Grade 12 / Bernoulli diagram and binomial distribution 45 exercises (including 44 corrected)

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ESESESESESESES ESESESESESESESESESESESESESESES SESESESESESESE SESESESESESESESESESESESESESESE 1. Independent repetitions of Bernoulli tests E.5382 Consider a trial admitting only two outcomes : one named ˇ succès ı and noted S probability 0.4 ; the other named ˇ échec ı and noted E . We decide to repeat this same trial three times. The probability tree shown opposite is obtained. These repetitions are assumed to be independent of each other. 1 Complete this probability tree? 2 a How many paths include 3 success? b Give the probability of getting three successes from this random experiment? 3 a How many paths include 0 success? b Give the probability of obtaining no successes from this random experiment? 4 a How many paths include 2 success? b Give the probability of obtaining exactly two successes from this random experiment? E.5383 Consider a trial with only two outcomes : one outcome with probability 0.3 , denoted S ; the other out-come is denoted E . Consider the random experiment consisting of four repetitions of the previous trial. This new random experiment is repre-sented by the choice tree below : 1 How many elementary events make up this random ex-periment? 2 Note X the random variable which, at each elementary event, counts the number of events S realized. Determine the following probabilities, rounded to the thousandth : a P X =0 b P X =1 c P X =2 2. Binomial coefficients E.5384 Here are the choice trees associated with repeating a Bernoulli test 3 and 4 times respectively: 1 For the Bernoulli test repeated three times, complete the table below : Number of succès 0 1 2 3 Number of issues 2 For the four-fold repetition of Bernoulli’s test, complete the table below : Number of succès 0 1 2 3 4 Nombre d’issues 3 Is there a method for obtaining the second table from the first? https://chingmath.fr chapExoCorrec/5382 sacados/5382 ESESESESESESES chapExoCorrec/5383 sacados/5383 ESESESESESESESESESESESESESESES chapExoCorrec/5384 sacados/5384 SESESESESESESE SESESESESESESESESESESESESESESE
E0E0S1E0S1E1E0S1E1S2E0S1E1S2E1E0S1E1S2E1S2E0S1E1S2E1S2E2E0S1E1S2E1S2E2S3E0S1E1S2E1S2E2S3E1E0S1E1S2E1S2E2S3E1S2E0S1E1S2E1S2E2S3E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5ESESESESESESESESESESESESESESES E.5203 The figure below represents the repetition of five Bernoulli trials où the two outcomes are S (success) and E (failure) . The subscript number on the fifth choice represents the num-ber of successes achieved in the chosen path. 1 Complete the table below : Nombre de succès 0 1 2 3 4 5 Nombre de chemins associés 2 Consider the same Bernoulli trial but repeated six times : a Give the number of paths achieving 4 success when a Bernoulli trial is repeated six times (you may complete the choice tree or reason) . b Complete the table below : Nombre de succès 0 1 2 3 4 5 6 Nombre de chemins associés 3. Binomial law E.6064 Let X follow a binomial distribution with parameters 15 and 0.35 . C’est-à-dire : X∼B (15 ; 0.35) Determine the exact value, then the value rounded to the thousandth of the following probabilities : a P X =5 b P X =7 c P X =9 E.4323 A player has a balanced cubic die whose faces are numbered from 1 to 6 . On each throw, he wins if he gets 2 , 3 , 4 , 5 or 6 ; he loses if he gets 1 . A game consists of 5 successive, independent throws of the die. Determine the exact probability that the player will lose 3 times during a game, then its value rounded to the hundredth. E.4157 An urn contains one white ball and two black balls. We carry out 10 successive draws of a ball with remittance (we draw a ball at random, note its color, put it back in the urn and start again) Indicate whether the following proposition is true or false, and give a justification for the answer chosen. Proposition: The probability of drawing exactly 3 white balls is : 3 × 1 3 3 × 2 3 7 E.4151 A random variable X follows a binomial distribution with parameters n and p où n equals 4 and p belongs to 0 ; 1 . Without justification, indicate whether each of the following propositions is true or false. Proposition 1: si P ( X =1)=8 ·P ( X =0) alors p = 2 3 . Proposition 2: si p = 1 5 alors P ( X =1)= P ( X =0) . 4. Binomial law and complementary events E.5387 Consider a random variable X following the binomial distribution with parameter n =15 and p =0.63 . 1 Using the calculator, determine the following binomial coefficients : a 15 13 b 15 14 c 15 15 2 Determine the exact value of the following probabilities, then rounded to the nearest 10 − 4 : a P X =13 b P X =14 c P X =15 3 Deduce the value, rounded to the nearest 10 − 4 , of the probability of the event X 12 . E.5408 Consider a random variable X following a binomial distribution with parameters 6 and 0.3 . 1 Determine the exact value of the following probabilities, then rounded to the nearest hundredth : a P X =0 b P X =1 2 Deduce the value, to the nearest thousandth, of : P X 2 https://chingmath.fr chapExoCorrec/5203 sacados/5203 E0E0S1E0S1E1E0S1E1S2E0S1E1S2E1E0S1E1S2E1S2E0S1E1S2E1S2E2E0S1E1S2E1S2E2S3E0S1E1S2E1S2E2S3E1E0S1E1S2E1S2E2S3E1S2E0S1E1S2E1S2E2S3E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5ESESESESESESESESESESESESESESES chapExoCorrec/6064 sacados/6064 chapExoCorrec/4323 sacados/4323 Extrait d'Antilles Juin 2011 chapExoCorrec/4157 sacados/4157 Extrait de Liban Juin 2010 chapExoCorrec/4151 sacados/4151 chapExoCorrec/5387 sacados/5387 chapExoCorrec/5408 sacados/5408
E.4213 During an epidemic in cattle, a test for this disease is set up. A study is carried out on this herd and the probability of the test being positive on an ani-mal from this herd is 0.058 . Five animals are chosen at random. The size of this herd means that the trials can be considered independent and the draws can be treated as if they were random draws. Note X the random variable which, for the five animals chosen, associates the number of animals with a positive test. 1 What is the probability law followed by X ? 2 What is the probability that at least one of the five ani-mals has a positive test? The exact value and the value approximated to the thousandth will be given. E.4214 Every year, a sporting competi-tion is organized to link two villages as quickly as possible. Several means of travel are possible : bicycle ; in roller ; on foot. The results for the different years are assumed to be indepen-dent of each other. On the basis of experience in previous years, we can assume that the probability of the winner hav- ing completed the route by bike is 2 3 . Calculate the probability that over the next six years the event will be won at least once by a non-cycling competitor ˇı. Also give the approximate value to the thousandth of this probability. E.4176 All results will be rounded to the nearest 10 − 2 . A company produces pens in large quantities. The probability of a pen having a defect is equal to 0.1 . Eight pens are taken from this production run, successively and with delivery. Note X the random variable that counts the number of pens with a defect among the eight pens taken. 1 It is assumed that X follows a binomial distribution. Give the parameters of this distribution. 2 Calculate the probability of the following events : a A : ˇ there are no pens with a défaut ı ; b B : ˇ there is at least one pen with a défaut ı ; c C : ˇ there are exactly two pens with a défaut ı. 5. Binomial law and distribution functions E.8210 Consider a random variable X follow-ing the binomial distribution with parameters 14 and 0.44 ( X∼B 14 ; 0.44 ) . k 0 1 2 3 4 5 6 7 P X k 0.0002 0.003 0.02 0.073 0.186 0.365 0.576 0.765 k 8 9 10 11 12 13 14 P X k 0.895 0.963 0.99 0.998 0.999 0.999 1 1 a Determine the probability of the event X =5 . b Give the value of : P X =8 + P X =9 2 Determine the probability value : P X > 5 3 Answer the questions below : a Give the probability that the variable X has a value of at least 7 . b Give the probability that the variable X for value at most 7 . 4 Determine the following probabilities : a P 3 X 6 b P 5 X 10 6. Binomial law with calculator: point values E.6065 Let X be a random variable following a binomial distribution with parameters 7 and 0.6 . That is : X ∼ B (7 ; 0.6) Using your calculator, complete the table below, with values rounded to the thousandth, to obtain the probability distri-bution of the random variable X : x 0 1 2 3 4 5 6 7 P X = x 7. Binomial law with calculator: cumulative values E.5407 Consider a random variable X following a binomial distribution with parameters 20 and 0.2 . Questions will be answered using the calculator. Results will be rounded to the nearest 10 − 3 : 1 Determine the value of the following probabilities : a P X =5 b P X =9 2 Determine the value of the following probabilities : a P X 5 b P X 9 https://chingmath.fr chapExoCorrec/4213 sacados/4213 Antilles-guyane Septembre 2010 chapExoCorrec/4214 sacados/4214 chapExoCorrec/4176 sacados/4176 sacados/8210 chapExoCorrec/6065 sacados/6065 chapExoCorrec/5407 sacados/5407
E.5426 It is assumed that a random variable X follows a binomial distribution with parameter n =22 and p =0.37 Using the calculator and without justification, give the prob-ability of P 3 X 7 rounded to the nearest 10 − 4 . E.5816 The manager of a tea shop buys 10 boxes of green tea from a wholesaler. It is assumed that the latter’s stock is large enough to model this situation by a random draw of 10 tins with a discount. A survey of the wholesaler’s green tea tins shows that 12 % of the tins have traces of pesticides in their tea. Consider the random variable X which associates with this sampling of 10 boxes, the number of boxes without traces of pesticides. 1 Justify that the random variable X follows a binomial distribution whose parameters we will specify. 2 Calculate the probability that the 10 boxes are pesticide-free. The exact value and the value rounded to 10 − 4 are given. 3 Give, to the nearest thousandth, the probability that at least 8 boxes have no traces of pesticides. E.5386 Consider a random variable X following a binomial distribution with parameter n =5 and p =0.6 . Probabilities will be rounded to the nearest thousandth. 1 Give the law of the variable X in the form of a table.. 2 Determine the following probabilities : a P X 1 b P X > 1 8. Binomial law - problems E.4153 Consider a questionnaire with five questions. For each of the five questions asked, three proposed answers are made ( A , B and C ) , only one of which is correct. A candidate answers all the questions posed by writing a five-letter word answer. For example, the word ˇ BBAAC ı means that the candidate answered B to the first and second questions, A to the third and fourth questions and C to the fifth question. 1 How many possible word-answers are there to this ques-tionnaire? 2 It is assumed that the candidate answers each of the five questions in this questionnaire at random. Calculate the probability of the following events : a E : ˇ the candidate has exactly one correct answer. ı. b F : ˇ the candidate has no answer exacte ı. c G : ˇ the candidate’s answer word is a palindrome ı. (Note that a palindrome is a word that can be read either left to right or right to left : for example, ˇ BACAB ı is a palindrome) E.3748 A watch factory manufactures a series of watches. During manufacture, two types of defect may appear, desig-nated a and b . 2 % of the watches manufactured have the defect a and 10 % the defect b . A watch is drawn at random from the production. The fol-lowing events are defined : A : ˇ the pulled watch has the defect a ı ; B : ˇ the pulled watch has the defect b ı ; C : ˇ the pulled watch has neither défauts ı ; D : ˇ the pulled watch has one and only one of the two défauts ı. The probability of the event C is equal to 0.882 . 1 Calculate the probability of the event D . 2 During manufacture, five watches are taken at random in succession. The number of watches manufactured is considered to be large enough to assume that the draws are made with a discount and are independent. Let X be the random variable which, for each sampling of five watches, associates the number of watches with none of the two defects a and b . We define the event : E : ˇ at least four watches have no defect ı. Calculate the probability of the event E . This will be approximated to the nearest 10 − 3 . https://chingmath.fr chapExoCorrec/5426 sacados/5426 chapExoCorrec/5816 sacados/5816 Extrait du Bac Asie Juin 2013 chapExoCorrec/5386 sacados/5386 chapExoCorrec/4153 sacados/4153 chapExoCorrec/3748 sacados/3748
0;85T0;15T0;01M0;05T0;95T0;99M E.5448 In a school, two associations offer extra-curricular activities for pupils : the sports association and the arts association. A study was carried out to examine the en-rolment of pupils in each of these two associations. Here are some of the results : 42 % of students signed up for the sports association; 35 % of students have joined the arts association; 32 % of students have enrolled in both associations. 1 A student is chosen at random from the school. The following events are considered : S : ˇ the student is registered with the association sportive ı ; A : ˇ the student is registered with the association artis-tique ı ; a From the data in the statement, give the value of the following probabilities : P ( S ) ; P ( A ) ; P ( S ∩ A ) b Determine the probability of the event : U : ˇ The student is enrolled in at least one of the two associations ı c Consider the event : E : ˇ The student is enrolled in one and only one of these two associations ı. Show that the probability of this event verifies : P ( E ) = 0.13 2 Groups of 32 students from this school are formed. It is assumed that the choice of students is made indepen-dently of the previously chosen students and does not alter the probability of the group. We are interested in the random variable X counting the number of students in such a group who are members of one and only one of these associations. a What probability law does the random variable X fol-low? b Determine the probability that there are at least two students in this group who have one and only one reg-istration in one of these two associations. Results will be rounded to the nearest 10 − 4 . E.5489 During an epidemic in cattle, it was discovered that if the disease was diagnosed early enough in an animal, it could be cured ; if not, the disease was fatal. A test was developed and tested on a sample of animals. The events are noted : M : ˇ the animal is a carrier of maladie ı ; T : ˇ the test is positif ı. Here is the probability tree obtained after studying the chep-tail: 1 An animal is chosen at random. a What is the probability that he is a carrier of the dis-ease and that his test is positive? b Show that the probability of his test being positive is 0.058 . 2 Five animals are chosen at random. The size of this herd means that the trials can be considered independent and the draws can be treated as if they were random draws. Note X the random variable which, for the five animals chosen, associates the number of animals with a positive test. a What is the probability law followed by X ? Justify. b What is the probability that at least one of the five animals has a positive test? Give the exact value and then the value rounded to the nearest thousandth. 3 The cost of caring for an animal that has reacted posi-tively to the test is 100 euros and the cost of slaughtering an animal not detected by the test and having developed the disease is 1 000 euros. The test is assumed to be free of charge. Based on the above data, the probability distribution of the cost to be incurred per animal undergoing the test is given by the following table : Coût 0 100 1 000 Probabilité 0.940 5 0.058 0 0.001 5 a Calculate the mathematical expectation of the random variable Z associating with an animal the cost to be incurred. b A breeder has a herd of 200 animals. If the entire herd is to be tested, how much money should he plan to spend? 9. Expectation of a binomial distribution E.4168 We have a well-balanced cubic die whose faces are numbered from 1 to 6 . We roll the well-balanced die three times in succession and denote by X the random variable giving the number of 6 obtained. 1 What probability law does the random variable X follow? 2 Give the exact value, then the value rounded to the thou- sandth of the probability P ( X =2) . 3 Give the expectation of the random variable X . https://chingmath.fr chapExoCorrec/5448 sacados/5448 chapExoCorrec/5489 sacados/5489 0;85T0;15T0;01M0;05T0;95T0;99M chapExoCorrec/4168 sacados/4168
PXkPYk0,10,20,30,40 E.4194 A transport company wants to optimize controls to limit the impact of fraud and the losses caused by this practice. The company carries out a study based on two journeys per day during the twenty working days of a month, i.e. a total of forty journeys. It is assumed that the checks are independent of each other and that the probability of any passenger being checked is equal to p . The fare for each trip is ten euros ; in the event of fraud, the fine is one hundred euros. Claude frauds systematically during the forty journeys sub-ject to this study. Let X i be the random variable that takes the value 1 if Claude is checked on the i -th trip and the value 0 otherwise. Let X be the random variable defined by: X = X 1 + X 2 + X 3 + · · · + X 40 1 Determine the probability distribution of X . 2 In this part, it is assumed that : p = 1 20 . a Calculate the mathematical expectation of X . b Calculate the probabilities : P ( X =0) ; P ( X =1) ; P ( X =2) c Calculate to the nearest 10 − 4 the probability that Claude will be checked at most twice. E.4215 A factory produces bags. We assume that the probability (rounded to two decimal places) that a bag is defective is equal to 0 ; 03 . A random sample of 100 bags is taken from one day’s pro-duction. Production is large enough that this sample can be considered a sampling with replacement of 100 bags. Con-sider the random variable X , which, for every sample of 100 bags, associates the number of defective bags. 1 Justify that the random variable X follows a binomial distribution, specifying the parameters. 2 What is the probability of the event ˇ at least one bag is defective ı? Round this probability to two decimal places. Interpret this result. 3 Calculate the mathematical expectation of the random variable X . Interpret this result in the context of the statement. E.4197 An association organizes moun-tain walks. Twelve guides each take a group of people out for the day, starting at sunrise. In summer, there are more requests than guides, and each group must register the day before the walk. But experience in recent years proves that the probability of each of the registered groups not showing up at the start of the walk is equal to 1 8 . It will be assumed that the groups entered present themselves independently of each other. A sum of 1 credit (the local currency) is requested from each group for the day. This sum is paid at the start of the walk. In the event où a group does not show up at the start, the association obviously does not earn the Credit that this group would have paid for the day. Annoyed by the number of unused guides, the association’s manager decides to take an extra booking each day. Obvi-ously if the 13 registered groups turn up, the 13 e group will be directed to a substitute activity. However, this substitute activity entails an expense of 2 Credit to the association. The probabilities requested will be rounded to the nearest 100 e . 1 What is the probability P 13 that on a given day there are no withdrawals, i.e. that the 13 groups registered the day before show up at the start of the walk? 2 Let R be the random variable equal to the cost of the substitution activity. Specify the law of the random variable R and calculate its mathematical expectation. 3 Show that the average gain obtained for each day is : 13 k =0 k · 13 k · 7 8 k · 1 8 13 − k − 2 · P 13 Calculate this gain. 4 Is the executive’s decision profitable for the association? 10. Distribution breakdown E.5427 1 Consider the random variable X following a binomial dis-tribution with parameters n =5 and p =0.5 . Draw up a table showing the probability distribution of the random variable X . 2 Consider the random variable Y following a binomial dis-tribution with parameters n =5 and p =0.3 . Draw up a table showing the probability distribution of the random variable Y . 3 In the graph below, complete the bar charts representing the law of each of these random variables : https://chingmath.fr chapExoCorrec/4194 sacados/4194 chapExoCorrec/4215 sacados/4215 chapExoCorrec/4197 sacados/4197 Extrait d'Antilles-guyane Septembre 2003 chapExoCorrec/5427 sacados/5427 PXkPYk0,10,20,30,40
02468100.050.10.150.20.250.3 02468100.050.10.150.20.250.3 E0E0S1E0S1E1E0S1E1S2E0S1E1S2E1E0S1E1S2E1S2E0S1E1S2E1S2E2E0S1E1S2E1S2E2S3E0S1E1S2E1S2E2S3E1E0S1E1S2E1S2E2S3E1S2E0S1E1S2E1S2E2S3E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5ESESESESESESESESESESESESESESES EESESEESESESESEESESESESESESEESESESESESESESESEESESESESESESESESESESEESESESESESESESESESESESESEESESESESESESESESESESESESESESEESESESESESESESESESESESESESESESESESESESESESESES E.5428 Of the two representations below, which one represents a binomial distribution with parameter n =10 and p =0.3 : 11. Reminders: binomial distribution E.5200 The figure below represents the repetition of five Bernoulli trials où the two outcomes are S (success) and E (failure) . The subscript number on the fifth choice represents the num-ber of successes achieved in the chosen path. 1 Complete the table below : Nombre de succès 0 1 2 3 4 5 Nombre de chemins associés 2 Consider the same Bernoulli trial but repeated six times : a Give the number of paths achieving 4 success when a Bernoulli trial is repeated six times (you may complete the choice tree or reason) . b Complete the table below : Nombre de succès 0 1 2 3 4 5 6 Nombre de chemins associés E.5201 The figure opposite represents the repe-tition of four Bernoulli trials où the two outcomes are S (success) and E (failure) . The following probabilities are assumed to be known : P ( S )= 1 3 ; P ( E )= 2 3 Consider the random variable X which counts the number of successes achieved after repeating these four Bernoulli trials. 1 a How many elementary events does the event X =3 } comprise? b Let ! ∈ X =3 } , show that : P ( ! ) = 2 81 c Justify that : P ( X =3)= 8 81 2 a Give the values taken by the random variable X . b Complete the table below to give the probability dis-tribution of the random variable X : x P ( X = x ) E.5202 The following questions can be answered using a calculator: 1 Give the value of the following binomial coefficients : a 15 3 b 24 3 c 54 12 d 51 51 2 Let X be a random variable such that X∼B (15 ; 0.3) . Give values to the nearest hundredth of the following probabilities : a P ( X =5) b P ( X =8) c P ( X =12) 3 Let X be a random variable such that X∼B (52 ; 0.3) . Give values to the nearest hundredth of the following probabilities : a P ( X 9) b P ( X 15) c P ( X 23) https://chingmath.fr chapExoCorrec/5428 sacados/5428 02468100.050.10.150.20.250.3 02468100.050.10.150.20.250.3 chapExoCorrec/5200 sacados/5200 E0E0S1E0S1E1E0S1E1S2E0S1E1S2E1E0S1E1S2E1S2E0S1E1S2E1S2E2E0S1E1S2E1S2E2S3E0S1E1S2E1S2E2S3E1E0S1E1S2E1S2E2S3E1S2E0S1E1S2E1S2E2S3E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5E0S1E1S2E1S2E2S3E1S2E2S3E2S3E3S4E1S2E2S3E2S3E3S4E2S3E3S4E3S4E4S5ESESESESESESESESESESESESESESES chapExoCorrec/5201 sacados/5201 EESESEESESESESEESESESESESESEESESESESESESESESEESESESESESESESESESESEESESESESESESESESESESESESEESESESESESESESESESESESESESESEESESESESESESESESESESESESESESESESESESESESESESES chapExoCorrec/5202 sacados/5202
E.5194 A Bernoulli test with parameter p =0.3 is repeated 10 times independently. Associated with this experiment is the variable X , which as-sociates the number of successes with each outcome of this experiment. 1 Determine the following probabilities : P ( X =0) ; P ( X =1) ; P ( X =2) 2 Determine the probability of the event X 3 . E.5195 An exam is based on a MCQ comprising 5 questions où each question offers four answer choices from which only one answer is correct. A student decides to randomly and independently complete each question on the questionnaire. 1 What is the probability of answering a question cor-rectly? We denote X the random variable counting the number of correct answers contained in the completed form. 2 Determine the following probabilities rounded to the nearest thousandth : a P ( X =5) b P ( X 3) 3 Using the calculator, determine the probability, rounded to the thousandth, that the student has at most 2 correct answers. E.5199 In a game, it is agreed that a game is won when the ball obtained is black and the probability of obtaining a black ball is 3 8 . A person plays ten independent games, returning the ball ob-tained after each game to the urn from which it came. Note X the random variable equal to the number of games won. 1 Calculate the probability of winning exactly three games. We will give the result rounded to the thousandth. 2 Calculate the probability of winning at least one game. Give the result rounded to the thousandth. 3 The following table is given : k 1 2 3 4 5 P ( X <k ) 0.0091 0.0637 0.2110 0.4467 0.6943 k 6 7 8 9 10 P ( X <k ) 0.8725 0.9616 0.9922 0.9990 0.9999 Let N be an integer between 1 and 10 . Consider the event : ˇ the person wins at least N parties ı. At what value of N is the probability of this event less than 1 10 ? 12. Conditional probability and binomial distribution E.3713 A bicycle repairer bought 30 % of his tire stock from a first supplier, 40 % from a second and the rest from a third. The first supplier produces 80 % tires without defects, the second 95 % and the third 85 % . 1 The repairer randomly takes a tire from his stock. a Construct a probability tree translating the situation, and show that the probability of this tire being defect-free is equal to 0.875 . b Knowing that the chosen tire is flawless, what is the probability that it comes from the second supplier? The result will be rounded to 10 − 3 . 2 The repairer chooses ten tires at random from his stock. It is assumed that the stock of tires is large enough to equate this choice of ten tires with a ten-tire toss. What is the probability that at most one of the tires se-lected will have a defect? We’ll give the value rounded to 10 − 3 . E.6951 For each question, a statement is pro-posed. Indicate whether it is true or false, justifying your answer. Any unjustified answer will be disregarded. Question 1 A and B are two events related to the same random event that verify: P A = 1 2 ; P A B = 3 10 ; P A B = 1 10 Assertion : The probability of the event A knowing that the event B is realized is equal to 1 4 Question 2 An urn A contains two blue balls and three red balls and an urn B contains one blue ball and three red balls. The balls are considered indistinguishable to the touch. A game consists of drawing a ball at random from the urn A , if the ball is blue the game is won. In the opposite case, the player draws a ball from the urn B and wins if the ball is blue. The balls drawn are returned to their respective urns at the end of the game. We are interested in a person who plays this game inde-pendently ten times in a row. Assertion : The probability of the player winning at least 2 times is less than 0.3 . 13. Non-symmetrical trees https://chingmath.fr chapExoCorrec/5194 sacados/5194 chapExoCorrec/5195 sacados/5195 chapExoCorrec/5199 sacados/5199 chapExoCorrec/3713 sacados/3713 chapExoCorrec/6951 sacados/6951
V1V2V3 F1F1F2F3 E.3725 Results will be given to 10 − 3 near. A company entrusts a telephone survey company with a sur-vey on the quality of its products. Each interviewer has a list of people to contact. On the first telephone call, the probability that the corre-spondent is absent is 0.4 . Knowing that the correspondent is present, the probability that he will agree to answer the questionnaire is 0.2 . 1 We note : A 1 the event ˇ the person is absent at the first appel ı ; R 1 the event ˇ the person agrees to answer the ques-tionnaire at the first appel ı. What is the probability of R 1 ? 2 When a person is absent on the first call, we phone her a second time, at a different time, and, then, the proba-bility that she is absent is 0.3 . And, knowing that she is present at the second call, the probability of her agreeing to answer the questionnaire is still 0.2 . If a person is absent during the second call, no further attempts are made to contact them. We note : A 2 the event ˇ the person is absent at the second ap-pel ı ; R 2 the event ˇ the person agrees to answer the ques-tionnaire at the second appel ı ; R the event ˇ the person agrees to answer the question-naire ı. Show that the probability of R is 0.176 (A tree may be used) . 3 Knowing that one person has agreed to answer the ques-tionnaire, what is the probability that the answer oc- curred on the first call? E.5833 With three identical valves V 1 , V 2 and V 3 , we make the hydraulic circuit shown opposite. The circuit is in working order if V 1 is in working order or if V 2 and V 3 are simultaneously. It is treated as a random experiment whether each valve is or is not in working order after 6 000 hours. Note: F 1 the event : ˇ the V 1 valve is in working order after 6 000 heures ı. F 2 the event : ˇ the valve V 2 is in working order after 6 000 heures ı. F 3 the event : ˇ the valve V 3 is in working order after 6 000 heures ı. E the event : ˇ the circuit is in working order after 6 000 heures ı. It is assumed that the events F 1 , F 2 and F 3 are both indepen-dent and each has a probability equal to 0.3 . 1 The probabilistic tree opposite represents part of the sit-uation. Reproduce this tree and place the probabilities on the branches. 2 Demonstrate that : P ( E )=0.363 . 3 Knowing that the circuit is in working order after 6 000 hours, calculate the probability that valve V 1 is in work-ing order at that time. Round to the nearest thousandth. 14. Probability: binomial E.4258 If X is a random variable following the binomial distribution of parameters 100 and 1 3 then : P ( X 1) = 1 − 2 3 100 E.4266 A pet shop has rare fish fry. The probability of buying a fry and the fish still being alive one month later is 0.92 . A person randomly and independently selects 5 two-month-old fry. What is the probability that one month later, only three will be alive? We’ll give an approximate value to the nearest 10 − 2 . E.4262 An urn contains balls that are indistinguishable to the touch. 20 % of the balls have the number 1 and are red. The others have the number 2 and of these, 10 % are red and the others are green. Let n be a natural number greater than or equal to 2 . We carry out n successive draws of a ball with delivery (after each draw the ball is returned to the urn) : 1 Express as a function of n the probability of obtaining at least one red ball bearing the number 1 during the n draws. 2 Determine the integer n at which the probability of ob-taining at least one red ball bearing the number 1 during the n draws is greater than or equal to 0.99 . https://chingmath.fr chapExoCorrec/3725 sacados/3725 Extrait France Septembre 2000 chapExoCorrec/5833 sacados/5833 V1V2V3 F1F1F2F3 chapExoCorrec/4258 sacados/4258 Extrait d'Antilles-guyane Septembre 2009 chapExoCorrec/4266 sacados/4266 Extrait de Nouvelle-Caledonie Mars 2008 chapExoCorrec/4262 sacados/4262
E.5526 An urn contains 10 white balls and n red balls, n being a natural number greater than or equal to 2 . A player is asked to draw balls from the urn. At each draw, all the balls have the same probability of being drawn. The player draws 20 times successively and with delivery a ball from the urn. The draws are independent. Determine the minimum value of the integer n so that the probability of obtaining at least one red ball during these 20 draws is strictly greater than 0.999 . E.4268 A gardener has two lots 1 and 2 , each containing a large number of bulbs producing tulips in a variety of colors. The probability of a bulb from lot 1 giving a yellow tulip is equal to 1 4 . The probability that a bulb from lot 2 will give a yellow tulip is equal to 1 2 . This gardener randomly selects a lot and plants 50 tulip bulbs. Let n be a natural number verifying 0 n 50 , we define the following events : A : ˇ the gardener has chosen the 1 ı lot; B : ˇ the gardener has chosen lot 2 ı ; J n : ˇ the gardener gets n tulips jaunes ı. 1 Show that : P B ( J n ) = 50 n · 2 − 50 2 Deduce the probability that the gardener will get n yel-low tulips. 3 Let p n be the conditional probability of the event A know-ing that J n is achieved. Establish that : p n = 3 50 − n 3 50 − n + 2 50 4 For what values of n has p n 0.9 ? How can this result be interpreted? 15. Unclassified financial years E.3737 A company A specializes in the mass production of an article; a quality check showed that each article produced by the company A could have two types of defect : a soldering defect with a probability equal to 0.03 and a defect on an electronic component with a probability equal to 0.02 . The test also showed that the two defects were independent. An item is said to be faulty if it has at least one of the two defects. 1 Show that the probability that an item manufactured by company A is defective is equal to 0.049 4 . 2 A department store receives 800 items from company A . Let X be the random variable that associates the number of defective items with this set of 800 items. a Define the law of X . b Calculate the mathematical expectation of X . For the company, what interpretation can be made of this ex-pectation? https://chingmath.fr chapExoCorrec/5526 sacados/5526 chapExoCorrec/4268 sacados/4268 chapExoCorrec/3737 sacados/3737 Extrait Antilles-Guyanes Juin 2003