MATH 131 (062) Finite Mathematics. Chapter 8: Probability. June 3, 2007. Dr. Raja Latif and Mohammad latif and Abdul Latif and Dr. Raja Mohammad Abdul Latif (a) At one time, licence plates were issued that consisted of three letters followed by three digits. How many diп¬Ђerent licence plates can be issued under this arrangement? (b) Later on, licence plates were issued that consisted of three digits followed by three letters. How many diп¬Ђerent licence plates can be issued under this arrangement? Solution. (a) The number of licence plate numbers that may be formed is (26) (26) (26) (10) (10) (10) , or 17, 576, 000. (b) The number licence plate numbers that may be formed is (10) (10) (10) (26) (26) (26) = 17, 576, 000. Example.360TAN20. Licence Plate Numbers: In recent years the state of California issued licence plates using a combination of one letter of the alphabet followed by three digits, followed by another three letters of the alphabet. How many diп¬Ђerent licence plates can be issued using this configuration? Solution. (26) (10) (10) (10) (26) (26) (26) = 316, 342, 000. Example.360TAN21. Exams: An exam consists of ten true-or-false questions. Assuming that every question is answered, in how many diп¬Ђerent ways can a student complete the exam? In how many ways the exam be completed if a penalty is imposed for each incorrect answer, so that a student may leave some questions unanswered? Solution. If every question is answered, there are 210 , or 1024, ways. In the second case, there are 3 ways to answer each question, and so we have 310 , or 59, 049, ways. Example.360TAN22. WARRANTY NUMBER: A warranty identification number for a certain product consists of a letter of the alphabet followed by a five-digit number. How many possible identification numbers are there if the first digit of the five-digit number must be nonzero? Solution. (26) (9) (10) (10) (10) (10) = 2, 340, 000. Example.360TAN24.TELEPHONE NUMBERS: (a) How many seven-digit telephone numbers are possible if the first digit must be nonzero? (b) How many international direct-dialing numbers are possible if each number consists of a three-digit area code (the first digit of which must be nonzero) and a number of the type described in part (a)? Solution. (a) If the first digit must be nonzero, then there are 9 possible choices for the first digit and 10 possible choices for the remaining 6 digits. Thus, the number of possible seven-digit numbers is given by (9) (10) (10) (10) (10) (10) (10) = 9, 000, 000. Contents 8.1-2:Basic Counting Principle and Permutations;Combinations and Other Counting Principle Example.360TAN9. Health-Care Plan Options: A new state employee is oп¬Ђered a choice of ten basic health plans, three dental plans, and two vision care plans. How many diп¬Ђerent health care plans are there to choose from each category? Solution. By the multiplication principle, we see that the number of ways a health-care plan be selected is (10) (3) (2) = 60. Example.360TAN10.Code Words: How many three code words can be constructed from the first ten letters of the Greek alphabet if no repetitions are allowed? Solution. Using the multiplication principle, we see that the number of three-letter code words that can be formed is (10) (9) (8) = 720, or 720 ways. Example.360TAN11. Social Security Number: A social security number has nine digits. How many Social Security numbers are possible? Solution. 109 = 1, 000, 000, 000. Example.360TAN12. Serial Numbers: Computers manufactured by a certain company have a serial number consisting of a letter of the alphabet followed by a four-digit number. If all the serial numbers of this type have been used, how many sets have already been manufactured? Solution. The number of sets that have already been manufactured is (26) (10) (10) (10) (10) = 260, 000. Example.360TAN14. Automobile Selection: An automobile manufacturer has three diп¬Ђerent subcompact cars in the line. Customers selecting one of these cars have a choice of three engine sizes, four body styles, and three color schemes. How many diп¬Ђerent selections can a customer make? Solution. The number of selections a customer can make is (3) (3) (4) (3) = 108. Example.360TAN17. ATM CARDS: To gain access to his account, a customer using an automatic teller machine (ATM) must enter a four-digit code. If repetition of the same four digits is not allowed (for example, 5555), How many combinations are there? Solution. The number of diп¬Ђerent selections is (10) (10) (10) (10) в€’ 10 = 10000 в€’ 10 = 9990. Example.360TAN19. Licence Plate Numbers: Over the years the state of California has used diп¬Ђerent combinations of letters of the alphabet and digits on its automobile licence plates. 1 (b) Using the multiplication principle, we find that the number of international directdialing numbers is given by (9) (10) (10) Г— 9, 000, 000 = 8, 100, 000, 000. Example.373TAN32. How many three-letter permutations can be formed from the first five letters of the alphabet? Solution. The number of 3в€’ letter permutations is 5! P (5, 3) = = 5.4.3 = 60. 2! Example.373TAN34. In how many ways can five people line up at a checkout counter in a supermarket? Solution. The number of diп¬Ђerent ways they can line 5! = 120. up is P (5, 5) = 0! Example.373TAN37. In how many ways can a member of a hiring committee select 3 of 12 job applicants for further consideration? Solution. The number of diп¬Ђerent ways the 3 candi12.11.10 12! = = 220. dates can be selected is C (12, 3) = 9!3! 3.2.1 Example.373TAN40. Find the number of distinguishable permutations that can be formed from the letters of the word PHILIPPINES. Solution. There are 11 letters in the word P HILIP P IN ES, 3P s, 1H, 3Is, 1L, 1N, 1E, and 1S. Therefore, we use the formula for the permutations n! 11! of n objects, not all distinct: = = n1 !n2 !...nr ! 3!3! 1, 108, 800. Example.373TAN42. BOOK SELECTION: A student is given a reading list of ten books from which he must select two for an outside reading requirement. In how many ways can he make his selections? Solution. The number of ways the student can select 10! 10.9 the two books is C (10, 2) = = = 45. 8!2! 2.1 Example.373TAN46. Seven people arrive at the ticket counter of a travel agency at the same time. In how many ways can they line up to purchase their tickets? Solution. The number of ways they can line up to purchase their ticket is P (7, 7) = 7! = 5040. Example.374TAN58. A student taking an examination is required to answer 10 out of 15 questions. (a) In how many ways can the 10 questions be selected? (b) In how many ways can the 10 questions be selected if exactly 2 of the first 3 questions must be answered? Solution. (a) The number of ways that the 10 questions can be selected from the 15 questions is 15! C (15, 10) = = 3003. 10!5! (b) The number of ways the 10 questions can be selected if exactly 2 of the first 3 questions must be answered 3! 12! is C (3, 2) .C (12, 8) = . = 1485. 1!2! 4!8! Example.381TAN33. In how many ways can six different compact discs be arranged on a shelf? Solution. The number of ways the compact discs can be arranged on a shelf is P (6, 6) = 6! = 720 ways. Example.381TAN36. How many three-digit numbers can be formed from the numerals in the set {1, 2, 3, 4, 5} if: (a) Repetitions of digits is not allowed? (b) Repetitions of digits is allowed? Solution. (a) If repetition is not allowed, the number 5! of three-digit numbers is P (5, 3) = = 60. 2! (b) If repetition is allowed, the number of three-digit numbers is (5) (5) (5) = 125. Example.381TAN37. Find the number of distinguishable permutations that can be formed from the letters of each word. (a) CINCINNATI (b) HONOLULU Solution. (a) Since there is repetition of the letters C, I, and N, we use the formula for the permutation of n objects, not all distinct, with n = 10, n1 = 2, n2 = 3, and n3 = 3. Then the number of permutations that can 10! be formed is given by = 50, 400. 2!3!3! 8! (b) Similarly as in (a) , = 5040. 2!2!2! Example.381TAN40. There are eight seniors and six juniors in the Math Club at Jeп¬Ђerson High School. In how many ways can a math team consisting of four seniors and two juniors be selected from the members of the Math Club? Solution. The math team can be selected in 8! 6! . = 1050 ways. C (8, 4) .C (6, 2) = 4!4! 4!2! Example.381TAN41. A sample of 4 balls is to be selected from an urn containing 15 balls numbered 1 to 15. If 6 balls are green, 5 are white, and 4 are black: (a) How many diп¬Ђerent samples can be selected? (b) How many samples can be selected that contain at least 1 white ball? Solution. (a) The number of samples can be selected 15! is C (15, 4) = = 1365. 4!11! 10! (b) There are C (10, 4) = = 210 4!6! ways of selecting 4 balls none of which are white. Therefore, there are 1365 в€’ 210, or 1155 ways of selecting 4 balls of which at least one is white. Example.381TAN42. From a shipment of 60 transistors, 5 of which are defective, a sample of 4 transistors is selected at random. (a) In how many diп¬Ђerent ways can the sample be selected? (b) How many samples contain 3 defective transistors? (c) How many samples do not contain any defective transistors? Solution. (a) The number of diп¬Ђerent ways the sample can be selected is given by 60! C (60, 4) = = 487, 365. 56!4! (b) The number of samples that contain 3 defective transistors is given by C (5, 3) .C (55, 1) 5! 55! = . = 10 (55) = 550. 3!2! 54!1! (c) The number of samples that do not contain any defective transistors is given by 55! C (55, 4) = = 341, 055. 51!4! 2 Example.391TAN27. GAME SHOWS: In a television show, the winner is asked to select three prizes from five diп¬Ђerent prizes, A, B, C, D and E. (a) Describe a sample space of possible outcomes (order is not important). (b) How many points are there in the sample space corresponding to a selection that includes A? (c) How many points are there in the sample space to a selection that includes A and B? (d) How many points are there in the sample space correspondingВЅto a selection that includes either A Вѕ or B? ABC, ABD, ABE, ACD, ACE, Solution.(a) ADE, BCD, BCE, BDE, CDE (b) 6 (c) 3 (d) 6 8.3: SAMPLE SPACES AND EVENTS Example.390TAN7-12: Let S = {1, 2, 3, 4, 5, 6} , E = {2, 4, 6} , F = {1, 3, 5} , and G = {5, 6} . (Q7 ) Find the event E в€Є F в€Є G. Solution.E в€Є F в€Є G = {2, 4, 6} в€Є {1, 3, 5} в€Є {5, 6} = {1, 2, 3, 4, 5, 6} (Q8 ) Find the event E в€© F в€© G. Solution. E в€© F в€© G = {2, 4, 6} в€© {1, 3, 5} в€© {5, 6} = О¦. c (Q9 ) Find the event (E в€Є F в€Є G) . c Solution. {1, 2, 3, 4, 5, 6} = О¦. (Q10 ) Find the event (E в€© F в€© G)c . Solution.(E в€© F в€© G)c = {1, 2, 3, 4, 5, 6} (Q11 ) Are the events E and F mutually exclusive? 8.4: PROBABILITY Solution. Yes, E в€© F = О¦, that is, E and F do not contain any common elements. Example.402TAN20. QUALITY CONTROL: One (Q12 ) Are the events F and G mutually exclusive? light bulb is selected at random from a lot of 120 bulbs, of Solution. No. 5 is an element of both sets. which 5% are defective. What is the probability that the Example.391TAN13-18. Let S be any sample space light bulb selected is defective? and E, F, and G be any three events associated with the Solution.The probability that a defective bulb is choexperiment. Describe the given events using the symbols 6 в€Є, в€©, and c . sen is = 0.05. 120 (Q13 ) The event that E and /or F occurs. Example.402TAN22. If a ball is selected at random Solution. E в€Є F from an urn containing three red balls, two white balls, (Q14 ) The event that both and F occur. and five blue balls, what is the probability that it will be Solution. E в€© F a white ball? (Q15 ) The event that G does not occur. Solution. The probability that it will be a white ball Solution. Gc 2 2 = = 0.2. is P (W ) = (Q16 ) The event that E but not F occurs. 3 + 2 + 5 10 c Solution. (E в€© F ) . (Q17 ) The event that none of Example.402TAN24. A pair of fair dice is cast. c the events E, F, and G occurs. (E в€Є F в€Є G) Solution. What is the probability that (a) The sum of the num(Q18 ) The event that E occurs but neither of the events bers shown uppermost is less than 5? (b) At least F or G occurs. Solution. (E в€© F c в€© Gc ) one space S = ВЅ 6 is cast? Solution. The sample Вѕ (1, 1) , (1, 2) , ..., (1, 6) , (2, 1) , (2, 2) , Example.391TAN20. Let S = {1, 2, 3} be a sample ..., (2, 6) , ..., (6, 1) , (6, 2) , ..., (6, 6) space associated with an experiment. (a) The required event is (a) List all events of this experiment. E = {(1, 1) , (1, 2) , (1, 3) , (2, 1) , (2, 2) , (3, 1)}and so (b) How many subsets of S contain the number 3? P (E) = P [{(1, 1)}] + P [{(1, 2)}] + ... + P [{(3, 1)}] Solution.(a) П†, {1} , {2} , {3} , {1, 2} , 1 1 1 1 {1, 3} , {2, 3} , {1, 2, 3} . (b) 4 (c) 4 = + + ... + = (six terms) 36 36 36 6 Example.391TAN22. An experiment consists of se(b) The required event is lecting a letter at random from the letters in the word F = {(1, 6) , (2, 6) , ..., (5, 6) , (6, 1) , ..., (6, 6)} , M ASSACHU SET T S and observing the outcomes. and so P (F ) = P [{(1, 6)}] + ... + P [{(6, 6)}] (a) What is an appropriate sample space for this ex1 1 1 11 periment? = + + ... + = (eleven terms) 36 36 36 36 (b) Describe the event вЂќthe letter selected is a vowelвЂќ. Example.402T7A2N25. TRAFFIC LIGHTS: What Solution. (a) {A, C, E, H, M, S, T, U } is the probability of arriving at a traп¬ѓc light when it is red (b) {A, E, U } if the red signal is flashed for 30 sec, and the green signal Example.391TAN23. An experiment consists of tossfor 45 sec? ing a coin and casting a die and observing the outcomes.(a) Solution. The probability of arriving at the traп¬ѓc Describe an appropriate sample space for this experiment. 30 30 = = 0.375. light when it is red is (b) Describe the event вЂќa head is tossed and an even num30 + 5 + 45 80 Example.403TAN37. Let S = {s1 , s2 , s3 , s4 , s5 , s6 } ber is castвЂќ. ВЅ Вѕ (H, 1) , (H, 2) , (H, 3) , (H, 4) , (H, 5) , (H, 6) , be the sample space associated with an experiment Sol.:(a) (T, 1) , (T, 2) , (T, 3) , (T, 4) , (T, 5) , (T, 6) having the following probability distribution (b) E = {(H, 2) , (H, 4) , (H, 6)} Outcome {s1 } {s2 } {s3 } {s4 } {s5 } {s6 } 1 1 1 1 1 1 Probability 12 4 12 6 3 12 3 1 2 = 3 3 (f ) P (Ac в€Є B c ) = P (Ac ) + P (B c ) в€’ P (Ac в€© B c ) = 13 11 1 2 + в€’ = . 24 24 3 3 Example.421TAN17. An exam consists of ten trueor-false questions. If a student guesses at every answer, what is the probability that he she will answer exactly six questions correctly? Solution. The number of elements in 10! , or 210 the sample space is 210 . There are C (10, 6) = 6!4! ways of answering exactly six questions correctly. There210 210 = 0.205. fore, the required probability is 10 = 2 1024 Example.421TAN19. QUALITY CONTROL: Two light bulbs are selected at random from a lot of 24, of which 4 are defective. What is the probability that: (a) Both of the light bulbs are defective? (b) At least 1 of the light bulbs is defective? Solution. (a) Let E denote the event that both C (4, 2) = of the bulbs are defective. Then P (E) = C (24, 2) В¶ Вµ В¶ Вµ 24! 4.3 4! / = = 0.022. 2!2! 22!2! 24.23 (b) Let F denote the event that none of the bulbs 20! 22!2! C (20, 2) = . = are defective. Then P (F ) = C (24, 2) 18!2! 24! 20 19 . = 0.6884. Therefore, the probability that at least 24 23 one of the light bulbs is defective is given by 1 в€’ P (F ) = 1 в€’ 0.6884 = 0.3116. Example.421T7A4N20. A customer at Farm 5 store picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotten. What is the probability that the sample contains 1 or more rotten oranges? Solution. Let E denote the event that there is at least one rotten orange in the sample. Then E is the event that there are no rotten oranges in the sample and C (56, 3) 166, 320 P (E c ) = = в‰€ 0.81, C (60, 3) 205, 320 and P (E) = 1 в€’ P (E c ) = 1 в€’ 0.81 = 0.19. Example.421TAN25. A student studying for a vocabulary test knows the meaning of 12 words from a list of 20 words. If the test contains 10 words from the study list, what is the probability that at least 8 of the words on the test are words that the student knows? Solution. The probability is given by C (12, 8) .C (8, 2) C (12, 9) C (8, 1) C (12, 10) + + = C (20, 10) C (20, 10) C (20, 10) (28) (495) + (220) (8) + 66 в‰€ 0.085. 184, 756 Example.418TAN4. Five people are selected at random. What is the probability that (a) none of the people in this group have the same birthday?(b) at least two of them have the same birthday? Solution.(a) We see that in a group of 5 people,the probability that none of the people 365.364.363.362.361 will have same birthday is = 0.973 365.365.365.365.365 (b) The probability that at least two of them will have the same birthday is 1 в€’ 0.973 = 0.027. Find the probability of the event: (a) A = {s1 , s2 } , (b) B = {s2 , s4 , s5 , s6 } , (c) C = S. 1 1 1 Solution.(a) P (A) = P (s1 ) + P (s2 ) = + = 12 12 6 (b) P (B) = P (s2 ) + P (s4 ) + P (s5 ) + P (s6 ) 1 1 1 1 5 = + + + = . (c) P (C) = 1. 4 6 3 12 6 Example.403TAN39. Consider the composition of a three-child family in which the children were born at diп¬Ђerent times. Assume that a girl is as likely as a boy at each birth. What is the probability that: (a) There are two girls and a boy in the family? (b) The oldest child is a girl? (c) The oldest child is a girl and the youngest child is a boy? Solution. Let G denote a female birth and let B denote a male birth. Then the eight equally likely outcomes of this experiment are GGG, GGB, GBG, BGG, BGB, BBG, GBB, BBB. (a) The event that there are two girls and a boy in the family is E = {GGB, GBG, BGG} . Since there are three favorable outcomes, P (E) = 3/8. (b) The event that the oldest child is a girl is F = {GGG, GGB, GBG, GBB} . Since there are 4 favorable outcomes, P (F ) = 1/2. (c) The event that the oldest child is a girl and the youngest child is a boy is G = {GGB, GBB} . Since there are two favorable outcomes, P (F ) = 1/4. Example.411TAN25. Let E and F be two events that are mutually exclusive and suppose P (E) = 0.2 and P (F ) = 0.5. Compute: (a) P (E в€© F ) (b) P (E в€Є F ) (c) P (E c ) (d) P (E c в€© F c ) Solution. (a) P (E в€© F ) = 0 since E and F are mutually exclusive.(b) P (E в€Є F ) = P (E) + P (F ) в€’ P (E в€© F ) = 0.2 + 0.5 = 0.7. (c) P (E c ) = 1 в€’ P (E) = 1 в€’ 0.6 = 0.4. (d) P (E c в€© F c ) = P [E c в€© F c ] = P [(E в€Є F )c ] = 1 в€’ P (E в€Є F ) = 1 в€’ 0.7 = 0.3. Example.412TAN28. Let S = {s1 , s2 , s3 , s4 , s5 , s6 } be the sample space associated with an experiment having the following probability distribution Outcome {s1 } {s2 } {s3 } {s4 } {s5 } {s6 } 1 1 1 1 1 1 Probability 3 8 6 6 12 8 If A = {s1 , s2 } and B = {s1 , s5 , s6 } , find: (a) P (A) , P (B) , (b) P (Ac ) , P (B c ) , (c) P (A в€© B) (d) P (A в€Є B) , (e) P (Ac в€© B c ) , (f ) P (Ac в€Є B c ) . 1 1 11 Solution.(a) P (A) = P (s1 ) + P (s2 ) = + = 3 8 24 1 1 1 13 P (B) = P (s1 ) + P (s5 ) + P (s6 ) = + + = 3 12 8 24 11 13 (b) P (Ac ) = 1 в€’ P (A) = 1 в€’ = 24 24 13 11 P (B c ) = 1 в€’ P (B) = 1 в€’ = 24 24 1 (c) P (A в€© B) = P (s1 ) = 3 (d) P (A в€Є B) = P (A) + P (B) в€’ P (A в€© B) 11 13 1 2 = + в€’ = 24 24 3 3 c (e) P (Ac в€© B c ) = P [(A в€Є B) ] = 1 в€’ 4 P (Ac в€© B) n (Ac в€© B) 1879 = = = c c P (A ) n (A ) 2610 0.720 (b) P (B|A) 6= P (B) , so A and B are not independent events. Example.439T7A5N38. MAIL DELIVERY: Suppose the probability that your mail will be delivered before 2 p.m. on a delivery day is 0.90. What is the probability that your mail will be delivered before 2 p.m. for :(a) Two consecutive delivery days? (b) Three consecutive delivery days? Solution. (a) These are independent events. Therefore, the probability is (0.9) (0.9) = 0.81. (b) The probability that the mail will be delivered before 2 p.m. is (0.9) (0.9) (0.9) = 0.729. Example.439T7A5N42. QUALITY CONTROL: Copykwik, Inc has four photocopy machines A, B, C, and D. The probability that a given ma1 chine will break down on a particular day is P (A) = , 50 1 1 1 P (B) = , P (C) = , P (D) = . Assuming in60 75 40 dependence, what is the probability on a particular day that: (a) All four machines will break down?(b) None of the machines will break down? (c) Exactly one machine will break down? Solution. The probability that on a particular day (a) all four machines will break down is P (A) P (B) P (C) P (D) 1 1 1 1 . . . = 0.0000001. (b) none of the ma= 50 60 75 40 chines will break down is P (Ac ) P (B c ) P (C c ) P (Dc ) = 49 59 74 39 . . . = 0.9270473 в‰€ 0.927 (c) Exactly one ma50 60 75 40 chine will break down is P (A) P (B c ) P (C c ) P (Dc ) + P (Ac ) P (B) P (C c ) P (Dc ) + P (Ac ) P (B c ) P (C) P (Dc ) + P (Ac ) P (B c ) P (C c ) P (D) = 1 59 74 39 49 1 74 39 49 59 1 39 . . . + . . . + . . . 50 60 75 40 50 60 75 40 50 60 75 40 49 59 74 1 + . . . = 0.0189193 + 0.0157127 50 60 75 40 +0.0125277 + 0.2377704 = 0.0709299 в‰€ 0.071. Example.452T7A7N13. In a group of 20 ballpoint pens on a shelf in the stationery department of the University Store, 2 are known to be defective. If the customer selects 3 of these pens, what is the probability that: (a) At least one is defective? (b) No more than one is defective? Solution. (a) The probability that none of the pens in C (18, 3) (18!) / (15!3!) the sample are defective is = = C (20, 3) (20!) / (17!3!) 68 18! 17! . = в‰€ 0.71579. 15! 20! 95 Therefore, the probability that at least one is defective is given by 1 в€’ 0.71579 в‰€ 0.284. (b) The probability that two are defective is given by C (2, 2) .C (18, 1) 18/20! 6 = = в‰€ 0.0158. ThereC (20, 3) 17!3! 380 fore, the probability that no more than 1 is defective is 374 6 = в‰€ 0.984. Example.452T7A7N14. given by 1 в€’ 380 380 Five people are selected at random. What is the probability that none of the people in this group were born at the same day of the week? 7.6.5.4.3 в‰€ 0.150. Solution. P (E) = 75 8.5: CONDITIONAL PROBABILITY & 8.6: INDEPENDENT P (B|Ac ) = Example.436T7A5N2. Let A and B be two events in a sample space S such that P (A) = 0.4, P (B) = 0.6, and P (A в€© B) = 0.3. Find: (a) P (A|B) (b) P (B|A) . P (A в€© B) 0.3 1 Solution. (a) P (A|B) = = = . P (B) 0.6 2 0.3 3 P (A в€© B) = = . (b) P (B|A) = P (A) 0.4 4 Example.436T7A5N3. Let A and B be two events in a sample space S such that P (A) = 0.6 and P (B |A|) = 0.5. Find P (A в€© B) .Solution.P (A в€© B) = P (A) P (B|A) = (0.6) (0.5) = 0.3.Example.436T7A5N10. If A and B are independent events and P (A) = 0.35 and P (B) = 0.45, find: (a) P (A в€© B) (b) P (A в€Є B) . Solution. (a) P (A в€© B) = P (A) P (B) = (0.35) (0.45) = 0.1575. (b) P (в€ЄB) = P (A) + P (B) в€’ P (A в€© B) = 0.35 + 0.45 в€’ 0.1575. Example.436T7A5N15. A pair of fair dice is cast. What is the probability that the sum of the numbers falling uppermost is less than 9, if it is known that one of the numbers is a 6? Solution. Let A denote the event that the sum of the numbers is less than 9 and let B denote the event that one of the numbers is a 6. Then, P (A|B) = P (A в€© B) 4/36 4 = = . Example.436T7A5N17. P (B) 11/36 11 A pair of fair dice is cast. Let E denote the event that the number landing uppermost on the first die is a 3 and let F denote the event that the sum of the numbers falling uppermost is 7. Determine whether E and F are independent events. Solution. E = {(3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 5) , (3, 6)} F = {(1, 6) , (6, 1) , (2, 5) , (5, 2) , (3, 4) , (4, 3)} ; Then 1 E в€© F = {(3, 4)} .Now, P (E в€© F ) = and this is equal 36 Вµ В¶Вµ В¶ 6 6 1 to P (E) .P (F ) = = . So E and F are 36 36 36 independent events. Example.439T7A5N34. STUDENT FINANCIAL AID: The accompanying data were obtained from the financial aid oп¬ѓce of a certain university: Finacial Aid NO Finacial Aid Total Ugrads 4, 222 3, 898 8, 120 Grads 1, 879 731 2, 610 Total 6, 101 4, 629 10, 730 Let A be the event that a student selected at random from this university is an undergraduate student and let B be the event that a student selected at random is receiving financial aid.(a) Find each of the following probabilities: P (A) , P (B) , P (A в€© B) , P (B|A) , and P (B|Ac ) . (b) Are the events A and B independent events? 8120 Solution.(a) P (A) = = 0.757; P (B) = 10730 6101 = 0.569; 10730 422 P (A в€© B) P (A в€© B) = = 0.393; P (B|A) = = 10730 P (A) n (A в€© B) 422 = = 0.520 n (A) 8120 5

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