Questions and Answers #2 Probability and Combinatorics

Questions and Answers Sheet 2 Probability and Combinatorics Question #1 A box contains 8 red marbles and 3 green ones. Six draws are made at random without replacement. True or false: the chance that the 3 green marbles are drawn equals 6! 8 3 3 ( ) ( ) 3!3! 11 3 11 Answer: Definition binomial probability: 𝑛! 𝑃(𝑋 = 𝑘) = ⋅ 𝑝𝑘 ⋅ (1 − 𝑝)𝑛−𝑘 𝑘!(𝑛−𝑘)! 6! We note that the expression probability at n =6 and 𝑝 = 8 3 3 3 ( ) ( ) is obtained by evaluating the formula of the binomial 3!3! 11 11 8 11 The four conditions for a binomial setting are binary, independent trials, fixed number of trials and probability of success is the same for each trial. Since the conditions for a binomial setting are not satisfied, the chance that the green marbles are drawn is not equal to 6! 8 3 3 3 ( ) ( ) and thus the statement is false. 3!3! 11 11 Question #2 How to calculate percentile? Is it possible to get 100 percentile? How do we calculate percentile? I think it should be calculated as: P = Total number of candidates L = Number of candidates whose marks are below yours 𝐿 Percentile = ( ) ⋅ 100 𝑃 That means if you are the highest in the range, and there are 200,000 candidates and if your marks are same as 7 others then; your percentile = (199992/200000) ⋅ 100 = 99.996 That means in any circumstances you can't have perfect 100 percentile. Answer: There are different definitions of ''percentile.'' Usually a percentile is computed by first ranking all the data with 1 the lowest, 2 the next lowest, up through N. (The possibility of ties is a complication but does not change the essential idea.) Ranking from lowest to highest instead of highest to lowest is arbitrary but it gives definite rules. However, it would be nice if the percentile computed from ranking in one direction agreed with the percentile from ranking in the other. You can convert a rank k into a percentile by: - Computing 100k/N. This value will range from 100/N through 100N/N=100. Note the asymmetry: the highest is 100 but the lowest is nonzero. - Computing 100(k-1)/N. This is the rule quoted in the question. The value will range from 0 through 100(N-1)/N<100. - You can make the range of percentiles symmetric. This means that the percentile corresponding to rank k is 100 minus the percentile corresponding the rank in reverse order, which is N+1-k. To do this, compute 100(k-1/2)/N. The values range from 100/(2𝑁) 𝑡𝑜 100(1 − 1/(2𝑁)). - There are other ways to make the range symmetric. For instance, compute 100(𝑘 − 1)/(𝑁 − 1). Values range from 0 through 100. (There are yet other rules, used especially for constructing probability plots, but you get the idea.) Either using the first rule, the last rule, or rounding to the nearest percent, can produce a 100th percentile. As I recall, Excel's function for computing percentiles can produce 100 for the top-ranked value. Question #3 How do you evaluate 𝐶46 ? Answer: 𝐶𝑟𝑛 = (𝑛!)/((𝑛 − 𝑟)! ∗ 𝑟!): . 𝐶46 = (6!)/((6 − 4)! ∗ 4!) = (6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1)/(2 ∗ 1 ∗ 4 ∗ 3 ∗ 2 ∗ 1) = 15 [Ans] Question #4 How do you simplify 𝑃52 ? Answer: 𝑃53 is the notation used for finding 'Permutation' Results 𝑃53 = (5!)/((5 − 3)!) = (5!)/(2!) = (5 ∗ 4 ∗ 3 ∗ 2 ∗ 1)/(2 ∗ 1) = 5 ∗ 4 ∗ 3 = 60 Question #5 Write v as a linear combination of 𝑢1 , 𝑈2 , and 𝑢3 , if possible. (Enter your answer in terms of 𝑢1 , 𝑈2 , and 𝑢3 . If not possible enter IMPOSSIBLE.) 𝑣 = (−1,7,2), 𝑢1 (3,2,8), 𝑢2 (2, −3,1), 𝑢3 (−2,3, −1) v=_____ Answer: Step 1 Since v is a lianer combination of 𝑢1 , 𝑢2 and 𝑢3 , so 𝑣 = 𝑐1 𝑢1 + 𝑐2 𝑢2 + 𝑐3 𝑢3 𝑐1 (3,2,8) + 𝑐2 (2, −3,1) + 𝑐3 (−2,3,1) = 𝑣 ([[3,2, −2], [2, −3,3], [8,1,1]][[𝑐1 ], [𝑐2 ], [𝑐3 ]] = [[−1], [7], [2]]) AC=V Step 2 So augmented matrix is [A:V] , ([[3,2, −2: −1], [2, −3,3: 7], [8,1,1: 2]]) Reduce the above augmented matrixinto raw echelon form, ([[3,2, −2: −1], [2, −3,3: 7], [8,1,1: 2]]) Apply 𝑅1 → 𝑅1 → 𝑅2 (−[[1,5, −5: −8], [2, −3,3: 7], [8,1,1: 2]]) Apply 𝑅2 → 𝑅2 − 2𝑅1 . 𝑅3 → 𝑅3 − 8𝑅1 (−[[1,5, −5: −8], [0, −13,13: 23], [0, −39,39: 66]]) Apply 𝑅3 → 𝑅3 − 3𝑅2 (−[[1,5, −5: −8], [0, −13,13: 23], [0,0,0: −3]]) Here rank of a matrix A=2 Rank of matrix [A:B]=3 So rank of (𝐴) ≠ 𝑟𝑎𝑛𝑘 of (A:B) That is system is consistent Step 3 Hence Since v is not a linear combination of 𝑢1 , 𝑢2 and 𝑢3 .