Questions and Answers #1 Probability and Combinatorics

Questions and Answers Sheet 1 Probability and Combinatorics Question #1 Suppose that the probability density function of the amount of milk deposited in a milk container is: 𝑓(𝑋) = 40.976 − 16𝑥 2 − 30𝑒 (−𝑥) for 1.95 ≤ 𝑥 ≤ 2.1 Liters. a) Calculate the probability that the actual amount of milk deposited is less than 2 liters. b) Calculate the Expected Value in Liters of this distribution. Answer: a) Total probability distribution value = 1 for 1.95 ≤ 𝑥 ≤ 2.1 Probability value: ∫1. 952. 1𝑓(𝑥)𝑑𝑥 = ∫1. 952 (40.976 − 16𝑥 2 − 30𝑒 (−𝑥) )𝑑𝑥 = |40.976𝑥 − 16𝑥 2 /2 − 30𝑒 (−𝑥) (−1)|1. 952 = (40.976 ∗ 2 − 16 ∗ 22 /2 + 30𝑒 (−2) ) − (40.976 ∗ 1.95 − 16 ∗ 1.952 /2 + 30𝑒 (−1.95) ) = 0.4681 = 46.81% b) Expected value = ∫1. 952. 1𝑥𝑓(𝑥)𝑑𝑥 = ∫1. 952. 1𝑥(40.976 − 16𝑥 6 − 30𝑒 (−𝑥) )𝑑𝑥 = |40.976𝑥 2 /2 − 16𝑥 3 /3 − 30(−𝑥𝑒 (−𝑥) − 𝑒 (−𝑥) )|1. 952. 1 = 1.39 litres Question #2 Suppose E(X) = 5 and E[X(X – 1)] = 27.5, find ∈ (𝑥 2 ) and the variance. Answer: We have the following properties: E(X + Y) = E(X) + E(Y) 𝑉(𝑋) = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2 So, if we have that E[X(X - 1)] = 27.5, we can write them using the first property as: E(X(X - 1)) = 27.5 𝐸(𝑋 2 − 𝑋) = 27.5 𝐸(𝑋 2 ) − 𝐸(𝑋) = 27.5 Then, replacing E(X) by 5 and solving for 𝐸(𝑋 2 ) we get: 𝐸(𝑋 2 ) − 5 = 27.5 𝐸(𝑋 2 ) = 27.5 + 5 𝐸(𝑋 2 ) = 32.5 Finally, using the second property and replacing E(X) by 5 and 𝐸(𝑋 2 ) by 32.5, we get that V(X) is equal to: 𝑉(𝑋) = 32.5 − 52 V(X) = 32.5 - 25 V(X) = 7.5 Finally answer: 𝐸(𝑋 2 ) = 32.5 V(X) = 7.5 𝐸[𝑋(𝑋 − 1)] = 𝐸(𝑋 2 ) − 𝐸(𝑋) 𝑉(𝑋) = 𝐸(𝑋 2 ) − [𝐸(𝑋)]2 Question #3 When an experiment is performed, one and only one of the events 𝐴1 , 𝐴2 and 𝐴3 will occur. Find 𝑃(𝐴1 ), 𝑃(𝐴2 ) and 𝑃(𝐴3 ) under each of the following assumptions: a) 𝑃(𝐴1 ) = 𝑃(𝐴2 ) = 𝑃(𝐴3 ) b) 𝑃(𝐴1 ) = 𝑃(𝐴2 ) and 𝑃(𝐴3 ) = 1/2 c) 𝑃(𝐴1 ) = 2𝑃(𝐴2 ) = 3𝑃(𝐴3 ) Answer: Here these events are mutually exclusive events.. a) 𝑃(𝐴1) = 𝑃(𝐴2 ) = 𝑃(𝐴3 ) 1 so, 𝑃(𝐴1 ) = 𝑃(𝐴2 ) = 𝑃(𝐴3 ) = 3 1 1 2 2 b) 𝑃(𝐴1 ) + 𝑃(𝐴2 ) = 1 − 𝑃(𝐴3 ) = 1 − = so, 𝑃(𝐴1 ) = 𝑃(𝐴2 ) = 1 4 c) 𝑃(𝐴1 ) = 2𝑃(𝐴2 ) = 3𝑃(𝐴3 ) we can write, 𝑃(𝐴2 ) = 𝑃(𝐴1 )/2 𝑃(𝐴3 ) = 𝑃(𝐴1 ) 3 now, 𝑃(𝐴1 ) + 𝑃(𝐴1 ) 2 + 𝑃(𝐴1 ) 3 =1 ⇒ 11/6 ∗ 𝑃(𝐴1 ) = 1 6 ⇒ 𝑃(𝐴1 ) = 11 using value of 𝑃(𝐴1 ) we find... 3 𝑃(𝐴2 ) = 𝑃(𝐴3 ) = 11 2 11 Question #4 The reaction 𝐴 + 𝐵 𝑦𝑖𝑒𝑙𝑑𝑠 𝐶 + 𝐷 𝑟𝑎𝑡𝑒 = 𝑘[𝐴][𝐵]2 has an initial rate of 0.00220M/s what will the initial rate be if [A] is halved and [B] is tripled? Answer must be in M/s What will the initial rate be if [A] is tripled and [B] is halved? Answer must be in M/s Answer: Step 1 The reaction is: The rate law for this reaction is written as: 𝑟 = 𝑘[𝐴][𝐵]2 Substitute initial rate of the reaction (𝑟)0.0220𝑚𝑠 −1 in rate law. 0.0220𝑀/𝑠 = 𝑘[𝐴][𝐵]2 An expression for initial rate is obtained using the rate law by substituting the value of initial rate. Step 2 When [A] is halved and [B] is tripled, the new concentrations are: [𝐴′ ] = [𝐴] 2 and [𝐵′ ] = [𝐵] × 3 The new rate law is written as: 𝑟 = 𝑘[𝐴′ ][𝐵′ ]2 Substitute, 1) 𝑟 = 𝑘 [𝐴] 2 [𝐴] 2 for [A'] and 3[B] for [B'] in new rate law as follows: (3[𝐵])2 9 = 𝑘[𝐴][𝐵]2 2 Since, 𝑘[𝐴][𝐵]2 = 0.0220𝑀𝑆 −1 Substitute, 0.0220𝑚𝑠 −1 for 𝑘[𝐴][𝐵]2 in equation (1), thus 9 𝑟 = × 0.0220𝑚𝑠 −1 2 = 0.099𝑚𝑠 −1 Rate of the reaction become 0.099𝑚𝑠 −1 When the initial concentrations of the reactants are changed by any factor, then the first step is to write the new concentrations in terms of the initial concentrations. Then, write the new rate law in terms of the initial concentrations. And then calculate the new rate law by substituting the values accordingly. Step 3 When [A] is tripled and [B] is halved, so the new concentrations are: [𝐴′′ ] = 3 × [𝐴] and [𝐵′′ ] = [𝐵] 2 The new rate law is written as: 𝑟 = 𝑘[𝐴′′ ][𝐵′′ ]2 Substitute, 3[A] for [A'] and 2) r = k (3  [ A])( [𝐵] 2 for [B'] in new rate law as follows: [ B] ) 22 3 = 𝑘[𝐴][𝐵]2 4 Since, 𝑘[𝐴][𝐵]2 = 0.0220𝑀𝑠 −1 Substitute, 0.0220𝑚𝑠 −1 for 𝑘[𝐴][𝐵]2 in equation (2), thus 3 𝑟 = × 0.0220𝑚𝑠 −1 4 = 0.0165𝑚𝑠 −1 Rate of the reaction become 0.0165𝑚𝑠 −1 When the initial concentrations of the reactants are changed by any factor, then the first step is to write the new concentrations in terms of the initial concentrations. Then, write the new rate law in terms of the initial concentrations. And then calculate the new rate law by substituting the values accordingly. Question #5 An insurance company looks at its auto insurance customers and finds that a) all insure at least one car, b) 85% insure more than one car, c) 23% insure a sports car, and d) 17% insure more than one car, including sports car. Find the probability that a customer selected at random insures exactly one car and it is not a sports car. What is the formula for an intersection of two events where one event is a complement? For example 𝑃(𝐴′ ∩ 𝐵), what will be its formula? For example this question. Answer: There isn't really a formula for the intersection of two events when one of the events is a complete. The basically need to use your common sense to determine the probability. It's often also easier to use the context to determine the required percentages and then rewrite them as probabilities. For example, 𝐴′ ∩ 𝐵 can be viewed as event B occurring but not both A and B occurring together and then you can calculate the probability as 𝑃(𝐴′ ∩ 𝐵) = 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) Question #6 Prove that there are no solutions in integers x and y to the equation 2𝑥 2 + 5𝑦 2 = 14. Answer: To prove: There exists no solutions in integers x and y to the equation 2𝑥 2 + 5𝑦 2 = 14 PROOF Since (−𝑎)2 = 𝑎2 for all integers and since the equation contains only squares of the unknown integers, it is safe to assume that the integers x and y are nonnegative. Since the squares are never negative and since the equation contains no differences (or negative signs), we only need to check all integers with squares smaller than 14. The only integers with squares smaller than 14 are 0, 1, 2 and 3: 02 = 0,12 = 1,22 = 4,32 = 9 Thus 𝑥 2 and 𝑦 2 can take on the values 0, 1, 4 and 9. Let us check if there is a possible combination that makes the equation true. 2(0) + 5(0) = 0 ≠ 14 2(0) + 5(1) = 5 ≠ 14 2(0) + 5(4) = 20 ≠ 14 2(0) + 5(9) = 45 ≠ 14 2(1) + 5(0) = 2 ≠ 14 2(1) + 5(1) = 7 ≠ 14 2(1) + 5(4) = 22 ≠ 14 2(1) + 5(9) = 47 ≠ 14 2(4) + 5(0) = 16 ≠ 14 2(4) + 5(1) = 13 ≠ 14 2(4) + 5(4) = 28 ≠ 14 2(4) + 5(9) = 53 ≠ 14 2(9) + 5(0) = 18 ≠ 14 2(9) + 5(1) = 23 ≠ 14 2(9) + 5(4) = 38 ≠ 14 2(9) + 5(9) = 63 ≠ 14 None of the possible combinations for 𝑥 2 and 𝑦 2 made the equation true and thus are no solution in integers x and y such that 2𝑥 2 + 5𝑦 2 = 15 Result: To prove it we only need to check for x and y from {0,1,2,3}, because those are only integers which squares are lesser than 14 (assuming that x, y are nonnegative without loss of generality). Question #7 Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter 𝜆 = 3 . (a) Find the probability that 3 or more accidents occur today. (b) Repeat part (a) under the assumption that at least 1 accident occurs today. Answer: 𝑃(𝑋 ≥ 3) = 1 − ∑2𝑖=0 =1−( 30 0!𝑒 3 + 31 1!𝑒 3 + 3𝑖 𝑖!𝑒 𝜆 32 2!𝑒 3 ) = 0.57681 P{( X  3) | ( X  1)} = (𝑋 ≥ 1) ⊂ (𝑋 ≥ 3) ⇒ P{ X  3)  ( X  1)} P{ X  1} 𝑃{(𝑋≥3)∩(𝑋≥1)} 𝑃{𝑋 ≥ 3} 0.57681 = 𝑃{𝑋 ≥ 1} 1 − 𝑃{𝑋 = 0} = 0.57681 1−𝑒 −3 = 0.60703 𝑃{𝑋≥1} = 𝑃{𝑋≥3} 𝑃{𝑋≥1} The probability that at least 3 accidents occur is 0.57681 while the probability that at least 3 accidents occur under the assumption that at least 1 accident accur today is 0.60703 Question #8 Given cyclic permutations: 𝜎 = (123), 𝜎2 = (45), then what are the inverse cycles 𝜎 −1 , 𝜎2−1 Answer: To find the inverse of a permutation just write it backwards. If 𝜏 = (1243)(67) then 𝜏 −1 = (76)(3421) which can then be rewritten as 𝜏 −1 = (1342)(67). For the question: (123) − 1 = (321) = (132). Question #9 Let P(n) be the statement that 𝑛! < 𝑛𝑛 where n is an integer greater than 1. a) What is the statement P (2)? b) Show that P (2) is true, completing the basis step of the proof. c) What is the inductive hypothesis? d) What do you need to prove in the inductive step? e) Complete the inductive step. f) Explain why these steps show that this inequality is true whenever n is an integer greater than 1. Answer: Let P(n) be 𝑛! < 𝑛𝑛 and n is an integer greater than 1. a) Replace n by 2 in the given expression for P(n). P(2) states 2! < 22 b) Basis step 2! = 2 < 4 = 22 We then note P(2) is true. c) Induction hypothesis Let P(k) be true. 𝑘! < 𝑘 𝑘 d) We need to prove that P(k+1) is also true. (𝑘 + 1)! < (𝑘 + 1)𝑘+1 e) Induction step. We need to prove that P(k+1) is also true. (𝑘 + 1)! = (𝑘 + 1) ⋅ 𝑘! < (𝑘 + 1) ⋅ 𝑘 𝑘 < (𝑘 + 1) ⋅ (𝑘 + 1)𝑘 = (𝑘 + 1)𝑘+1 = (𝑘 + 1)𝑘+1 We then note that P(k+1) is also true. f) Conclusion Vy the principle of mathematical induction, P(b) is true for all positive integers n greater than 1. Question #10 Suppose that A and B are independent events such that the probability that neither occurs is a and the probability of B is b. Show that 𝑃(𝐴) = 1−𝑏−𝑎 1−𝑏 Answer: Given information: 1. A and B are independent events Therefore 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵) 2. P(B)=b 3. 𝑃(𝐴 ∪ 𝐵) = 1 − 𝑃(𝐴 ∪ 𝐵) = 𝑎 Now, we have 𝑎 = 1 − 𝑃(𝐴 ∪ 𝐵) = 1 − (𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)) = 1 − 𝑃(𝐴) − 𝑃(𝐵) + 𝑃(𝐴)𝑃(𝐵) = 1 − 𝑃(𝐴) − 𝑏 + 𝑃(𝐴) ⋅ 𝑏 Hence, (1 − 𝑏)𝑃(𝐴) = 1 − 𝑏 − 𝑎 𝑃(𝐴) = 1−𝑏−𝑎 1−𝑏