A. S. BEN GHORBAL Page 2 of 8

33 EXERCISE 1. (05 MARKS) Fill-in-the-Blank Questions 1. A well-defined collection of objects constitutes a of interest. ANSWER: population 2. statistics involves summarizing and describing important features of the data. ANSWER: Descriptive 3. A variable x almost results from counting, in which case possible values are 0, 1, 2, or some subset of these integers. ANSWER: discrete 4. The number of traffic citations a person received during the last year is not an example of a variable. ANSWER: continuous 5. A histogram is if the left half is a mirror image of the right half. ANSWER: symmetric Dr. A. S. BEN GHORBAL Page 3 of 8

34 EXERCISE 2. (03 MARKS) Multiple-Choice Questions 1. Which of the following are examples of a variable? A. Gender of a high school graduate B. Number of major credit cards a person has C. Type of automobile transmission D. All of the above ANSWER: D 2. Which of the following statements are false? A. A variable is continuous if its set of possible values either is finite or else can be listed in an infinite sequence in which there is a first number, a second number, and so on. B. Continuous variables arise from making measurements. C. The frequency of any particular observation of a discrete variable x is the number of times that value occurs in the data set. D. In theory, the relative frequencies should sum to 1.0, but in practice the sum may differ slightly from 1.0 due to rounding. E. None of the above. ANSWER: A 3. Which of the following is not a measure of location? A. The mean B. The median C. The mode D. The variance E. All of the above ANSWER: D Dr. A. S. BEN GHORBAL Page 4 of 8

35 4. Which of the following is not a measure of variability? A. Variance B. Standard deviation C. Mean absolute deviation D. Median E. All of the above ANSWER: D 5. Which of the following is used as a divisor in the sample variance s 2, where n is the sample size? A. n+1 B. n C. n-1 D. n-2 ANSWER: C 6. i 2 Given that n 10, x 25, and x 512, thenthe sample standard deviation is i A B C D. None of the above answers is correct ANSWER: C Dr. A. S. BEN GHORBAL Page 5 of 8

36 EXERCISE 3. (04 MARKS) The accompanying data specific gravity values for various wood types used in construction Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display. ANSWER: By reordering the data we get the following data table Using the normal stem-and Leaf where leaf is hundredths and stem is tenths L L L H 5 Another method of denoting the pairs of stems having equal values is to denote the stem by L, for low and the second stem by H, for high. Using this notation, the stem-and-leaf display would appear as follows: 3L 1 stem: tenths 3H leaf: hundredths 4L H 5L 144 5H 58 6L 2 6H L 7H 5 The stem-and-leaf display on the previous page shows that.45 is a good representative value for the data. In addition, the display is not symmetric and appears to be positively skewed. The spread of the data is =.44, which is.44/.45 =.978 or about 98% of the typical value of.45. This constitutes a reasonably large amount of variation in the data. The data value.75 is a possible outlier. Dr. A. S. BEN GHORBAL Page 6 of 8

37 EXERCISE 4. (04 MARKS) The cumulative frequency and cumulative relative frequency for a particular class interval are the sum of frequencies and relative frequencies, respectively, for that interval and all intervals lying below it. Compute the cumulative frequencies and cumulative relative frequencies for the following data: ANSWER: Class Frequency Relative Frequency Cumulative Frequency Cumulative Relative Frequency 60 under under under under under under under under Dr. A. S. BEN GHORBAL Page 7 of 8

38 EXERCISE 5. (04 MARKS) Consider the following observations on shear strength of a joint bonded in a particular manner: a. Determine the value of the sample mean. b. Determine the value of the sample median. Why is it so different from the mean? c. Calculate the upper quartile, lower quartile and the IQR. ANSWER: a. The sum of the n = 11 data points is , so x = /11 = b. The sample size (n = 11) is odd, so there will be a middle value. Sorting from smallest to largest: The sixth value, 36.6 is the middle, or median, value. mean differs from the median because the largest sample observations are much further from the median than are the smallest values. c. Q 1? The position of the lower quartile Q 1 is 0.25(n + 1) = 0.25 (11 + 1) = = 3. Then the lower is the 3 rd value and is equal to Q 2? The second quartile Q 2 is equal to the median Q 2 = Q 3? The position of the upper quartile Q2 is 0.75(n + 1) = 0.75 (11 + 1) = = 9. Then the lower is the 9 th value and is equal to IQR? The Dr. A. S. BEN GHORBAL Page 8 of 8

39 SOLUTIONS MIDTERM 1 SEMESTER: SECOND YEAR: 1429/1430 COURSES: MAT 301 & MAT 141 & STA 111 DATE: 26/04/1430 (22/04/2009) DURATION: 2 HOURS Instructors: Drs. A. MUSTAFA, A. BARKAOUI, B. CHOURAR & A. S. BEN GHORBAL Name Section Student ID Your Signature TOTAL MARKS SCORE EXERCISE EXERCISE EXERCISE EXERCISE EXERCISE EXERCISE TOTAL IMAMU Page 1 of 11

40 EXERCISE 1. (6 MARKS) A sample of 36 mice was used to investigate the use of iron in Fe+ form as a dietary supplement. The iron was given orally and was radioactively labeled so that the exact percentage of iron retained could be measured accurately. The measurements were ) Find the mean, variance and standard deviation of this data. SOLUTIONS: We have x i ORDER THE DATA So we get xi xi - x ( x - x) i Mean: x = Variance: s = Standard Deviation: s = 7.09 x i ORDER THE DATA continued xi xi - x ( x - x) i IMAMU Page 2 of 11

41 2) Produce a stem-and-leaf display. SOLUTIONS: 1 st METHOD: We have Since the data is so skewed, a sensible approach is to round to the nearest whole number, and draw a double stem-leaf plot, the first value corresponds to leaf values of 0-4, and the second values corresponds to values of nd METHOD: We have (Note: the first four zeroes correspond to the values 0.1, 0.2, 0.2 and 0.3; 1.50 the 0.7 is represented as a 1.) Since the data is so skewed, a sensible approach is to round to the nearest whole number, and draw a double stem-leaf plot, the first value corresponds to leaf values of 0-4, and the second values corresponds to values of IMAMU Page 3 of 11

42 3) Find the median and the quartiles. SOLUTIONS: We recall that n = 36. So the median is the average between the data at 18 th and 19 th positions (after arranging the data in the increasing way!) = 4.05 Now for the quartiles we have Position of Q 1 is in between 9 th and 10 th data (since Q Position of Q 2 is the median so = = Q 2 = 4.05 Position of Q 3 is in between 27 th and 28 th data (since Q = = n = = 9.25 ) so 4 4 ( n ) = 27.75) so ) Find IQR SOLUTIONS: We have IQR = Q - Q = = There are 3 outliers (21.00, 24.00, and 29.10) in the data set. IMAMU Page 4 of 11

43 0.50 EXERCISE 2. (3 MARKS) Suppose that we are given the following data Construct the frequency, relative frequency and cumulative frequency based on class intervals having as length 50. SOLUTIONS: Class Interval Frequency [0, 50) 9 [50, 100) 19 [100, 150) 11 [150, 200) 3 [200, 250) 2 [250, 300) 2 [300, 350) 1 [350, 400) 2 [400, 450) 1 [450, 500) 1 [500, 550) Relative Frequency TOTAL Cumulative Frequency Cumulative Relative Frequency = = = = = = = = = = = IMAMU Page 5 of 11

44 EXERCISE 3. PART A (4 MARKS) A jar contains 2 red, 2 green and 1 blue beads. Two beads are drawn without replacement. Use a tree diagram to illustrate the outcomes and figure out the probability of drawing at least one red bead. SOLUTIONS Here is a "tree diagram" for this problem. The fractions in parentheses give the probabilities a bead of the indicated color being drawn at each stage. For example, the figure (2/5) after "Red" in the "First Draw" column comes from the fact that at this stage there are 2 red beads out of 5 beads all together in the jar. The figure (1/4) in the top box in the "Second Draw" column comes from the fact that now, after one red has been removed, there is only 1 red of 4 beads First Draw Second Draw Outcome Probability Red (1/4) RR (2/5)(1/4)=1/10 Red (2/5) Green (2/4) RG (2/5)(2/4)=1/5 Blue (1/4) RB (2/5)(1/4)=1/10 Red (2/4) GR (2/5)(2/4)=1/5 Green (2/5) Green (1/4) GG (2/5)(1/4)=1/10 Blue (1/4) GB (2/5)(1/4)=1/10 Blue (1/5) Red (2/4) BR (1/5)(2/4)=1/10 Green (2/4) BG (1/5)(2/4)=1/10 The event "at least one bead is red" is the sum of the probabilities of the outcomes in this event is 7/10, in fact ( at least one bead is red) = ( ) + ( ) + ( ) + ( ) + ( ) P P RR P RG P RB P GR P BR = = IMAMU Page 6 of 11

45 PART B A jar contains 2 red, 2 green and 1 blue beads. Two beads are drawn with replacement. Use a tree diagram to illustrate the outcomes and figure out the probability of drawing at least one red bead. SOLUTIONS: Here is the "tree diagram" for this problem First Draw Second Draw Outcome Probability Red (2/5) RR æ2ö æ2ö 4 ç ç = è5 ø è5 ø 25 Red (2/5) Green (2/5) RG æ2ö æ2ö 4 ç ç = è5 ø è5 ø 25 Blue (1/5) RB æ2ö æ1ö 2 ç ç = è5 ø è5 ø 25 Red (2/5) GR æ2ö æ2ö 4 ç ç = è5 ø è5 ø 25 Green (2/5) Green (2/5) GG æ2ö æ2ö 4 ç ç = è5 ø è5 ø 25 Blue (1/5) GB æ2ö æ1ö 2 ç ç = è5 ø è5 ø 25 Red (2/5) BR æ1ö æ2ö 2 ç ç = è5 ø è5 ø 25 Blue (1/5) Green (2/5) BG æ1ö æ2ö 2 ç ç = è5 ø è5 ø 25 Blue (1/5) BB æ1ö æ1ö 1 ç ç = è5 ø è5 ø 25 The event "at least one bead is red" is the following sum ( at least one bead is red) = ( ) + ( ) + ( ) + ( ) + ( ) P P RR P RG P RB P GR P BR = = IMAMU Page 7 of 11

46 EXERCISE 4. (3 MARKS) Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Visa Credit Card and B be the analogous event for a Master Card. Suppose that ( ) ( ) ( ) P A = 0.5, P B = 0.4 and P AÇ B = ) Compute the probability that the selected individual has at least one of the two types of cards). SOLUTIONS: We have P ( The selected individual has at least one of the two types of cards) ( B) 1.50 ( ) ( ) ( ) = P AÈ = P A + P B -P AÇB = = ) Find the probability that the selected individual has neither type of card. SOLUTIONS: We have P ( The selected individual has neither of the two types of cards) P( ( A c B) ) P( A B) 1.50 = È = 1- È = = 0.35 IMAMU Page 8 of 11

47 EXERCISE 5. (4 MARKS) A box in a certain supply room contains four 40-W light bulbs, five 60-W light bulbs, and six 75-W light bulbs. Suppose that three light bulbs are randomly selected. 1. Find the probability that exactly two of the selected light bulbs are rated 75-W. SOLUTIONS: We have P ( Selecting exactly two 75-W light bulbs) æ6ö æ9ö ç ç = è ø è ø = =» æ15ö ç è 3 ø 2. Find the probability that all three of the selected light bulbs have the same rating. SOLUTIONS: We have P ( All three of the selected light bulbs have the same rating) ( Three 40-W light bulbs) P( Three 60-W light bulbs) + P( Three 75-W light bulbs) = P + æ4ö æ5ö æ6ö ç + ç + ç = è ø è ø è ø = =» æ15ö ç è 3 ø IMAMU Page 9 of 11

48 3. Find the probability that one light bulb of each type is selected. SOLUTIONS: We have P ( One light bulb of each type is selected) æ4ö æ5ö æ6ö ç ç ç = è ø è ø è ø =» æ15ö 455 ç è 3 ø 4. Suppose now that light bulbs are to be selected one by one until a 75- W is found. Find the probability it is necessary to examine at least six light bulbs. SOLUTIONS: To examine exactly one, a 75 watt bulb must be chosen first. (6 ways to accomplish this). To examine exactly two, we must choose another wattage first, then a 75 watt. ( 9 6 ways). Following the pattern, for exactly three, ways; for four, ; for five, P ( examine atleast 6 bulbs) = 1 P( examine 5 or less) P( ) ép( ) P( ) K P( ) ù = 1- examine exactly 1 or 2 or 3 or 4 or 5 = 1- ë One + Two + + Five û é ù = 1- ê ë ú û = [ ] = = IMAMU Page 10 of 11

49 EXERCISE 6. (3 MARKS) A single fair die is rolled and then a fair coin is flipped twice. 1) Describe a sample space S giving all outcomes from this experiment. SOLUTIONS: We have ( ) { 1,2,3,4,5,6 } {, } {, } ( ) ì( H H)( H H)( H H)( H H)( H H)( H H) ï ï( 1, HT, )( ; 2, HT, )( ; 3, HT, )( ; 4, HT, )( ; 5, HT, )( ; 6, HT, ); ï( 1, T, H)( ; 2, T, H)( ; 3, T, H)( ; 4, T, H)( ; 5, T, H)( ; 6, T, H) ; ï( 1, TT, )( ; 2, TT, )( ; 3, TT, )( ; 4, TT, )( ; 5, TT, )( ; 6, TT, ); S= HT HT Þ ns = = 1,, ; 2,, ; 3,, ; 4,, ; 5,, ; 6,, ; ü ï ï =í ý ï ï î þ 2) Assume all outcomes in question 1) have the same probability. Find the probability of the following events: A: 6 is rolled and at least one head turns up ; { } ( ) A= 6, H, H ; 6, H, T ; 6, T, H Þ n A = 3. SOLUTIONS: We have ( ) ( ) ( ) n( A) 3 1 Then, P( A) = = = n( S) 24 8 B: an even number is rolled and head turns up on the second toss ; SOLUTIONS: B= 2, H, H ; 4, H, H ; 6, H, H ; 2, T, H ; 4, T, H ; 6, T, H. { } We have ( ) ( ) ( ) ( ) ( ) ( ) n( B) 6 1 Then, P( B) = = = n( S) C: at least one head turns up and a number less than 5 is rolled. SOLUTIONS: We have ìï( H H) ( H H) ( H H) ( H H) ( H T) ( H T) C ïî( 3, H, T)( ; 4, H, T)( ; 1, T, H)( ; 2, T, H)( ; 3, T, H)( ; 4, T, H) ; nc ( ) 12 1 Then, P( C) = = = 0.75 n( S) ,, ; 2,, ; 3,, ; 4,, ; 1,, ; 2,, ; üï =í ý ïþ IMAMU Page 11 of

50

51

52

53

54

55

56

57

58

59