Exam 1 with Solution for First Year Interest Group Seminar | N 1, Exams of Health sciences

Download Exam 1 with Solution for First Year Interest Group Seminar | N 1 and more Exams Health sciences in PDF only on Docsity! Examination 1 Solutions CS 336 1. The important issue is the logic you used to arrive at your answer. 2. Use extra paper to determine your solutions then neatly transcribe them onto these sheets. 3. Do not submit the scratch sheets. However, all of the logic necessary to obtain the solution should be on these sheets. 4. Comment on all logical flaws and omissions and enclose the comments in boxes 1. [5] Suppose all rolls of a six-side die are equally likely. What is the probability the roll is a six given that it is not one? The probability of not rolling a one is 5/6. The probability of rolling a six and not rolling a one is 1/6. Thus, the probability of rolling a six given the roll is not a one is 5/1 6/5 6/1 = . 2. a. [10] Present a combinatorial argument that for all : 1≥n 12 1 −=     ∑ = n n k k n (Note: The summation begins with 1=k .) Consider the cardinality of the set of non-empty subsets of a set A of n elements. For each element of A, there are two options: either be present in a subset or not. Thus there are 2 total subsets but one of these is empty so there are 2 non- empty subsets of A. Alternatively, let k indicate the cardinality of the subset. Since we are counting non-empty subsets, k ranges from 1 to n. For a fixed value of k, there are ways of selecting the k subset elements from the n total elements of A. Adding this to include all possible cases of k, we obtain and this must equal . n 1−n       k n ∑ =      n k k n 1 12 −n b. [10] Present a combinatorial argument that for all integers k and n satisfying nk ≤≤3       − − +      − − +      − − +      − =      3 3 2 3 3 1 3 3 3 k n k n k n k n k n (Hint: Consider three special elements.) Consider the number of subsets of size k of a set B of cardinality n. Since n , we may select three elements b of B and let C = B . Thus C has cardinality n-3 and B = C∪ . We know there are such subsets. Alternatively, to select k elements of B for a subset there are four options: all k come from C, k-1 come from C and the kth is either b or , k-2come from C and the k-1st and kth are exactly two of b or , or k-3 come from C and all of and b are present. For the first option, there are   possibilities since all k come from C. For the second option, there are 3 possibilities, since k-1 elements are selected from C and one from the three of or b . For the third option, there are possibilities, since k-2 elements are selected from C and one from the three of or is not selected. Lastly, if k-3 come from C and all of b and b are present, then there are   options. The total is and this must equal   3≥ 321 ,, bb ,,{ 21 bb    ,2b 3b    +   − − 2 3 k n },,{~ 321 bbb       k n 3b     − k n 3       − − 1 3 k n ,, 21 bb       k n }3b ,1 − − 3 3 ,, 21 b 3b  − − 3 3 k n ,2b    ,, 21 bb    − 3 k n 3 ,    k n 3    − − 2 3 3 k n ,1b 3    +   3 1 3 k n , 21 b +   3 − − 3. [10] How many distinct permutations are there of the letters in “mississippi”? There must be a total of 11 positions in any such permutation. We may select the four positions to hold the s’s in ways, four positions from the remaining seven to hold the i’s in ways, two positions from the remaining three to hold the p’s in ways, and the m must be in the remaining position. Thus, there are ways of permuting the letters. Written as a multinomial, this is .       4 11       4 7 4 11       2 3       4 11    4             2 3 4 7 =   24 11       124