Grade 11 / Conditional probabilities 57 exercises (including 56 corrected)

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5points0point3points0point AsAsAsAsRRRRDDDDVVVV10101010999988887777♥♦♠♣ ChingQuizz : 8 exercises available for Quizz assessment : 1. Probability reminder E.4791 Here is a table showing the proba-bility distribution of a six-sided dice : x i 1 2 3 4 5 6 p i 0.15 0.1 0.08 0.17 0.22 0.28 Determine the probability of each of the events below : 1 A : ˇ The number obtained is greater than or equal to 4 ı. 2 B : ˇ The number obtained is pair ı. E.10199 Consider a rigged die with 6 faces. The random experiment consists in throwing the die and con-sidering the value of the top face of the die. For k an integer between 1 and 6 , consider the event F k de-fined by ˇ the value obtained is k ı The only information on the die is : The incomplete table of the probability law of this ran-dom experiment : A F 1 F 2 F 3 F 4 F 5 F 6 P A 0.11 0.07 0.2 0.15 The probability of getting an even number is 0.4 . Copy and complete the probability distribution table for this random experiment, justifying your approach. E.4147 A game involves throwing darts at a target. The target is divided into four sectors, as shown in the figure below : It is assumed that the throws are in-dependent and that the player hits the target every time. The player throws a dart. Note: p 0 the probability of obtaining 0 point ; p 3 the probability of obtaining 3 points ; p 5 the probability of obtaining 5 points. Knowing that p 5 = 1 2 · p 3 and that p 5 = 1 3 · p 0 , determine the val-ues of p 0 , p 3 and p 5 . E.3701 Consider A six-sided die num-bered from 1 to 6 . Its probability distribution is given in table : Face 1 2 3 4 5 6 Probabilité p 1 p 2 p 3 p 4 p 5 p 6 where each of the numbers p i is proportional to the index i . 1 Determine the probability law of this random experi-ment. 2 Calculate the probabilities of the events : A : ˇ obtain a number strictly lower than 3 ı ; B : ˇ get a multiple of 2 ı ; C = A ∪ B . 2. Reminders: opposite event E.3702 Proposition: let A be an event and A its opposite event. We have : P A =1 −P A One card is drawn at random from a deck of 32 cards. Consider the following events : A : ˇ the card drawn is a pique ı ; B : ˇ the card drawn is a figure ı. 1 Describe the outcomes of this ran-dom experiment. 2 Calculate the probabilities of the events : A ; B ; A ∩ B ; A ∪ B 3 Calculate the probability of the event : C : ˇ the card drawn is neither a spade nor a figure ı https://chingmath.fr chapExoCorrec/4791 sacados/4791 chapExoCorrec/10199 sacados/10199 chapExoCorrec/4147 sacados/4147 5points0point3points0point chapExoCorrec/3701 sacados/3701 Extrait du MathX - Didier chapExoCorrec/3702 sacados/3702 Extrait du MathX - Didier AsAsAsAsRRRRDDDDVVVV10101010999988887777♥♦♠♣
R-RRR-BBR-BBB-RRB-BBB-BBB-RRB-BBB-BBRBB 012 ou plusTotal026221273547125073233462 ou plus603314107Total5723181101000Retardsle1ermoisRetardsle2emois E.3711 An urn contains two blue balls and one red ball, all identical to the touch. 1 One ball is drawn and then re-turned to the urn before a second ball is drawn. The outcome tree for this random experiment is shown opposite : We will admit that we are in a situation of equiprobabil-ity. a Determine the probability of the following events : A : ˇ The first ball drawn is rouge ı. B : ˇ The two balls drawn are the same couleur ı C : ˇ The first ball drawn is red or the second ball drawn is bleu ı. b Determine the probability of the following events : A ; B ∪ C ; A ∩ B ; B ∩ C 2 We change the rules of this game as follows : there is no more handing over; the first ball drawn is discarded from the game. a Construct the outcome tree linked to this new game. b Calculate the new probabilities of the events listed in question a and b of the previous question. c Of the draws that had a blue ball on the first draw, what is the probability of getting a blue ball on the second draw? 3. Reminders: union formula E.4809 Proposition: let A and B be two events. We have the equality: P A ∪ B = P A + P B −P A ∩ B The works council of a Parisian company wants to organize a weekend in the provinces. A survey is carried out among the 1 200 employees of this company to find out their choice of means of transport (the only means of transport proposed are train, plane or coach) The results of the company’s employee survey are listed in the following table : Train Avion Autocar Total Femme 468 196 56 720 Homme 150 266 64 480 Total 618 462 120 1200 An employee of this company is randomly interviewed (it is assumed that all employees have the same chance of being in-terviewed) . F the event : ˇ the employee is a woman ı ; T the event : ˇ the employee chooses the train ı. 1 Calculate the probabilities P ( F ) , P ( T ) then determine the probability that the employee will not choose the train (results will be given in decimal form) 2 a Determine the probability of the event F ∩ T . b Deduce the probability of the event F ∪ T . E.7560 Let Ω ; P be a probabilized space où A and B be two events of Ω such that : P ( A ) = 0.42 ; P ( B ) = 0.19 Knowing that P A ∪ B =0.43 , determine the probability of the event A ∩ B . E.3703 In one school, 60 students are tak-ing part in the sports discovery day: 45 have signed up for taekwondo and 24 for judo. Knowing that 6 of them haven’t signed up for any of these activities, determine the probability that a young person en-countered at random in the center practices today: 1 taekwondo ; 2 judo ; 3 none of the above ; 4 taekwondo or judo ; 5 taekwondo and judo. 4. Introduction to conditional probabilities E.4191 1 000 people were asked the following question : ˇ How many times have you arrived late for work in the last two months? ı. The responses have been grouped in the following table : https://chingmath.fr chapExoCorrec/3711 sacados/3711 R-RRR-BBR-BBB-RRB-BBB-BBB-RRB-BBB-BBRBB chapExoCorrec/4809 sacados/4809 chapExoCorrec/7560 sacados/7560 chapExoCorrec/3703 sacados/3703 chapExoCorrec/4191 sacados/4191 012 ou plusTotal026221273547125073233462 ou plus603314107Total5723181101000Retardsle1ermoisRetardsle2emois
AB P(ABP(A·PA(BPA(BBP(ABP(A·PA(BPA(BBP(AAP(ABP(A·PA(BPA(BBP(ABP(A·PA(BPA(BBP(AA Demi-pensionnaireGarçon ::::::D:::DG::::::D:::DG An individual from this population is chosen at random. Probabilities are rounded to the nearest thousandth. 1 Determine the probability that the individual had at least one delay in the first month. 2 a Of the individuals who had no delays in the first month, what is the probability of randomly selecting an individual who had at least one delay in the second month? b Of the individuals who were at least one late in the second month, what is the probability of choosing an individual who was not late in the first month? E.5193 Consider a set Ω and two of its parts A and B represented below and whose elements are represented by crosses : Equiprobably, one element is chosen at random. 1 a What is the probability that the element drawn be-longs to A ? b Knowing that an element has been drawn from B , what is the probability that this element also belongs to A ? 2 a Determine the following probabilities : P ( B ) ; P ( A ∩ B ) b Give the value of the quotient : P ( A ∩ B ) P ( B ) . What do we notice? E.3728 The game involves shaking and tip-ping a bottle to get one of its parts out. Here are the contents of the bottle: 1 Determine the probability of the following evèvents : a A : ˇ The output element is a carré ı ; b B : ˇ Item output is rayé ı ; c A ∩ B : ˇ The element output is a square rayé ı. 2 a Determine the value of the quotient : P ( A ∩ B ) P ( A ) b The value 2 3 represents what probability? ˇ the probability of having a striped element among the square elements? ı or ˇ the probability of having a square among the striped elements ı. 3 a Determine the value of the quotient : P ( A ∩ B ) P ( B ) b Complete the sentence below : ˇ The probability of the elements . . . . . . among the elements . . . . . . has a probability of 1 3 ı 5. Representation of a random experiment E.3628 Definition: consider two events A and B with P ( A ) =0 . The probability of the event B knowing A , denoted P A ( B ) , has the value : P A B = P A ∩ B P A Note: the random experiment can then be translated into the probability tree : In a statistical study, consider a class of 24 students where each student is represented by a box in the graph below : Two characters are studied in the individuals of this study : G : ˇ the student is a garçon ı ; D : ˇ the student is halfpensionnaire ı. The random experiment consists of choosing a student at ran-dom from the class and watching whether or not these two criteria are met : 1 Without justification, answer the following questions : a What is the probability of choosing a boy? b Knowing that a boy has been chosen, what is the prob- https://chingmath.fr chapExoCorrec/5193 sacados/5193 AB chapExoCorrec/3728 sacados/3728 chapExoCorrec/3628 sacados/3628 P(ABP(A·PA(BPA(BBP(ABP(A·PA(BPA(BBP(AAP(ABP(A·PA(BPA(BBP(ABP(A·PA(BPA(BBP(AA Demi-pensionnaireGarçon ::::::D:::DG::::::D:::DG
ClasseBDemi-pensionnaireGarçon ::::::G:::GD::::::G:::GD AB BBABBA AABAAB ABE∑ectifs:81ABE∑ectifs:54ABE∑ectifs:27ABE∑ectifs:18 BBABBA AABAAB 0;40;7B0;3BA0;60;2B0;8BA ability that he will be a half-boarder? c Knowing that a girl has been chosen, what is the prob-ability that she will be a half-boarder? 2 Complete the probability tree above. E.9450 In a statistical study, consider a class of 24 students where each student is represented by a box in the graph below : Two characters are studied in the individuals of this study : G : ˇ the student is a garçon ı ; D : ˇ the student is halfpensionnaire ı. The random experiment consists of choosing a student at ran-dom from the class and watching whether or not these two criteria are met. Complete the probability tree above. E.9451 Consider the universe Ω and its two events A and B are represented : The law of equiprobability is used. Complete the two probability trees : E.9517 In a random experiment, consider two events A and B that realize the partition of the universe shown below : Statistical data for each of these two groups are shown. Since the experiment consists in randomly drawing an individual from the universe, complete the two probability trees associ-ated with this random experiment : 6. Using conditional probability E.8291 Consider the following probability tree : 1 By reading this tree, give the probabilities below : a P A B b P A B 2 Determine the probabilities below : a P A ∩ B b P A ∩ B https://chingmath.fr chapExoCorrec/9450 sacados/9450 ClasseBDemi-pensionnaireGarçon ::::::G:::GD::::::G:::GD chapExoCorrec/9451 sacados/9451 AB BBABBA AABAAB chapExoCorrec/9517 sacados/9517 ABE∑ectifs:81ABE∑ectifs:54ABE∑ectifs:27ABE∑ectifs:18 BBABBA AABAAB chapExoCorrec/8291 sacados/8291 0;40;7B0;3BA0;60;2B0;8BA
B15BA3413B23BA AAEAAE E.130 A toy company specializes in making dolls that talk and walk. Each doll can have two faults, and only two: a mechanical fault and an electrical fault. A statistical study shows that : 8% of the dolls have the mechanical defect ; 5% of the dolls have the electrical defect ; 2% of the dolls have both defects. Daily production is 1000 dolls. 1 Copy and complete the table below describing daily pro-duction : poupées avec défaut mécanique Poupées sans défaut mécanique total Poupées avec défaut électrique Poupées sans défaut électrique total 80 1 000 In the remainder of the exercise, each numerical re-sult will be given in decimal form. 2 One doll is taken at random from one day’s production. a Let A be the event ˇ the doll taken is without défaut ı. Calculate the probability of A . b Let B be the event ˇ the doll taken has at least one défaut ı. Show that the probability of B is 0.11. c Let C be the event ˇ the sampled doll has only one dé-faut ı. What is the probability of C ? d What is the probability that the removed headstock has the mechanical fault knowing that it has an elec-trical fault? E.4255 An urn contains five balls that are indistinguishable to the touch : two green and three red. Two successive draws of one ball are made according to the following rule : ˇif the ball drawn is red, we put it back in the urn ; if it’s green, we don’t put it backı. Construct a probability tree illustrating this situation. E.3727 The scene takes place at the top of a cliff by the sea. To find a beach and go swimming, tourists can only choose between two beaches, one to the east and the other to the west. A tourist finds himself at the top of the cliff on two con-secutive days. On the first day, he chooses one of the two directions at random. On the second day, it is assumed that the probability of him choosing a direction opposite to the one taken the day before is 0.8 . For t =1 or t =2 , we note E t the event : ˇ The tourist heads east on the t -th jour ı and O t the event : ˇ The tourist heads west on the t -th jour ı. 1 Draw up a probability tree describing the situation. 2 Determine the following probabilities : P ( E 1 ) ; P E 1 ( O 2 ) ; P ( E 1 ∩ E 2 ) . 3 Calculate the probability that this tourist will visit the same beach on two consecutive days. 7. Conditional probabilities and probability of the opposite event E.2094 Proposition: let A be an event and A its opposite event. We have : P A =1 −P A Corollary : let A and B be two events. We have : P A B = 1 − P A B Consider the following incomplete probability tree : Then P ( A ∩ B ) the probability of the event A ∩ B is equal to : a 21 20 b 1 5 c 20 21 d 1 12 E.2093 In this MCQ, each of the three proposed statements must be copied onto the copy and completed with the chosen answer. Only one choice is correct. No justification is re-quired. A correct answer is worth one point, a wrong answer takes away a quarter point, no answer is scored 0. If the total number of points obtained on this exercise is negative or zero, the exercise is graded zero. The following tree represents the data for a probability exer-cise. The probability of an event H is noted P ( H ) . We know that : P ( E ) = 0.3 ; P E ( A ) = 0.1 ; P E ∩ A = 0.14 1 The probability of E ∩ A is equal to : a 0.4 b 0.03 c 0.33 d 0.1 2 The probability of A knowing E is equal to : https://chingmath.fr chapExoCorrec/130 sacados/130 ?? - 2004 - 7 points - au choix chapExoCorrec/4255 sacados/4255 chapExoCorrec/3727 sacados/3727 Extrait France Septembre 2006 chapExoCorrec/2094 sacados/2094 B15BA3413B23BA chapExoCorrec/2093 sacados/2093 AAEAAE
:::::::::B0;2PAB0;08BA0;6:::PAB0;24B::::::BA 0;60;2BBAB0;3BA ::::::B:::BA::::::B:::BA 0;30;6B0;4BA0;70;5B0;5BA a 0.7 b 0.14 c 0.2 d 1.1 E.4405 Consider the following incomplete probability tree : Justifying your answers, determine the following probabili-ties : a P A b P A B c P A ∩ B d P A B e P A B f P A ∩ B E.10232 Every working day, Paul has to get to the station to take the train to work. For this, he takes his bike two times out of three, and if he doesn’t take his bike, he takes his car. When he takes his bike to the station, Paul misses the train only once in 50 whereas, when he takes his car to the station Paul misses his train once in 10 . Consider a random day on which Paul will be at the station to catch the train to work. Note: V the event ˇ Paul takes his bike to gare ı ; R the event ˇ Paul misses his train ı. 1 Construct the probability tree for this random experi- ment. 2 Without justification, give the following probabilities in the form of irreducible fractions : a P V R b P V R c P V ∩ R E.10329 Consider a random experiment and the two events A and B such that : P A = 0.6 ; P A B = 0.2 ; P A B = 0.3 Justifying each of your results, determine the following prob-abilities: a P A B b P B c P B A E.10573 In a probabilized space Ω ; P , consider the events A and B achieving: P A = 0.6 ; P A B = 0.2 ; P A ∩ B = 0.28 Leaving the steps of your reasoning, copy and complete the probability tree below : 8. Total probability formula E.4260 Definition: be A 1 , A 2 , . . . , A n events. These events are said to form a partition of Ω if we have the following two properties : For any i;j ∈ 1 ; 2 ; : : : ; n and i = j , A i ∩ A j = ∅ (they are said to be two by two disjoint) A 1 ∪ A 2 ∪ : : : ∪ A n = Ω Proposition: Total probability formula Let A 1 ; A 2 ; : : : ; A n be a partition of Ω . For any event B , we have : P = P A 1 ∩ B + P A 2 ∩ B + · · · + P A n ∩ B Corollary : For any event A and B , we have : P B = P A ∩ B + P A ∩ B Consider the following probability tree : 1 Determine the probabilities below : a P A ∩ B b P A ∩ B 2 Deduce the probability of the event B . E.2339 In a probabilistic space, consider the two events A and B verifying the following conditions : P ( A ) = 0.64 ; P A ( B ) = 0.3 ; P A ( B ) = 0.5 1 Construct a probability tree representing this situation. 2 a Determine the probabilities of the following events : P ( A ∩ B ) ; P ( A ∩ B ) b Using the total probability formula, determine the probability of the event B . https://chingmath.fr chapExoCorrec/4405 sacados/4405 :::::::::B0;2PAB0;08BA0;6:::PAB0;24B::::::BA chapExoCorrec/10232 sacados/10232 chapExoCorrec/10329 sacados/10329 0;60;2BBAB0;3BA sacados/10573 ::::::B:::BA::::::B:::BA chapExoCorrec/4260 sacados/4260 0;30;6B0;4BA0;70;5B0;5BA chapExoCorrec/2339 sacados/2339
AADAAD 5835B25BA3813B23BA E.137 Throughout the exercise, results are rounded to 10 − 4 The results of a survey of vehicles on the road in France show that : 88 % of vehicles inspected have brakes in good condition ; of vehicles inspected with brakes in good condition, 92 % have lighting in good condition ; among vehicles inspected with defective brakes, 80 % have lighting in good condition. One of the vehicles involved in the survey is chosen at random. The choices are equiprobable. We note F the event ˇ the vehicle inspected has brakes in good condition état ı. We note E the event ˇ the controlled vehicle has lighting in good condition ı. E and F denote the opposite events of E and F . 1 Describe this situation using a tree. 2 a Determine the probability P ( F ) of the event F . b What is the probability P F ( E ) , probability that the lighting is not in good condition, knowing that the brakes are not in good condition? c Show that the probability P ( E ∩ F ) of the event E ∩ F is equal to 0.8096. d What is the probability that the vehicle has good light-ing? e Any driver of a vehicle involved in the survey with faulty brakes or lighting, must have their vehicle re-paired. Calculate the probability of a driver needing repairs to his brakes or lighting. E.138 At the farm ˇ La ferme de la Poule Pondeuse ı, every day, eggs of two different sizes are produced : 60% of eggs are medium and 40% of eggs are large. Eggs are classified into two categories: those of ordinary qual-ity and those of superior quality. It has been noted that : 50 % average eggs are of ordinary quality, 20% of large eggs are ordinary quality An egg is chosen at random. Choosing an egg at random from the day’s production means that we are in a model with equiprobability. The following events are defined : M : ˇ the egg is moyen ı G : ˇ the egg is gros ı O : ˇ the egg is quality ordinaire ı S : ˇ the egg is quality supérieure ı 1 Give the following probabilities : P ( G ) : probability that the egg is large, P G ( S ) : probability that the egg is of superior quality knowing that it is large. 2 Show that the probability of taking a large, high-quality egg is equal to 0.32. 3 Calculate the probability P ( M ∩ S ) that the egg is medium and of superior quality, then the probability P ( S ) of the event S . E.134 A manufacturing line produces disposable razors in very large quantities. At the end of the line, each razor undergoes a control test by an automatic machine. The machine rejects razors with defects. Sometimes, however, the test does not detect a defect and the razor is allowed to pass, or on the contrary, a razor with no defect is rejected. A statistical study made on a very large number of razors has in fact shown that : when the shaver is correctly manufactured, the test con-firms this and accepts the item in 998 cases out of 1000. if the razor has a manufacturing defect, the test detects this and rejects the razor in 985 cases out of 1000. out of 1000 shavers manufactured, 980 have no defect. We choose a razor at random. In the following, we note : D : the event ˇ the razor has no manufacturing defect ı, D : the opposite event of D , A : the event ˇ the test accepts rasoir ı A : the opposite event of A 1 Describe each of the following events with a sentence : D ∩ A ; D ∩ A ; D ∩ A , D ∩ A 2 Using the statement, give the following probabilities : P D ( A ) (probability of A knowing that D is realized) P D A 3 a Copy and complete the following probability tree with the correct results b What is the probability that a razor will be accepted after the control test? Give the rounding to an accu-racy of 10 − 4 . E.8558 Consider a random experiment and two of its events A and B yielding the probability tree below : Establish that : P B = 1 2 https://chingmath.fr chapExoCorrec/137 sacados/137 chapExoCorrec/138 sacados/138 chapExoCorrec/134 sacados/134 AADAAD chapExoCorrec/8558 sacados/8558 5835B25BA3813B23BA
BBABBA ::::::Po::::::Po:::M::::::Po::::::Po:::M 9. Condition reversal E.5555 Consider a random experi-ment and two of its events A and B . We have the information on these two events : P ( A ) = 0.55 ; P A ( B ) = 0.95 ; P A ( B ) = 0.1 1 Using the information in the statement, complete the probability tree below : 2 Calculate P A ∩ B the probability of the event A ∩ B . 3 Show that : P ( B )=0.5675 4 Calculate P B ( A ) , the probability of the event A knowing the event B realized. Give a value rounded to 10 − 4 . E.4160 A and B are two events re-lated to the same event that verify: P ( A ) = 0.4 ; P A ( B ) = 0.7 ; P A ( B ) = 0.1 Indicate whether the following statement is true or false, jus-tifying your answer. Assertion: The probability of the event A , knowing that the event B is realized, is 14 41 E.6090 On a probability space, con-sider two events A and B whose following probabilities are known : P A = 1 4 ; P A B = 1 3 ; P A B = 1 2 1 Translate this situation into a probability tree. 2 Determine the probability of the event A ∩ B . 3 Demonstrate that the probability of the event B is 11 24 . 4 What is the probability of the event B knowing that the event A is realized? E.6760 A market gardener specializes in strawberry production. The market gardener produces his strawberries in two green-houses noted A and B : 55 % of the strawberry flowers are in the A greenhouse, and 45 % in the B greenhouse. In the A greenhouse, the probability of each flower giving fruit is equal to 0.88 ; in the B greenhouse, it is equal to 0.84 . For each of the following propositions, indicate whether it is true or false, justifying the answer. An unjustified answer will not be taken into account. Proposition 1: The probability that a strawberry flower, chosen at random from this farm, will give a fruit is equal to 0.862 . Proposition 2: We find that a flower, chosen at random from this farm, pro-duces a fruit. The probability that it is located in the green-house A , rounded to the thousandth is equal to 0.439 . E.123 In one European country, 12 % of sheep are affected by a disease. A screening test for this disease has just been released, but it is not totally reliable. A study has shown that when the sheep is sick the test is positive in 93 % of cases ; when the sheep is healthy, the test is negative in 97 % of cases. A sheep is chosen at random and tested for the disease. We note M the event ˇ the sheep is malade ı. We note Po the event ˇ the test is positive ı. 1 Copy and complete the following probability tree : 2 Calculate the probabilities of the following events A , B , C : A : ˇThe sheep is sick and the test is positifı B : ˇThe sheep is healthy and the test is positifı C : ˇThe sheep is sick and the test is négatifı. 3 Deduce that the probability of the event Po is equal to 0.138 . What is the probability of the test being negative? 4 In this question, results will be rounded to the thou-sandth. a Knowing that a sheep has a positive test, what is the probability that it is not sick? b Knowing that a sheep has a negative test, what is the probability that it is sick? https://chingmath.fr chapExoCorrec/5555 sacados/5555 BBABBA chapExoCorrec/4160 sacados/4160 chapExoCorrec/6090 sacados/6090 chapExoCorrec/6760 sacados/6760 Extrait Asie Juin 2016 chapExoCorrec/123 sacados/123 ::::::Po::::::Po:::M::::::Po::::::Po:::M
B2N2B1B2N2N1 E.139 Reminders : we note : P ( A ) the probability of an event A , ˇ A ∩ B ı the intersection of two events A and B . P B ( A ) the probability of an event A occurring, know-ing that an event B (of non-zero probability) has already occurred. We have : P B ( A ) = P ( A ∩ B ) P ( B ) . There are two urns numbered 1 and 2. Urn 1 contains one white and one black ball. Urn 2 contains two black balls and one white ball. The following random experiment is performed : a ball is drawn at random from urn 1 and placed in urn 2, then a ball is drawn at random from urn 2. All draws are assumed to be equiprobable. Note: N 1 the event : ˇ The ball drawn from urn 1 is noire ı ; B 1 the event : ˇ The ball drawn from urn 1 is blanche ı ; N 2 the event : ˇ The ball drawn from urn 2 is noire ı ; B 2 the event : ˇ The ball drawn from urn 2 is blanche ı 1 Give the values of P ( B 1 ) and P ( N 1 ) . 2 Show that : P B 1 ( B 2 )= 1 2 . In the same way give the values of : P B 1 ( N 2 ) ; P N 1 ( B 2 ) ; P N 1 ( N 2 ) . 3 Complete the probability tree : 4 Calculer P ( B 1 ∩ B 2 ) . 5 Show that P ( B 2 )= 3 8 then calculate P ( N 2 ) . 6 Knowing that a white ball has just been drawn from urn 2, what is the probability that a white ball was previously drawn from urn 1? E.4145 An Internet user wishes to make a purchase via the Internet. Four sales sites, one French, one German, one Canadian and one Indian, present the equip-ment he wishes to purchase. The experiment has shown that the probability of him using each of these sites verifies the following conditions (the initials of the countries country ini-tials denote events ˇ purchase takes place in country ı) : P ( F ) = P ( A ) ; P ( F ) = 1 2 ·P ( C ) ; P ( C ) = P ( I ) 1 Calculate the four probabilities P ( F ) , P ( A ) , P ( C ) and P ( I ) . 2 On each of the four sites, the surfer can buy a supplement for his material. His previous experiments lead him to formulate the conditional probabilities of this event, de-noted S , as follows : P F ( S ) = 0.2 ; P A ( S ) = 0.5 ; P C ( S ) = 0.1 ; P I ( S ) = 0.4 a Determine : P ( S ∩ A ) . b Show that : P ( S )= 17 60 . c The surfer finally bought a supplement. Determine the probability that he bought it on the Canadian site. E.4170 On my street, it rains one evening in four. If it rains, I take my dog out with a probability equal to 1 10 ; if it doesn’t rain, I take my dog out with a probability equal to 9 10 . Knowing that I’ve taken my dog out, what’s the probability of it raining? https://chingmath.fr chapExoCorrec/139 sacados/139 B2N2B1B2N2N1 chapExoCorrec/4145 sacados/4145 chapExoCorrec/4170 sacados/4170 Extrait de Liban Juin 2009
0;34...B...BA......B...BA BBABBA AABAAB 0;540;7B...BA......B0,8BA ::::::A...AB......A:::AB E.7166 A marathon is a running sporting event. In this exercise, all approximate results will be given to the nearest 10 − 3 . A study of the Tartonville Marathon shows that : 34 % of runners finish the race in less than 234 minutes ; among runners who finish the race in less than 234 min-utes, 5 % are over 60 years old ; among runners who finish the race in more than 234 min-utes, 84 % are less than 60 years old. A runner is selected at random and the following events are considered : A : ˇ the runner finished the marathon in less than 234 minutes ı ; B : ˇ the runner has less than 60 ans ı. Recall that if E and F are two events, the probability of the event E is denoted P E and that of E knowing F is denoted P F E . In addition, E denotes the opposite event of E . 1 Copy and complete the probability tree below associated with the situation in the exercise: 2 a Calculate the probability that the chosen person will have completed the marathon in less than 234 minutes and be over 60 years old. b Check that : P B ≈ 0.123 c Calculate P B A and interpret the result in the con-text of the exercise. E.10342 The librarian at a high school wants to buy the Harry Potter saga novels. She investigates whether the subject interests the students : 10 % of the students have read the 7 e volume ; 90 % of students who have read the 7 e tome have seen the 7 e film; 55 % of students who have not read the 7 e tome have seen the 7 e film. 1 Represent the situation using a weighted probability tree. 2 A student is chosen at random, what is the probability that he has only read 7 e tome? 3 What is the probability that a randomly chosen student has seen the 7 e film? 4 Student has seen the 7 e film, what is the probability that he has read the 7 e volume? (the value will be rounded to the nearest thousandth) 10. Tree construction with condition inversion E.9448 Consider the two events A and B achieving the following conditions : P A ∩ B = 0.05 ; P A ∩ B = 0.15 P A ∩ B = 0.35 ; P A ∩ B = 0.45 Complete the following two probability trees : E.3601 In a probability space, consider two events A and B . Here’s a probability tree made with these two events : 1 Complete the probabil-ity tree representing this random experiment. 2 Justify each of the following values (rounded to the near-est thousandth) : a P ( A ∩ B ) = 0.378 b P ( A ∩ B ) = 0.092 c P ( B ) = 0.47 d P B ( A ) ≈ 0.804 3 Determine the following probabilities : a P ( A ∩ B ) d P ( B ) f P B ( A ) 4 Construct the probability tree opposite, completing it with the values of the prob-abilities rounded to the thou-sandth : https://chingmath.fr chapExoCorrec/7166 sacados/7166 0;34...B...BA......B...BA chapExoCorrec/10342 sacados/10342 From Maud Sauver chapExoCorrec/9448 sacados/9448 BBABBA AABAAB chapExoCorrec/3601 sacados/3601 0;540;7B...BA......B0,8BA ::::::A...AB......A:::AB
BBABBA AABAAB A2R2R2A2A1R1R1A1 V2N2V1V2N2R2N1 E.5832 In a probability space, consider two events A and B . The following probabilities are known : P ( A ) = 0.3 ; P A ( B ) = 0.8 ; P A ( B ) = 0.6 Complete, if necessary with values rounded to the nearest hundredth, the two probability trees above. 11. Non-symmetrical trees E.9446 A company commissions a telephone survey company to conduct a survey on the qual-ity of its products. Each interviewer has a list of people to contact. On the first phone call, the probability of the caller be-ing absent is 0.4 . Knowing that the correspondent is present, the probability that he will agree to answer the questionnaire is 0.2 . 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 on the second call, no further at-tempts are made to contact them. For i ∈ 1 ; 2 , we note : A i the event ˇ the person is absent at i ième appel ı ; R i the event ˇ the person agrees to answer the question-naire at the i ième appel ı ; 1 Complete the probability tree : 2 Show that the probability of achieving the event ˇ the person answered questionnaire ı is 0.176 . E.4264 In this exercise, all results will be given as irreducible frac-tions. There is a perfectly balanced A cube die with one green, two black and three red faces and a second balanced B cube die with four green and two black faces. The game proceeds as follows : the die B is rolled: if the face obtained is green, the die is rolled again B and the color of the face obtained is noted ; if the face obtained is black, we roll the die A and note the color of the face obtained. 1 a Complete the probability tree below : b What is the probability of getting a green face on the second throw, knowing that you got a green face on the first throw? 2 Show that the probability of obtaining two green faces is equal to 4 9 . 3 What is the probability of getting a green face on the second throw? https://chingmath.fr chapExoCorrec/5832 sacados/5832 BBABBA AABAAB chapExoCorrec/9446 sacados/9446 A2R2R2A2A1R1R1A1 chapExoCorrec/4264 sacados/4264 V2N2V1V2N2R2N1
F2F2F2F2F2F2F1F1F1F1MM EEAEEA B1B2B2B1 E.10577 In this maternity ward, among twin births, it is estimated that’there are 30 % monozygotic twins (called ˇidentical twinsı who are necessarily of the same sex: two boys or two girls) and therefore 70 % dizygotic twins (called ˇfraternal twins,ı who may be of different sexes: two boys, two girls, or a boy and a girl) . In the case of twin births, it is assumed that, as with ordinary births, the probability of being a girl at birth is equal to 0.49 and that of being a boy at birth is equal to 0.51 . In the case of ’a twin birth of dizygotic twins, it is also as-sumed that the sex of the second twin is independent of the sex of the first twin. A twin birth in this maternity ward is chosen at random and the following events are considered : M : ˇ the twins are monozygotic ı ; F 1 : ˇ the first newborn is a girl ı ; F 2 : ˇ the second newborn is a girl ı. We note P ( A ) the probability of event ’ A and A the’contrary event of A . 1 Copy and complete the probability tree above 2 Show that the probability that both newborns are girls is 0.315 07 . 3 The two newborns are twins. Calculate the probability that they are monozygotic. 12. With a little algebra E.9535 A chocolate company has two production lines. On the C 1 line, 2 % of the chocolate plaque-ttes have a defect on their amballage whereas on the C 2 line, 5 % of chocolate wafers have a defect on their packaging. At the end of the day, the inspector draws a chocolate wafer at random from the stock produced during the day. The events are noted : A : ˇ the chocolate wafer was produced on the 1 ı chain E : ˇ the chocolate wafer has a defect emballage ı and we note x the probability of drawing a chocolate wafer from the chain C 1 . 1 Copy and complete the probability tree below. 2 Knowing that the inspector has a probability of 0.97 of drawing a chocolate wafer with no packaging defects, de-termine the probability of the event A . E.9449 A showman wants to build a game where participants have 5 % chance of winning. To do this, he has an urn containing 40 balls of which n are blue (where n is an integer) . The participant wins if he draws a blue ball. If on the first try the ball drawn is not blue, he does not return the drawn ball to the urn and withdraws a second ball. We note B the event ˇ the drawn ball is blue ı. 1 Complete the probability tree : 2 Note G the event ˇ the participant has gagné ı. a Show that : P G = 79 · n − n 2 1560 b Determine the number n of blue balls contained in the urn. https://chingmath.fr chapExoCorrec/10577 sacados/10577 Annales 2023 F2F2F2F2F2F2F1F1F1F1MM chapExoCorrec/9535 sacados/9535 EEAEEA chapExoCorrec/9449 sacados/9449 B1B2B2B1
x2xB1−2xBA1−x3xB1−3xBA 0;xxB1−xBA0;3−x0;3−xB0;xBA 0;30;10;5C0;5CB0;90;6C0;4CBA0;70;80;24C0;76CB0;20;53C0;47CBA AA491332122686934250611058VVVVPP :::::::::V:::VP::::::V:::VPA:::::::::V:::VP::::::V:::VPA E.6741 In a random experiment, con-sider two events A and B that allow the probability tree below to be constructed and such that there exists a real number x verifying: P A = x ; P A B = 2 x ; P A B = 3 x 1 In this question, it is assumed that P B = 29 100 . Determine the value of x . 2 In this question, it is assumed that P B A = 1 5 . Determine the value of x . E.10233 An urn contains blue and red balls. The associated random experiment consists of : a first ball is drawn from the urn and set aside then a second ball is drawn from the urn. We note : B 1 : the first ball drawn is blue ; B 2 : the second ball drawn is blue. We know that : initially, the urn contains 21 blue balls ; the probability that the first ball is blue knowing that the second ball is blue has the value 5 6 . Determine the number of balls initially contained in the urn. E.9516 In a probability space, con-sider two events A and B whose probability tree, given below, whose probabilities depend on a real parameter x belonging to the interval 0 ; 0.3 : Determine the value of parameter x so that : P B =0.1 . 13. Deepening: a succession of tests E.8322 We consider a random experiment and three such events A , B and C allowing us to construct the following probability tree : 1 Noting the values on the probability tree, give the prob-abilities: a P A B b P A ∩ B C c P A ∩ B C 2 Determine probabilities : a P A ∩ B b P A ∩ B ∩ C c P A ∩ B ∩ C E.7249 We consider a population with three criteria: age, whether or not they own their own home, and whether or not they own a car. To organize this population, we consider the following three classes of individuals : A : ˇ the individual is 50 years old or plus ı P : ˇ the individual owns his logement ı. V : ˇ the individual owns a voiture ı. Here is the table of numbers obtained from the study : Consider the random experiment involving the random selec-tion of an individual from the study population. Considering the events associated with the classes in the statistical study, complete the probability tree below : 14. Deepening: with suites https://chingmath.fr chapExoCorrec/6741 sacados/6741 x2xB1−2xBA1−x3xB1−3xBA chapExoCorrec/10233 sacados/10233 chapExoCorrec/9516 sacados/9516 0;xxB1−xBA0;3−x0;3−xB0;xBA chapExoCorrec/8322 sacados/8322 0;30;10;5C0;5CB0;90;6C0;4CBA0;70;80;24C0;76CB0;20;53C0;47CBA chapExoCorrec/7249 sacados/7249 AA491332122686934250611058VVVVPP :::::::::V:::VP::::::V:::VPA:::::::::V:::VP::::::V:::VPA
AnAnAnAnAnAn E.6066 A player starts a video game and plays several successive games. It is assumed that : the probability of him winning the first game is 0.1 ; if he wins one game, the probability of winning the next is equal to 0.8 ; if he loses one game, the probability of winning the next is equal to 0.6 . We note, for any non-zero natural number n : G n the event ˇ the player wins the n -th partie ı ; p n the probability of the event G n . We therefore have : p 1 =0.1 Using the total probability formula, show that for any non-zero natural number n : p n +1 = 1 5 · p n + 3 5 . E.6742 In a probabilized space Ω ; P . Con-sider a sequence of events A n verifying the following rela- tions : P A 0 ) = 0.4 ; P A n A n +1 = 0.6 P A n A n +1 = 0.4 for any n ∈ N Note: p n = P A n . 1 Complete the probability tree opposite. 2 a Establish that : p n +1 = 0.2 · p n + 0.4 b We define the sequence q n by: q n = p n − 0.5 ∀ n ∈ N Establish that the se-quence q n is a geomet-ric sequence of reason 0.2 . c Deduce the expression of the sequence p n as a func-tion of n . 2 Determine limit: lim n ↦→ + ∞ P A n . https://chingmath.fr chapExoCorrec/6066 sacados/6066 chapExoCorrec/6742 sacados/6742 AnAnAnAnAnAn