Dnyansandhya IIT MATHS Past Paper PDF 10-11-2024

kailas suralkar Dnyansandhya IIT MATHS Anand Nagar, Jamner Contact : 9730289357 12 Date : 10-11-2024...

kailas suralkar Dnyansandhya IIT MATHS Anand Nagar, Jamner Contact : 9730289357 12 Date : 10-11-2024 TEST ID: 37 Time : 01:30:00 MATHEMATICS Marks : 100 14.PROBABILITY DISTRIBUTIONS ,15.BINOMIAL DISTRIBUTION ,7.LINEAR PROGRAMMING ,8.PROBABILITY DISTRIBUTION,9.BERNOULLI TRIALS AND BINOMIAL DISTRIBUTION,9.LINEAR PROGRAMMING PROBLEMS Single Correct Answer Type 8 32 c) d) 1. For a Binomial variate X with n = 6, 81 81 if P(X = 2) = 9 P(X = ), then its variance is 7. In eight throws of a die 1 or 3 is considered a 8 1 success. Then the standard deviation of the a) b) 9 4 success is 9 d) 4 16 8 4 2 c) a) b) c) d) 8 9 3 3 3 2. A rifleman is firing at a distant target and has 8. A die has four black faces and two faces only 10% chance of hitting it. The least marked 3. The chance of getting a total of 12 in number of rounds, he must fire in order to 5 throws is have more than 50% chance of hitting it at 1 4 2 1 2 4 a) 5𝐶4 ( ) ( ) b) 5𝐶4 ( ) ( ) least once is 3 3 3 3 a) 5 b) 7 c) 9 d) 11 1 5 1 4 5 c) 5𝐶4 ( ) d) 5𝐶4 ( ) ( ) 3. The mean and the variance of a binomial 6 6 6 distribution are 4 and 2 respectively. Then, the 9. It is observed that 25% of the cases related to probability of 2 successes is child labor reported to the police station are 28 219 solved. If 6 new cases are reported, then the a) b) 256 256 probability that atleast 5 of them will be 128 37 c) d) solved is 256 256 4. If the records of a hospital show that 10% of 1 6 19 a) ( ) b) 4 1024 the cases of a certain disease are fatal. If 6 19 19 patients are suffering from the disease, then c) d) 2048 4096 the probability that only three will die is 10. If two coins are tossed five times, then the a) 8748× 10−5 b) 1458× 10−5 probability of getting5 heads and 5 fails is c) 1458× 10−6 d) 41× 10−6 63 1 5. Assuming that for a husband-wife couple, the a) b) 256 1024 chances of their child being a boy or a girl are 2 9 c) d) the same, the probability of their two children 205 64 being a boy and a girl is 11. Below are the constraints of an LPP. 5 ≤ 𝑥 ≤ 1 1 1 10, 5 ≤ 𝑦 ≤ 10. a) b) 1 c) d) 4 2 8 Determine the vertices of the feasible region 6. The probability of India winning a test match formed by them 2 against England is. Assuming independence a) (5, 5), (10, 5), (10, 10), (15, 10) 3 from match to match, the probability that in a 7 b) (5, 5), (10, 10), (10, 15), (5, 10) match series India's third win occurs at the c) (5, 5), (10, 5), (10, 10), (5, 10) fifth match, is d) (5, 5), (15, 10), (10, 10), (5, 10) 8 16 12. The maximum value of P = 6𝑥 + 11𝑦 subject a) b) to the constraints 27 81 2𝑥 + 𝑦 ≤ 104 Page|1  𝑥 + 2𝑦 ≤ 76 And 𝑥 ≥ 0, 𝑦 ≥ 0 is a) 240 b) 540 c) 440 d) None of these 13. The solution set of the inequation 2𝑥 + 𝑦 > 5 is a) Half plane that contains the origin b) Half plane not containing the origin Whole XY-plane except the points lying on c) the line 2𝑥 + 𝑦 = 5 d) None of these Let Z = 3x - 4y be the objective function. 14. The area of the feasible region for the following Then , maximum and minimum of Z occurs at constraints 3𝑦 + 𝑥 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 will be a) (5, 0), (0, 8) b) (6, 5), (4, 10) a) Bounded b) Unbounded c) (6, 8), (0, 8) d) (4, 10), (0, 8) c) Convex d) Concave 21. The maximum value of the objective function 15. Inequality 𝑥 + 5𝑦 ≤ 6 lies 𝑍 = 3𝑥 + 2𝑦 for linear constraints 𝑥 + 𝑦 ≤ 7, a) In origin side of 𝑥 + 5𝑦 = 6 2𝑥 + 3𝑦 ≤ 16, 𝑥 ≥ 0, 𝑦 ≥ 0 is a) 16 b) 21 b) In non-origin side of 𝑥 + 5𝑦 = 6 c) 25 d) 28 c) In either side of 𝑥 + 5𝑦 = 6 22. The solution of the linear inequalities 2𝑥 + 𝑦 ≥ d) None of these 8 and 𝑥 + 2𝑦 ≥ 10 is 16. Solve the linear programming problem. Maximise Z = x + 2y Subject to constraints x - y ≤ 10, 2x + 3y ≤ 20 and x ≥ 0, y ≥ 0 a) Z = 10 b) Z = 30 a) c) Z = 40 d) None of these 17. For the constraint of a linear optimizing function 𝑧 = 𝑥1 + 𝑥2 , given by 𝑥1 + 𝑥2 ≤ 1, 3𝑥1 + 𝑥2 ≥ 3 and 𝑥1 , 𝑥2 ≥ 0 a) There are two feasible regions b) There are infinite feasible regions b) c) There are no feasible regions d) None of these 18. The shaded region for the inequality x + 5y ≤ 6 is a) to the non-origin side b) to the either side of of x + 5y = 6 x + 5y = 6 c) to the origin side of x d) to the neither side of + 5y = 6 x + 5y = 6 19. For an LPP, minimise 𝑍 = 2𝑥 + 𝑦 subject to c) constraints 5𝑥 + 10𝑦 ≤ 50, 𝑥 + 𝑦 ≥ 1, 𝑦 ≤ 4 and 𝑥, 𝑦 ≥ 0, then 𝑍 is equal to a) 0 b) 1 c) 2 d) 12 20. The feasible solution for a LPP is shown as d) None of these below : 23. Which of the following statements is correct? a) Every L.P.P has an optimal solution b) A L.P.P has a unique solution c) If a L.P.P has two optimal solutions, then it has an infinite number of optimal solutions d) A L.P.P has two optimal solutions 24. The coordinate of the point at which minimum Page|2  value of Z = 7x - 8y, subject to the constraints x P(X = 𝑥) k 3k 5k 2k k + y - 20 ≤ 0, y≥ 5, x ≥ 0 is attained, is Then the value of P(X ≥ 2) is a) (20, 0) b) (15, 5) 1 2 3 1 a) b) c) d) c) (0, 5) d) (0, 20) 3 3 4 4 25. The maximum value of Z = x + 3y such that 2x 34. For a random variable X, V (X) = 4 and + y ≤ 20, x + 2y ≤ 20, x ≥ 0, y≥ 0 is E(X 2 ) = 13, the value of E(X) is a) 10 b) 60 a) 3 b) 4 c) 5 d) 6 c) 30 d) None of these 35. A coin is tossed three times. If denotes the 26. The maximum value of 𝑍 = 5𝑥 + 4𝑦, Subject to absolute difference between the number of 𝑦 ≤ 2𝑥, 𝑥 ≤ 2𝑦, 𝑥 + 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 is...... heads and the number of rails, then P(X =1)= a) 14 b) 12 1 2 c) 13 d) 18 a) b) 2 3 27. The maximum value of 𝑧 = 40𝑥 + 50𝑦 subject 1 3 to the constraints 3𝑥 + 𝑦 ≤ 9, 𝑥 + 2𝑦 ≤ 8 and c) d) 6 4 𝑥 ≥ 0, 𝑦 ≥ 0 is 36. The probability distribution of a discrete a) 0 b) 120 c) 230 d) 200 random variable X is given by 28. Which of the following is not a vertex of the X −1 0 1 2 positive region bounded by the inequalities P(X) 1 1 1 1 2𝑥 + 3𝑦 ≤ 6, 5𝑥 + 3𝑦 ≤ 15 and 𝑥, 𝑦 ≥ 0 3 6 6 3 a) (0, 2) b) (0, 0) Then, the value of 6E(X 2 ) − Var (X) is c) (3, 0) d) None of these 12 113 19 1 a) b) c) d) 29. In case of a linear programming problem, 113 12 12 2 feasible region is always 37. Let X denotes the sum of the numbers obtained a) a convex set b) a concave set when two fair dice are rolled. The variance and c) a bounded convex set d) a bounded concave standard deviation of X are set 31 31 35 35 a) and √ 6 b) and √ 6 30. The variables x and y in a linear programming 6 6 problem are called 17 17 d) None of these a) decision variables b) linear variables c) and √ 6 6 c) optimal variables d) None of these 38. A class ha 15 students whose ages are 14, 17, 31. A random variable X is defined by 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 1 3 with probability = 20 yr. One student is selected in such a manner 3 that each has the same chance of being of 1 X= 4 with probability = chosen and the age X of the selected student is 4 5 recorded. Mean, variance and standard {12 with probability = 12 deviation (SD) of X, are respectively Then E(X) a) 17.53, 4.8 and 2.19 b) 2.19, 4.8 and 17.53 a) 6 b) 7 c) 5 d) 8 c) 17.53, 2.19 and 4.8 d) None of these 32. If the probability distribution of a random 39. In a dice game, a player pays a stake of ₹1 for variable X is as given below each throw of a dice. She receives ₹5, if the dice shows a 3 , ₹ 2 , if the dice shows a 1 or 6 and 𝑿=𝒙 −2 −1 0 1 2 3 nothing otherwise. What is the player's 1 1 3 expected profit per throw over a long series of 𝑷(𝑿 = 𝒙) 𝑘 2𝑘 𝑘 10 5 10 throws? Then, mean and variance are a) 0.50 b) 0.20 4 54 4 54 c) 0.70 d) 0.90 a) 𝑎𝑛𝑑 b) 𝑎𝑛𝑑 40. A random variable X has the probability 5 25 10 25 4 27 d) None of these distribution given below c) 𝑎𝑛𝑑 15 25 𝑿 1 2 3 4 5 33. The random variable X has the following probability distribution 𝑷(𝑿 = 𝒙) 𝐾 2𝐾 3𝐾 2𝐾 𝐾 𝑥 0 1 2 3 4 Its variance is Page|3  16 4 For the events 𝐸 = {𝑋 is a prime number } and a) b) 3 3 𝐹 = {𝑋 < 4}, 𝑃(𝐸 ∪ 𝐹) is 5 10 a) 0.77 b) 0.87 c) d) 3 3 c) 0.35 d) 0.50 41. If the random variable X takes the values 47. Anil's company estimates the net profit on a 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥10 with probabilities P(X = 𝑥i ) = new product, it is launching, to be ₹3,000,000 ki, then the value of k is equal to during the first year. If it is successful, ₹ 1 1 1 a) b) c) d) 10 1,000,000 if it is moderately successful and a 10 15 55 loss of ₹ 1,000,000 if it is 'unsuccessful'. The 42. For a random variable X, if E(X) = 5 and company assigns the following probabilities to V(X) = 6, then E(X 2 ) is equal to first year prospects for the product, successful: a) 19 b) 31 c) 61 d) 11 0.15, moderately successful 0.25 and 43. If three fair coins are tossed, where X = unsuccessful 0.60. Then, the standard number of heads obtained, then E(X) is 1 3 2 3 deviation of first year net profit for the product a) b) c) d) (in million) is 2 2 3 4 44. The mean and variance of heads in three tosses a) 1.48 b) 12.40 of a coin are c) 13.8 d) None of these 3 48. Two cards are drawn simultaneously (or a) Mean = 1.5, variance = 4 3 successively without replacement) from a b) Mean = 1.75, variance = 2 well-shuffled pack of 52 cards. Find the 3 standard deviation of the number of kings c) Mean = 1.65, variance = 16 3 drawn is d) Mean = 1.8, variance = 5 a) √0.1392 b) √0.6462 45. If the probability mass function of a discrete c) √0.7672 d) √0.5431 random variable X is 49. If the random variable 𝑋 takes the values C P(𝑥) = 3 ; 𝑥 = 1, 2, 3 𝑥1 , 𝑥2 , 𝑥3 , … , 𝑥10 with probabilities 𝑃(𝑋 = 𝑥𝑖 ) = 𝑥 = 0 ; otherwise 𝑘𝑖, then the value of 𝑘 is equal to 1 1 Then E(X) = a) b) 343 294 297 251 10 4 a) b) c) d) 1 7 297 251 294 294 c) d) 55 12 46. A random variable X has the probability 50. A coin is tossed four times. Let X denote the distribution number of heads occurring. Then the expected 𝑿 1 2 3 4 5 6 7value is8 a) 1.7 b) 2 c) 1.63 d) 1.5 𝑷(𝑿) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 Page|4  kailas suralkar Dnyansandhya IIT MATHS Anand Nagar, Jamner Contact : 9730289357 12 Date : 10-11-2024 TEST ID: 37 Time : 01:30:00 MATHEMATICS Marks : 100 14.PROBABILITY DISTRIBUTIONS ,15.BINOMIAL DISTRIBUTION ,7.LINEAR PROGRAMMING ,8.PROBABILITY DISTRIBUTION,9.BERNOULLI TRIALS AND BINOMIAL DISTRIBUTION,9.LINEAR PROGRAMMING PROBLEMS : ANSWER KEY : 1) c 2) b 3) a 4) b 29) a 30) a 31) b 32) a 5) c 6) b 7) c 8) a 33) b 34) a 35) d 36) b 9) d 10) a 11) c 12) c 37) b 38) a 39) a 40) b 13) b 14) b 15) a 16) d 41) c 42) b 43) b 44) a 17) c 18) c 19) b 20) a 45) b 46) a 47) a 48) a 21) b 22) c 23) c 24) d 49) c 50) b 25) c 26) a 27) c 28) d Page|5  kailas suralkar Dnyansandhya IIT MATHS Anand Nagar, Jamner Contact : 9730289357 12 Date : 10-11-2024 TEST ID: 37 Time : 01:30:00 MATHEMATICS Marks : 100 14.PROBABILITY DISTRIBUTIONS ,15.BINOMIAL DISTRIBUTION ,7.LINEAR PROGRAMMING ,8.PROBABILITY DISTRIBUTION,9.BERNOULLI TRIALS AND BINOMIAL DISTRIBUTION,9.LINEAR PROGRAMMING PROBLEMS : HINTS AND SOLUTIONS : Single Correct Answer Type 1 1 1 2−1 1 1 (c) P(1 boy, 1 girl) = 2 C1 ( ) ( ) = 2 2 2 Given, n = 6 and P(X=2) =9P(X=4) 6 (b) ⇒ 6𝐶2 𝑝2 𝑞4 = 96𝐶4 𝑝4 𝑞2 ⇒ 9𝑝2 = 𝑞 2 Consider the following events 1 A = India wins 2 matches in first 4 matches ∴𝑝=3 𝑞 B = India wins fifth match We know that, p + q = 1 𝑞 3 1 C = Getting any outcome in sixth and seventh ⇒ + 𝑞 = 1 ⇒ 𝑞 = 𝑎𝑛𝑑 𝑝 = ∴ Variance = 3 4 4 match 1 3 9 2 𝑛𝑝𝑞 = 6. 4. 4 = 8 Given probability of India winning, 𝑝 = 3 and not 3 (a) winning India 𝑞 = 3 1 Let n and p be the parameters for the distribution. 2 2 1 2 2 Then, np = 4 and npq =2 ∴ 𝑃(𝐴) = 4𝐶2 ( ) ( ) , 𝑃(𝐵) = 𝑎𝑛𝑑𝑃(𝐶) = 1 3 3 3 1 ⇒ 𝑝 = 𝑞 = 2 𝑎𝑛𝑑 𝑛 = 8 ∴ Required probability = 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) Let X denotes the number of successes. Then, =P (A) P (B) P (C) 1 𝑟 1 8−𝑟 1 8 [∴ A, B, C are independent event] 𝑃(𝑋 = 𝑟) = 8𝐶𝑟 ( ) ( ) = 8𝐶𝑟 ( ) 2 2 1 2 2 16 2 2 2 = 4𝐶2 ( ) ( ) × × 1 = 1 8 3 3 3 81 ∴ 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑃(𝑋 = 2) = 8𝐶2 ( ) 8 (a) 2 2 28 Probability of success, P= 6 = = 256 1 2 𝑎𝑛𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒, 𝑞 = 3 4 (b) 3 Since, the probability of person die, due to 1 4 2 ∴ Required probability = 5𝐶4 (3) (3) suffering from a disease is 10%. 10 1 9 9 (d) ∴ 𝑝 = 100 = 10 𝑎𝑛𝑑 𝑞 = 10 1 3 Here, n= 6, p = 25%= 4 , 𝑞 = 1 − 𝑝 = 4 Total number of patients, n = 6 Let X be the random variable, which denotes the 1 3 9 3 ∴ Required probability= 6𝐶3 (10) (10) number of cases that police station are solved. 6.5.4 1 9×9×9 ∴ According to question, = × × 𝑃(𝑋 = 5) + 𝑃(𝑋 = 6) 3.2.1 1000 1000 2 = 6𝐶5 (𝑝)5 (𝑞)6−5 + 6𝐶6 (𝑝)6 (𝑞)6−6 = 5 × 729 10 1 5 3 1 1 6 3 0 = 1458 × 10−5 = 6𝐶1 ( ) ( ) + 6𝐶0 ( ) ( ) 4 4 4 4 5 (c) 1 19 = [6 × 3 + 1] = The chance of being a boy or a girl, (4)6 4096 1 1 p = 2 and q = 2 10 (a) 1 Total child = 2, Number of boys = 1 Probability of getting head, 𝑝 = 2 Page|6  1 and probability of getting tail, 𝑞 = 2 Substituting 𝑥 = 0, 𝑦 = 0, we get 0 ≤ 6 which is true Here, n = 10 ∴ it lies in origin side ∴ Required probability = P(X =5) 16 (d) 1 5 1 5 10! 1 10 = 10𝐶5 ( ) ( ) = ×( ) Given, maximise 𝑍 = 𝑥 + 2𝑦 Subject to constraints 2 2 5! 5! 2 𝑥 − 𝑦 ≤ 10,2𝑥 + 3𝑦 ≤ 20, 𝑥 ≥ 0, 𝑦 ≥ 0 205 63 = = 1024 256 11 (c) Plot the lines 𝑥 = 5, 𝑥 = 10, 𝑦 = 5 and 𝑦 = 10 on the graph as shown below Since (0, 0) lies within the region 𝑥 ≤ 10, 𝑦 ≤ 10 shade below the line 𝑥 = 10 and 𝑦 = 10 Since (0, 0) does not lie in the region 𝑥 ≥ 5, 𝑦 ≥ 5, Here, 𝑂𝐴𝐵 is the required feasible region whose 20 shade above the line 𝑥 = 5 and 𝑦 = 5 corner points are 𝑂(0,0), 𝐴(10,0) and 𝐵 (0, ). 3 The common shaded region is the feasible region. Corner points 𝑍 = 𝑥 + 2𝑦 The vertices of the feasible region are (5, 5), (10, 5), (10, 10) and (5, 10) 𝑂(0,0) 𝑍=0 12 (c) 𝐴(10,0) 𝑍 = 10 The equalities are 20 20 40 2𝑥 + 𝑦 ≤ 104 ….(i) 𝐵 (0, ) 𝑍 =0+2× = 𝑥 + 2𝑦 ≤ 76 ….(ii) 3 3 3 By putting 𝑥 = 0 and 𝑦 = 0, we gets 40 A(52, 0), B(0, 104), C(76, 0), D(0, 38) Hence, maximum value of 𝑍 is. 3 Solving (i) and (ii), we have P (44, 16) 17 (c) ∴ z at O(520, 0) = 6(52) + 11(0) Clearly from graph there is no feasible region = 312 z at A(44,16) = 6(44) + 11(16) = 262 + 176 = 440 z at P(0, 38) = 6(0) + 11(38) = 418 ∴ Maximum value of z is 440 14 (b) 18 (c) Given inequality is x + 5y ≤ 6 On putting x = 0 and y = 0 we get 0 + 5.0 = 0 < 6 (true) So given inequality represents region to the origin side of line x + 5y = 6. 19 (b) The feasible region is 𝐴𝐵𝐶𝐷𝐸𝐴, whose corner Hence, it is unbounded points are 𝐴(0,1), 𝐵(1,0), 𝐶(10,0), 𝐷(2,4) and 15 (a) 𝐸(0,4). 𝑥 + 5𝑦 ≤ 6 Now, table for object function 𝑍 = 2𝑥 + 𝑦 is given Page|7  is 15 which occurs at point (5, 0) and the minimum value of Z is-32 which occurs at point (0, 8). 21 (b) The feasible region is OABCO. Corner points Value of 𝑍 = 2𝑥 + 𝑦 Now, 𝑍 = 3𝑥 + 2𝑦 At 𝑂(0,0), 𝑍 = 0 16 32 𝐴(0,1) 𝑍=1 At 𝐴 (0, 3 ) , 𝑍 = 3 𝐵(1,0) 𝑍=2 At 𝐵(5,2), 𝑍 = 15 + 4 = 19 At 𝐶(7,0), 𝑍 = 21 𝐶(10,0) 𝑍 = 20 (Maximum) ∴ Maximum value of 𝑍 is 21. 22 (c) 𝐷(2,4) 𝑍=8 Given inequalities are 2𝑥 + 𝑦 ≥ 8 and 𝑥 + 2𝑦 ≥ 𝐸(0,4) 𝑍=4 10 ∴ 2𝑥 + 𝑦 = 8 ….(i) Hence, minimum value of Z is 1 at A(0, 1). And 𝑥 + 2𝑦 = 10 ….(ii) 20 (a) By putting 𝑥 = 0 and 𝑦 = 0 in equations (i) and (ii), We get A(4, 0), B(0, 8), C(10, 0) and D(0, 5) Given objective function is Z = 3x - 4y. Corner points Corresponding value of 𝑍 = 3𝑥 − 4𝑦 ∴ The solution set is as shown in the figure (0,0) 0 24 (d) Feasible region is ABCA (0,8) −32 (Minimum) (4,10) −28 (6,8) −14 (6,5) −2 (5,0) 15 (Maximum) From the above table. The maximum value of Z is and Z = 7x -8y Page|8  Now, at A(0, 5), Z = 7(0) - 8(5) = -40 0, (0,0) 5×0+4×0=0 At B(15, 5), Z = 7(15) - 8(5) = 65 At C(0, 20), Z = 7(0) - 8(20) = 0160 (Minimum) 𝐴(1,2) 5 × 1 + 4 × 2 = 13 Hence, minimum value of Z is attained at point (0, 20). 5 × 2 + 4 × 1 = 14 𝐵(2,1) 25 (c) (maximum) Given, objective function 𝑍 = 𝑥 + 3𝑦. Here 27 (c) feasible region is 𝑂𝐴𝐵𝐶𝐷. Whose corner points The equalities are 3𝑥 + 𝑦 = 9 ….(i) are And 𝑥 + 2𝑦 = 8 ….(ii) 20 20 𝑂(0,0), 𝐶(0,10), 𝐵 ( , ) , 𝐴(10,0) By putting 𝑥 = 0 and 𝑦 = 0 in (i) and (ii), we have 3 3 A(3, 0), B(0, 9), C(8, 0), D(0, 4) ∴ The feasible region is as shown in fig. Corner points Value of 𝑍 = 𝑥 + 3𝑦 𝑂(0,0) 𝑍 = 0 + 3(0) = 0 𝐴(10,0) 𝑍 = 10 + 3(0) = 10 Solving (i) and (ii) 20 20 20 20 20 60 80 ∴ P(2, 3) 𝐵( , ) 𝑍= + 3( ) = + = 3 3 3 3 3 3 3 ∴ O(0, 0), A(3, 0), P(2, 3) and D(0, 4) are corner 𝐶(0,10) 𝑍 = 0 + 3(10) = 30 (Maximum) points of OAPD ∴ z at A(3, 0) = 40(3) + 50(0) = 120 ∴ Maximum value of Z is 30 at C (0, 10). z at O(0, 0) = 40(0) + 50(0) = 0 26 (a) z at D(0, 4) = 40(0) + 50(4) = 200 We have, 𝑍 = 5𝑥 + 4𝑦 z at P(2, 3) = 40(2) + 50(3) = 80 + 150 = 230 Subject to constraints Maximum value is z = 230 𝑦 ≤ 2𝑥, 𝑥 ≤ 2𝑦, 𝑥 + 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 28 (d) On taking given constraints as equations, we get the following graph. Here (0, 2), (0, 0) and (3, 0) all are vertices of feasible region 29 (a) Intersecting point of line 𝑦 = 2𝑥 and 𝑥 + 𝑦 = 3 is In case of a linear programming problem feasible 𝐴(1,2) and intersecting point of line 𝑥 = 2𝑦 and region is always a convex set. 𝑥 + 𝑦 = 3 is 𝐵(2,1). 30 (a) Here, 𝑂𝐴𝐵𝑂 is the required feasible region whose The variables x and y in a linear programming corner points are 𝑂(0,0), 𝐴(1,2) and 𝐵(2,1). problem are called decision variables. 31 (b) Corner 𝑧 = 5𝑥 + 4𝑦 1 1 5 points E(X) = 3 × + 4 × + 12 × =7 3 4 12 Page|9 32 (a) 1 1 2 1 E(X) = ∑ 𝑥i ∙ P(𝑥i ) = − + 0 + + = The given distribution is a probability 3 6 3 2 distribution. Var(X) = ∑ 𝑥i2 ∙ P(𝑥i ) − [E(X)]2 ∴ P(x = -1) + P(x = -1) + P(x = 0) + P( x = 1) + (−1)2 12 23 1 2 P(x = 2) + P(x = 3) = 1 = +0+ + −( ) 1 1 3 3 6 3 2 ⇒ 10 + 𝑘 + 5 + 2𝑘 + 10 + 𝑘 = 1 1 1 4 1 4 = + + − 3 6 3 4 4𝑘 = 11 1 19 10 = − = 1 ⇒ 𝑘 = 10 6 4 12 ∴ 6 E(X 2 ) − Var(X) 1 1 Mean = (−2) + 𝑘(−1) + (0) + 2𝑘(1) 10 5 = 6 ∑ 𝑥i2 ∙ P(𝑥i ) − Var(X) 3 1 1 4 19 + (2) + 𝑘(3) = 6( + 0 + + ) − 10 3 6 3 12 2 6 =− − 𝑘 + 0 + 2𝑘 + + 3𝑘 19 113 10 10 4 4 1 = 11 − = = 10 + 4𝑘 = 10 + 4 × 10 12 12 8 4 37 (b) = = Let X denotes the sum of the numbers obtained 10 5 1 1 when two fair dice are rolled. So, X may have Variance = 10 (−2)2 + 𝑘(−1)2 + 5 (0)2 + 3 values, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11or 12. 2𝑘(1)2 +10 (2)2 + 𝑘(3)2 − (𝑚𝑒𝑎𝑛)2 (as 1 can't be the sum of two numbers on fair 4 12 4 2 dice) = +k+0+2k+ + 9𝑘 − ( ) 10 10 5 1 16 16 16 1 16 54 P(X=2)=P[{1, 1}]=36 = + 12𝑘 − = + 12 × − = 10 25 10 10 25 25 2 P(X=3)=P([{1, 2), (2, 1)}]= 36 33 (b) 3 𝑃(𝑋 = 4) = 𝑃[{(1, 3), (2, 2), (3, 1)}] = Since, ∑ Pi (X = 𝑥) = 1 36 4 ∴ k + 3k + 5k + 2k + k = 1 P(X=5)𝑃[{(1, 4), (2, 3), (3, 2), (4, !)}] = 36 1 𝑃(𝑋 = 6) = 𝑃[{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}] ∴ 12 k = 1 ∴ k = 12 5 Now, P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) = 36 = 5k + 2k + k 𝑃(𝑋 = 7) 1 2 6 = 8k = 8 ( ) = = 𝑃[{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}] = 12 3 36 34 (a) P(X=8)=P[{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}]=36 5 We know that, 4 V(X) = E(X 2 ) − [E(X)]2 P(X=9)=𝑃[{(3, 6), (4, 5), (5, 4), (6, 3)}] = 36 ∴ 4 = 13 − [E(X)]2 3 𝑃(𝑋 = 10) = 𝑃[{(4, 6), (5, 5), (6, 4)}] = ∴ [E(X)]2 = 13 − 4 = 9 36 2 ∴ E(X) = 3 𝑃(𝑋 = 11) = 𝑃[{(5, 6), (6, 5)}] = 35 (d) 36 1 Given, a coin is tossed three times and x = 𝑃(𝑋 = 12) = 𝑃[{(6, 6)}] = 36 absolute difference between the number of heads 𝑿 2 3 4 5 6 7 8 9 and number of tails. Now, X = 1 when exactly two head or two tail 1 2 3 4 5 6 5 4 comes 𝑷(𝑿) 36 36 36 36 36 36 36 36 3 3 ∴ P(x = 1) = 8 + 8 Mean of X = Σ X P (X) 3 ∴ 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑡𝑤𝑜 ℎ𝑒𝑎𝑑 = 8 ] [2 × 1 + 3 × 2 + 4 × 3 + 5 × 4 + 6 × 5 + 7 × 6 [ 3 +8 × 5 + 9 × 4 + 10 × 3 + 11 × 2 + 12 × 1] 𝑎𝑛𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙𝑠 = = 8 36 36 (b) P a g e | 10  252 39 (a) = 36 Let X be the money won in one throw. =7 Money lost in 1 throw = 1 Variance X = Σ 𝑋 2 P(X)- (Mean)2 Also, probability of getting 3 = 6 1 = Probability of getting 1 or 6 is [22 ×1+32 ×2+42 ×3+52 ×4+62 ×5+72 ×6+82 ×5+92 ×4+102 ×3+ 112 ×2+122 ×1] 36 1 1 2 1974 1974−1764 + = = − 49 = 6 6 6 36 36 Probability of getting any other number i.e. 2 or 4 210 35 = 36 = 6 or 5 35 1 1 1 3 Hence, SD = √Variance = √ = + + = 6 6 6 6 6 38 (a) Then, probability distribution is Here, total number of students =15 𝒙 5 2 0 The ages of students in ascending order are 14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20, 21 1 2 3 𝑷(𝑿 = 𝒙) 2 1 6 6 6 Now, P(X=14)= 15 , 𝑃(𝑋 = 15) = 15, 2 3 P(X=16)= 15 , 𝑃(𝑋 = 17) = 15 The expected money that player can won 1 2 5 4 9 𝑃(𝑋 = 18) = , 𝑃(𝑋 = 19) = 𝐸(𝑋) = + + 0 = = 1.5 15 15 6 6 6 3 1 Then, player's expected profit = ₹ 1.5 - ₹ 1=0.50 𝑃(𝑋 = 20) = , 𝑃(𝑋 = 21) = 15 15 40 (b) Therefore, the probability distribution of random Given distribution is variable X is as follows X is as follows 𝑿 1 2 3 4 5 X 14 15 16 17 18 19 𝑷(𝑋20= 𝑥)21 𝑘 2𝑘 3𝑘 2𝑘 𝑘 2 2 Number of Students 2 1 2 3 1 2∴ Variance= 3 1 Σ𝑋𝑖 𝑝 − (Σ𝑋𝑖 𝑝) =(1k + 8k + 27k + 32k + 25k)−(𝑘 + 4𝑘 + 9𝑘 + 2 1 2 3 1 28𝑘 + 5𝑘) 3 2 1 𝑃(𝑋) 15 15 15 15 15 15 15 − 15 = (93𝑘) (27𝑘)2 The third row gives the probability distributed of 1 = (93 × ) X. 9 Mean of X = Σ X P(X) 1 2 1 14 × 2 + 15 × 1 + 16 × 2 + 17 × 3 + 18 × 1 + 19 × 2 − (27 × ) [∴ Σ𝑝 = 1, 𝑠𝑜 𝑘 = ] 9 9 = +20 × 3 + 21 × 1 93 93 − 81 12 4 15 = −9= = = 9 9 9 3 28 + 15 + 32 + 51 + 18 + 38 + 60 + 21 42 (b) = 15 We know that, 263 = V(X) = E(X 2 ) − [E(X)]2 15 =17.53 ∴ 6 = E(X 2 ) − (5)2 Variance X = Σ 𝑋 2 P(X)- (Mean)2 ∴ E(X 2 ) = 25 + 6 = 31 [(14)2 ×2+(15)2 ×1+(16)2 ×2+(17)2 ×3+(18)2 ×1+(19)2 ×2 43 (b) 2 2 = +(20) ×3+(21) ×1] - X can take values 0, 1, 2 and 3 15 P(X = 0) = Probability of not getting heads 263 2 ( ) 1 15 = 392+225+512+867+324+722+1200+441 263 2 8 = 15 - ( 15 ) P(X = 1) = Probability of getting one head 4683 263 2 3 = − ( ) = 312.2 – 307.4 = 4.8 = 15 15 8 SD of X = √Variance = √4.8 = 2.19 P(X = 2) = Probability of getting two heads P a g e | 11  3 216 49 294 = = × = 8 251 36 251 P(X = 3) = Probability of getting three heads 46 (a) 1 P(E) = P(X = 2) + P( X =3) + P(X = 5) + P(X = = 8 7) 1 3 3 1 =0.23 + 0.12 + 0.20 + 0.07 = 0.62 ∴ E(X) = (0) + (1) + (2) + (3) P(F) = P(X =1) + P(X = 2) + P(X = 3) 8 8 8 8 3 3 3 3 + 6 + 3 12 3 =0.15 + 0.23 + 0.12 = 0.5 =0+ + + = = = 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) 8 4 8 8 8 2 44 (a) =0.23 + 0.12 = 0.35 Let X denotes the number of heads in three tosses ∴ P(E ∪ 𝐹) = 𝑃(𝐸) + 𝑃(𝐹) − 𝑃(𝐸 ∩ 𝐹) =0.62 + 0.5 - 0.35 = 0.77 of a coin. Then X can assume values 0, 1, 2 and 3 47 (a) P(X = 0) = Probability of getting no heads The probability distribution of net profit (x) of the 1 = P(TTT) = new product in the first year is given to be 8 P(X = 1) = Probability of getting one heads Profit (in ₹ million) 𝑋 3 1 −1 3 = P(THH, HTT, THT) = Probability 𝑃(𝑋) 0.15 0.25 0.60 8 P(X = 2) = Probability of getting two heads Therefore, expected value of profit is given by 3 𝐸(𝑋) = Σ𝑋𝑃(𝑋) = 3 × 0.15 + 1 × 0.25 − 1 × 0.60 = P(HHT, HTH, THH) = 8 = ₹0.10 million = ₹1,00,000 P(X = 3) = Probability of getting three heads 𝐸(𝑋 2 ) = Σ𝑋 2 𝑃(𝑋) 1 = P(HHH) = = 9 × 0.15 + 1 × 0.25 + 1 × 0.6 = ₹2.20 million 8 Var(𝑋) = 𝐸(𝑋 2 ) − {𝐸(𝑋)}2 = 2.20 − (0.10)2 Mean = ∑ 𝑥i pi 1 3 3 1 = 2.19 = 0 × + 1 × + 2 × + 3 × = 1.5 𝑆𝐷 = 𝜎 = √2.19 = ₹1.48 million 8 8 8 8 𝑥i pi pi 𝑥i pi 𝑥i 2 48 (a) 0 1 0 0 Let X denote the number of kings in a draw of two 8 cards. X is a random variable which can assume 1 3 3 3 the values 0, 1 or 2. 8 8 8 Then, P(X=0)=P(no card is king) 2 3 6 12 48𝐶2 48 × 47 188 8 8 8 = = 3 1 3 9 52𝐶2 52 × 51 221 8 8 8 Then, P(X=1)=P(exactly one card is king) 1.5 3 4𝐶1 × 48𝐶1 2) 2 2 = Variance = E(X − [E(X)] = 3 − (1.5) = 0.75 52𝐶2 45 (b) = 4×48×27 = 32 52×51 221 C C C P(1) = 3 ; P(2) = 3 ; P(3) = 3 Then, P(X=2)=P(both cards are king) 1 2 3 4𝐶2 4×3 1 Now, P(1) + P(2) + P(3) = 1 = = C C C 52𝐶2 52 × 51 221 ∴ 3+ 3+ 3=1 𝑛 1 2 3 𝜇 = 𝐸(𝑋) = ∑ 𝑋𝑖 𝑝𝑖 1 1 1 ⇒ C( + + ) = 1 𝑖=1 1 8 27 188 32 1 34 216 + 27 + 8 =0× +1× +2× = ⇒ C( )=1 221 221 221 221 216 𝑛 𝑛 2 216 ⇒C= 𝑣𝑎𝑟(𝑋) = (∑ 𝑋𝑖2 𝑝𝑖 ) − (∑ 𝑋𝑖 𝑝𝑖 ) 251 𝑖=1 𝑖=1 C C C ∴ E(X) = (1) 3 + (2) 3 + (3) 3 188 32 1 34 2 1 2 3 (02 × + 12 × + 22 × )−( ) 1 1 36 + 9 + 4 221 221 221 221 = C (1 + + ) = C ( ) 36 1156 4 9 36 = − 221 48841 P a g e | 12  6800 4 = = 0.1392 = 48841 16 ∴ Standard deviation, 𝜎 = √𝑉𝑎𝑟(𝑋) = √0.1392 P(X = 4) = Probability of getting four heads 49 (c) 1 = As we know, the sum of all the probabilities in a 16 1 4 6 4 probability distribution is one. ∴ E(X) = 0 × +1× +2× +3× +4 16 16 16 16 ∴ P(X =x1) + P(X = x2)+…..+P9X =x10) = 1 1 ⇒ 1k + 2k + 3k +……+ 10k = 1 × 16 10(10+1) 1 ⇒ 𝑘 =1⇒𝑘= =2 2 55 50 (b) X can take values 0, 1, 2, 3 and 4 P(X = 0) = Probability of not getting heads 1 = 16 P(X = 1) = Probability of getting one head 4 = 16 P(X = 2) = Probability of getting two heads 6 = 16 P(X = 3) = Probability of getting three heads P a g e | 13