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概率论试题及答案(Probability theory questions and answers)Paper OneFill in the blanks (2 points per day, a total of 10 points)1. is the three random events, at least two can be expressed as _.2. throw a dice, said odd point, said the number is not greater than 3, said _.Two events of the 3. known exclusive content, is _.4. to two random events, and, is _.5. is the three random events, and at least, the probability of occurrence for a _.Two, single choice (four options every day in only one correct answer, please fill in the number of correct answers in parentheses. Every day 2 points, a total of 20 points)1. take two balls from a bag containing 2 red balls and 2 white balls, and remember take 2 white balls, then ().(A) 2 red balls (B) were taken and 1 white balls were taken(C) did not take the white ball (D) and took at least 1 red balls2. pairs of flip a coin test, appear positive known as ().(A) random events (B) necessary events(C) the impossible event (D) sample space3. set A and B as random events, then ().(A) A (B) B(C) AB (D) Phi4. set and is any two probability is not zero mutually exclusive event, then the following conclusion is definitely correct ().(A) and mutual exclusion (B) and non mutual exclusion(C) (D)5. to two random events, and then the following formula is correct ().(A) (B)(C) (D)6., independent of each other; ().(A) (B)(C) (D)The 7. set is three random events, and there are.(A) 0.1 (B) 0.6(C) 0.8 (D) 0.78. for a series of independent trials, the probability of success for each trial was p, and the probability of failure 3 times before the success of 2 was.(A) P2 (1 - P) 3 (B) 4 P (1 - P) 3(C) 5, P, 2 (1 - P) 3 (D) 4, P 2 (1 - P) 39. A, B two and random events, the following formula is correct, ().(A) (B)(C) (D)10. when the event A and B occur at the same time, the event C must occur, then ().(A) P (A B) = P (C) (B) P (A) + P (B) - P (C) = 1(C) P (A) + P (B) - P (C) = 1 (D) P (A) + P (B) = P (C)Three, calculation and application (every day 8 points, a total of 64 points)Equipped with 5 white ball 1. bags, 3 black balls. Take two from each other.The probability of getting two balls of different colors.2.10 there are 3 keys to open the door lock. Take two things at this time.The probability of opening a door.3. there are 6 students living in a dormitory,Ask 4 of them to have their birthdays in the same month.Of the 4.50 products, 46 were qualified and 4 were defective, and 3 were extracted at one time,The probability of getting at least one defective product.5. to process a part, it takes three steps to assume that the defective rate of the first, second, third processes is 0.2, 0.1, 0.1, respectively, and that any process is defective and has nothing to do with the other processes.Find the defective rate of this kind of parts.6., the qualified rate of a known product is 0.95, and the first grade rate of qualified products is 0.65.Ask for the first rate of the product.7., a box of 100 products, of which the number of defective products from 0 to 2 is possible. When unpacking inspection, 10 pieces are taken out randomly, and if the defective goods are found, the product is rejected. If the case is known to have passed the acceptance,The probability that there is no defect in it.8., a factory product, according to a process of processing, according to the process of processing, two kinds of processing products out of the pass rate was 0.8 and 0.9. It is now being tested from 5 of the products in the factory,The probability of having at most one defect.Four, the certificate (6 points)Set up,. ProvePaper OneReference answerFill in the blanks1. or2., the number of points appears exactly 5ThreeMutually exclusivebe4. 0.6soFiveAt least one, that isAgain fromsoTwo, individual choiceOne2. A3. AUsing the operational properties of setsFourMutually exclusivesoFivesoSixMutual independenceSevenAndbeEight9. B10. BThe P (A) + P (B) - P (C) = 1Three, calculation and application questions1. solutions:Set the indication that the two balls scored are different in colorAnd the total sample pointsso2. solutions:If you can lock the door, thenso3. solutions:It is stated that 4 birthdays are on JanuaryAnd the total sample points areso4. solutions:To indicate at least one defective product, because of its complexity, consider adverse events = no defective goods are taken.Contains sample points of. And the total sample points areso5. solutions:Set up a part for a defective productBy the request, but is more complex, considering adverse events take a part for the genuine, said by the three processes are qualified,beTherefore6. solutions:It means the product is the best, which means that the product is qualifiedObviouslyThereforeThat is, the primary grade of the product is7. solutions:There is a defective item in the box,Also set the box products through acceptance by the total probability formulaTherefore8. solutions:According to the passage, the products of this factory pass rate,Therefore, the rate of defective products isIt says take back and take 5, at most one defective item is takenbeFour. Proof questionsProve,Knowledge of the nature of probabilityalsoAndsoPaper TwoFill in the blanks (2 points per day, a total of 10 points)1. if the probability distribution of random variables, and then _.The 2. set of random variables, and then _.3. random variables, is _.4. random variables, is _.5. if the probability distribution of the random variable isIs _.Two, single choice (four options for each question is only one right answer, please fill in the number of correct answers in parentheses. Every day 2 points, a total of 20 points)1. set and is the distribution function of two random variables, for the distribution function of a random variable in the following set of numerical values should be taken ().(A) (B)(C) (D)2. the probability density of the set of random variables is.(A) (B)(C) (D)3. the following function is a random variable whose distribution density is ().(A) (B)(C) (D)4. the following function is a random variable whose distribution density is ().(A) (B)(C) (D)5. the probability density of the random variable is, and the probability density is ().(A) (B)(C) (D)6. sets obey two distributions, then ().(A) (B)(C) (D)7. sets, then ().(A) (B)(C) (D)8. the distribution density of the random variables is.(A) 2 (B) 1(C) 1/2 (D) 49. for a random variable, if it is, it can be determined not to obey.(A) two item distribution (B) exponential distribution(C) normal distribution (D) Poisson distribution10. to be a random variable subject to a normal distribution.(A) 9 (B) 6(C) 4 (D) -3Three, calculation and application (every day 8 points, a total of 64 points)There are 12 table tennis balls in the 1. boxes, 9 of which are new balls and the 3 ones are old ones. Take it back, take one at a time until you get the new one.Find the probability distribution of the number of extraction.2. of the 6 workers in the workshop are working independently. It is known that each person needs a small crane within 1 hours of 12 minutes.(1) what is the most probable value of the number of small cranes needed at the same time?(2) if there are only 2 small cranes in the workshop, what is the probability of delaying work due to the shortage of small cranes?3. the lifetime of a certain electronic component is a random variable whose probability density isFind (1) constant;(2) if the 3 elements are connected on a single line, the probability that the line will work normally after 150 hours is calculated.4. the life of a battery (unit: H) is a random variable, and.(1) the probability that such a battery will last more than 250 hours;(2) the probability of battery life is not less than 0.9.5. set random variables.Probability density.6., if the random variable obeys Poisson distribution, namely.Beg.7. the probability density of the random variables is.Summing.8. a car runs along a street and needs to pass through three intersections without traffic lights,Each signal lamp is red or green, and the other signals are red or green. Each other signal is red or green, and the two signals display the same time. The number of crossings that indicate that the car has not encountered a red light.Find (1) the probability distribution;(2).Four, the certificate (6 points)The exponential distribution of random variables subject to parameter 2 is established.Proof: uniformly distributed over the interval.Paper TwoReference answerFill in the blanks1.6By the nature of the probability distributionNamely,Get.TwoThen,3. 0.5Four5., 0.25By question setting, you can setThat is010.5, 0.5beTwo, individual choice(1.)By the nature of the distribution functionHowever, after verification, only satisfaction, election(2.)By the nature of probability density, there are(3.)By the nature of probability density, there are(4.)By the nature of the density function, there are(5.)Is a single reduction function, whose inverse function is, and the derivative is obtainedBy the formula, the density is(6.)By subject to two distributions, thenIt is also known by the nature of variance,(7.)Therefore8. (A) is defined by the normal distribution density9. (D)If the star, can only choose the Poisson distribution.10. (D)Dreams X normal distribution N (-1, 2), EX = -1* E (2X - 1) = -3Three, calculation and application questions1. solutions:Number of times to extractThere is only one old ball, so the possible value is.:By classical model, there arebe12342. solutions:A representation of the same time the number of required small crane, is a random variable, the problem is,SoThe most probable value of (1) is the maximum probability(2)3. solutions:(1) available(2) the necessary and sufficient condition of the normal operation of the series line is that each component can work normally, and the work of the three components is independent of each otherandso4. solutions:(1)(check the normal distribution table)(2) by C.Look-up table.5. solutions:The corresponding function increases monotonically, and its inverse function is, and the derivative is obtained,By question set againIt is known by the formula:6. solutions:Then,andSet by questionThat isAvailablesoCheck Poisson distribution table,7. solutions:Defined by mathematical expectation,andso8. solutions:(1) the possible value is available and by youThat is0123(2) the mathematical expectation of the function of the discrete random variable hasFour. Proof questionsProve:KnownAgain from continuous, monotone, existence inverse functionAndAt that time, thensoThat isTest paper threeFill in the blanks (please fill in the blanks directly on the line. Every day 2 points, a total of 10 points)1., the joint distribution law of two dimensional random variables is established,Is _, _.2. set random variables and independent of each other, and their probability distributions are,Is _.3. if the random variables are independent of each other, and,Obey the _ distribution.4. are known to be independent and identically distributed, andIs _.5. the mathematical expectation of the random variable is the variance, and the Chebyshev inequality is_.Two, single choice (in four options for each question is only one right answer, please fill in the number of correct answers in parentheses. Every day 2 points, a total of 20 points)1. if the joint probability density of two dimensional random variables is then the coefficient (.)(A) (B)(C) (D)2. set two independent random variables and obey the normal distribution respectively, and the following conclusions are correct(A) (B)(C) (D)3. set the joint distribution density of random vectors (X, Y), then ()(A) (X, Y) obey exponential distribution (B), X and Y are not independent(C) X and Y are independent of each other (D) cov (X, Y) = 04. random variables are independent and obey uniform distribution on the interval 0,1, the following is subject to the uniform distribution of random variables (a).(A) (B)(C) (D)5., the random variable and the random variable are independent and identically distributed, andIn the following categories, the establishment is ()(A) (B) (C) (D);6. both the expectation and the variance of the set of random variables exist(A) (B)(C) (D)7. if the random variable is a linear function, and the random variable exists mathematical expectation and variance,Coefficient of correlation ()(A) (B) (C) (D);The 8. is a two dimensional random variable, and the necessary and sufficient condition is that the random variable is uncorrelated(A)(B)(C)(D)The 9. set is a random variable independent of each other,Yes, yes (yes)(A) (B)(C) (D)10. set as independent and identically distributed random variables sequence, and Xi (I = 1,2),. The exponential distribution of the parameter is lambda, and the density function of the normal distribution N (0, 1) is, and ()Three, calculation and application (every day 8 points, a total of 64 points)1. put the 2 balls randomly into 3 boxes, and indicate the number of balls placed in the first box, which indicates the number of boxes with the ballFind the joint probability distribution of two dimensional random variables2. the joint probability density of two dimensional random variables is(1) the determined value;(2) seekingThe joint density of 3. sets is(1) for edge density and;(2) judge whether or not they are independent of each otherThe joint density of 4. sets isThe probability density of5. set, and are independent of each otherJoint probability density of (1);(2);(3)The joint probability density of 6. sets isSum7., 100 artillery bombardment of enemy positions. The mathematical expectation of each projectile hitting the target is 4, and the standard deviation is 1.5.Ask for the probability of 380 to 420 shells hitting the target in the 100 artillery bombardment8. when sampling the quality of the product, if the number of defective products is found to be more than 10, the products will not be acceptedHow many products should be inspected to allow the rate of defective products to be 10%? The probability of this product being rejected is 0.9.Four, the certificate (6 points)The mathematical expectation of a random variable exists, and it is proved that the covariance of a random variable with any constant is zeroTest paper threeReference answerFill in the blanksOneThe relation between the joint distribution law and the edge distribution is obtained by the nature of the joint distribution lawTwoThreeThe sum of independent normal variables still obeys normal distributionAnd,LFourFiveTwo, individual choice1. (B)fromThat isStar selection (B).2. (B)The problem is known,Standardization will therefore be madeStar selection (B).3. (C)Star selection (C).4. (C)Dreams are random variables are independent and uniformly distributed on an interval 0,1, thenStar selection (C).5. (A)Star selection (A).6. (A)The nature of knowledge by the expectation of imprisonmentStar selection (A).7. (D)Star selection (D).8. (B)The necessary and sufficient condition is thatThat isbeStar selection (B).9. (C)Star selection (C).10. (A)Xi (I = 1,2),. An exponential distribution that obeys the parameter lambdasoStar selection (A).Three, calculation and application questions1. solutionThe apparent possibility is that the possible value isNotice that the ball is randomly placed into a box, and there is a method of seed lettingThe joint distribution law is2. solution(1) by the nature of probability densityAvailable(2) establishment;3. solution(1)That isNamely,(2) at that timeTherefore, random variables are independent of each other4. solutionThe distribution function we seek first, obviously, the value of the random variable is not negative, thereforeThen,Then,The probability density is5. solution(1) independent of each otherThe joint density is(2)(3)6. solutionThereforeBy symmetryso.7. solutionSets the number of shells that represent the target of the first bombardment,By question set, there are,The number of shells hit by the secondary bombardment,Because they are independent and identically distributed, they are known by the central limit theoremApproximately obeys a normal distributionTherefore8. solutionThe product shall be inspected, and the number of defective items shall be determined,Here, you can think that the larger, then by the di - Laplasse theorem known,Approximately obeys a normal distributionAccording to the passage, there isThat isOrLook-up tableTherefore, at least one product should be inspected to meet the design requirementsFour. Proof questionscardThe definition of covariance and the nature of mathematical expectation are obtained
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