Probability and combinatorics #1 - Questions and Answers

Question 1 How many subsets with an odd number of elements does a set with 10 elements have? Answer Definitions An r-combination of a set of n elements is a subset that contains r of the n elements. The number of r-permutations of a set with n elements is C(n,r) = n!/ (r!(n - r)!) with n! = n(n-1)...2.1 Solution: We are interested in the number od subset with an odd number of elements from a set with 10 elements. n = 10 r = 1,3,5,7, or 9 A subset of r elements from a set with n elements is an r-combinations and thus we are intersted in the number of r-combinations of a set with n=10 elements that contain an odd number of elemets. Number of subsets = ∑ C(10,r) where r ∈ {1,3,5,7,9} = C(10, 1) + C(10,3) + C(10,5) + C(10,7) + C(10,9) = 10!/(1!(10-1)!) + 10!/(3!(10-3)!) + 10!/(5!(10-5)!) + 10!/(7!(10-7)!) + 10!/(9!(10-9)!) = 10!/(1!9!) + 10!/(3!7!) + 10!/(5!5!) + 10!/(7!3!) + 10!/(9!1!) = 10 + 120 + 252 + 120 + 10 = 512 Thus there are 512 subsets with an odd number of elemets of a set with 10 elements. Question 2 How many combinations of three items out of a group of six can be chosen? List all of the combinations of objects B, D, and F after identifying them with the letters A, B, C, D, E, and F. Answer 1 Permutation definition (order matters): P_n,r = n!/(n-r)! Combination definition (order is not crucial): C_n,r = n!/ (r!(n - r)!) with n! = n(n-1)....2.1 Given: n=6 r=3 P_6,3 = 6!/(6-3)! = 6!/3! = 654 = 120 All the methods that we can choose three results from the set of three items (B,D, and F): BBB, BBD, BDB, DBB, BBF, BFB, FBB, BDF, BFD, DDD, DDB, DBD, BDD, DDF, DFD, FDD, DBF, DFB, FFF, FFB, FBF, BFF, FFD, FDF, DFF, FBD, FDB Answer 2 To calculate the number of combinations of three items chosen from a group of six, we can use the formula for combinations, which is given by: C(n,k) = n! / (k!(n-k)!) Here, n represents the total number of items in the group (6 in this case), and k represents the number of items to be chosen (3 in this case). Substituting the values into the formula, we have: C(6,3) = 6! / (3!(6-3)!) Simplifying further: C(6,3) = 6! / (3! * 3!) Since n! = n(n-1)(n-2).....21, we can calculate: 6! = 654321 = 720 Similarly, 3! = 321 = 6 Substituting these values into the equation: C(6,3) = 720/ (6*6) C(6,3) = 720 / 36 C(6,3) = 20 Therefore, there are 20 different combinations of three items that can be chosen from a group of six. Now, let's list all the combinations of objects B, D, and F, using the letters A, B, C, D, E, and F. The possible combinations are: A, B, D A, B, F A, C, D A, C, F A, D, E A, D, F B, C, D B, C, F B, D, E B, D, F C, D, E C, D, F C, E, F D, E, F These are the 14 combinations of objects B, D, and F when represented by the letters A, B, C, D, E, and F. Answer 3 The number of combinations of three items chosen from a group of six can be calculated using the formula for combinations, which is given by (n choose k) = n! / (k!(n-k)!), where n is the total number of items and k is the number of items chosen. In this case, we have n = 6 and k = 3. Plugging these values into the formula, we get (6 choose 3) = 6! / (3!(6-3)!) = 6!/(3!3!) = (654)/(32*1) = 20. Therefore, there are 20 combinations of three items that can be chosen from a group of six. To list all of the combinations of objects B, D, and F, we can use the following combinations: BDF BFD DBF DFB FBD FDB These are the six possible combinations of objects B, D, and F. Question 3 The reaction A + B → C + D rate = k[A][B]² has an intial rate of 0.00220 M/s what will the initial rate be if [A] is halved and [B] is tripled? Answer must be in M/s What will the intial 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) 0.0220M/s in rate law. 0.0220 M/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, [A]/2 for [A’] and 3[B] for [B’] in new rate law as follows: r = k([A]/2)(3[B])² r = k * (1/2)[A] * 9[B]² r = 9/2 * k[A][B]² Since, k[A][B]² = 0.0220M/s Substitute, 0.0220M/s for k[A][B]² in equation (1), thus r = 9/2 * 0.0220M/s r = 0.099 M/s Rate of the reaction become 0.099M/s 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 [B]/2 for [B’] in new rate law as follows: 2) r = k(3*[A])([B]/2)² r = k * 3[A] * 1/4[B]² r = 3/4 * k[A][B]² Since, k[A][B]² = 0.0220M/s Substitute, 0.0220M/s for k[A][B]² in equation (2), thus r = 3/4 * 0.0220M/s r = 0.0165 M/s Question 4 Show that the equation has no solutions with integers x and y 2x² + 5y² = 14. Answer Step 1 To prove: The equation has no integer solutions for x or y 2x² + 5y² = 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. The equation has no differences and the squares are never negative (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 x² and y² 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 x² and y² made the equation true and thus are no solution in integers x and y such that 2x² + 5y² = 14 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 5 A gambling book recommends the following "winning strategy" for the game of roulette. It advises a $1 single wager from the player. When the color red appears (probability 18/38), the gambler should take her $1 profit. If the gambler loses this bet (which has probablity 20/38 of occurring), she should make additional $1 bets on red on each of the next two spins of the roulette wheel and then quite. Let X denote the gamblers Answer a) Remember that when you bet $1 and lose the bet, you lose the $1. Therefore, you have to earn $2 just to walk out with a net gain of $1. So the answer is: P{X>0} = (18/38) + (20/38)(18/38)(18/38) = 0.5918 b) To get this answer, you really need to do part first. III