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Chapter1_1

Course: M 362k, Spring 2008
School: University of Texas
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NOTES LECTURE ON PROBABILITY I 1. Combinatorial Analysis Probability concerns how much possibility an event can occur in a random experiment. For those random experiments with finite number of outcomes, say N , each being equally like, the probability of an event E is defined classically by N (E) , P (E) := N where N (E) denotes the number of outcomes resulting in the occurrence of the event E. Example 1.1....

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NOTES LECTURE ON PROBABILITY I 1. Combinatorial Analysis Probability concerns how much possibility an event can occur in a random experiment. For those random experiments with finite number of outcomes, say N , each being equally like, the probability of an event E is defined classically by N (E) , P (E) := N where N (E) denotes the number of outcomes resulting in the occurrence of the event E. Example 1.1. Consider flipping a fair coin, and consider the event that the heads 1 come up. Then N = 2 and N (E) = 1, so the probability of this event is P (E) = 2 . Example 1.2. Consider tossing two dice, and denote by (i, j), i, j = 1, , 6, the outcome where i appears on the first die and j appears on the second die. Let E denote the event consisting of those outcomes (i, j) with i + j = 8. Then 5 N = 6 6 = 36 and N (E) = 5. Thus P (E) = 36 . In order to obtain the probability, we need to calculate the total number of possible outcomes for a random experiment; calculate the number of outcomes of an event in the random experiment. This requires the techniques from combinatorics. 1.1. The counting principle. Proposition 1.1 (The basic counting principle). Suppose that two experiments are to be performed. If the first experiment has m possible outcomes, and the second experiment has n possible outcomes, then there are mn possible outcomes of the two experiments. Proof. The proof follows from enumerating all the possible outcomes. Indeed, suppose the m outcomes of the first experiment are a1 , a2 , , am and the n outcomes of the second experiment are b1 , b2 , , bn . then all the possible outcomes of the two experiments are (a1 , b1 ), (a1 , b2 ), , (a1 , bn ) (a2 , b1 ), (a2 , b2 ), , (a2 , bn ) (am , b1 ), (am , b2 ), , (am , bn ) where (ai , bj ) denotes the outcome that the first experiment has outcome ai and the second has outcome bj . The above table has m rows, and each row has n elements. So there are mn elements in total. The basic counting principle has the following generalization which can be proved by induction. 1 2 Proposition 1.2 (The generalized counting prnciple). Suppose r experiments are to be performed. The experiment 1 has n1 possible outcomes, the experiment 2 has n2 possible outcomes, , and experiment r has nr possible outcomes. then there are n1 n2 nr possible outcomes of the r experiments in total. Example 1.3. The 7-space license plates are designed by occupying the first 3 places by letters and last 4 by numbers. (i) How many different possibilities are there? (ii) If repetition among letters and numbers are prohibited, how many different designs are there? Solution. By the generalized counting principle, we have (i) There are 26 26 26 10 10 10 10 = 175, 760, 000 possibilities. (ii) There are 26 25 24 10 9 8 7 = 78, 624, 000 different designs. 1.2. Permutations. A permutation is an ordered list without repetition. Suppose we have n different objects. Then by the generalized counting principle, there are n (n - 1) (n - 2) 3 2 1(= n!) different permutations of these n objects. Example 1.4. Ten books will be put on a bookshelf. Of these, 4 are math books, 3 are chemistry books, 2 are history books, and 1 is a language book. (i) How many different arrangements are possible? (ii) If the books on same subjects are required to put together, how many different arrangments are there? Solution. (i) There are 10! = 3, 628, 800 different arrangements. (ii) We first divide the bookshelf into 4 parts, there are 4! possibilities to label them as math, chemistry, history and language. Then for each case, there are 4! to ways put math books, 3! ways to put chemistry books, 2! ways to put history books, and 1! way to put language book. By the generalized counting principle, there are 4!4!3!2!1! = 6912 different arragements. Example 1.5. How many different signals, each consisting of 9 flags hung in a line, can be made from a set of 4 white flags, 3 red flags, and 2 blue flags, if all flags of the same color are identical? Solution. If we distiguish the flags with the same color, there are 9! permutations. Since we identify the flags of the same color, different permutations may represent the same signal. By the counting principle, each signal corresponds to 4!3!2! 9! permutations. So the total number of different signals is 4!3!2! = 1260. Remark 1.1. Suppose there are n objects, of which n1 are aloke, n2 are alike, , nr are alike, where n1 + n2 + + nr = n. Then there are n! n1 !n2 ! nr ! different permutations. 3 1.3. Combinations. A combination is an unordered colllection of unique elements. Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important. A k-combination is a subset of S with k elements. Let us calculate the number of k-combinations from a set S with n elements. First, assume the order is an issue. Then there are n ways to select item 1, (n-1) ways to select item 2, , (n - k + 1) ways to select item k. So there are n (n - 1) (n - k + 1) ways in total. However, since the order is irrelevant, there are some repetitions. In fact, each group of k elements is counted k! times. So the number of k-combinations is n! n(n - 1) (n - k + 1) = . k! k!(n - k)! Definition 1.1. We will use the notation n! n , = k k!(n - k)! which represents the number of k-combinations selected from n objects. Example 1.6. From a group of 5 women and 7 men, we need to select 2 women and 3 men to form a committee. (i) How many different committee are there? (ii) If 2 of the men refuse to serve together, how many different committee are there? Solution. (i) By the counting principle there are 5 2 7 3 = 350 different committee. (ii) Since 2 of men refuse to serve together, there are 5 3 selections for men. So there are Proposition 1.3. n k = + 5 2 n-1 k-1 5 2 + 5 2 = 30 30 = 300 different committee. + n-1 k , where 1 k n - 1. First proof. By definition, we have n-1 k-1 + n-1 k (n - 1)! (n - 1)! + (k - 1)!(n - k)! k!(n - k - 1)! (n - 1)![k + (n - k)] = k!(n - k)! n! = k!(n - k)! n = . k = 4 Second proof. Recall that n represents the number of k-combinations of a set k containing n objects. This number can be calculated in a different way. Let us fix n-1 an object, say the first one, then there are k-combinations that contain k-1 n-1 the first object, and there are also k-combinations that do not contain k n-1 n-1 the first object. So there are + k-combinations in total. k-1 k Theorem 1.1. (Binomial Theorem) n (x + y)n = k=0 n k xk y n-k . Proof. We use the induction argument. It is clear for n = 1: 1 k=0 1 k xk y 1-k = 1 0 x0 y + 1 1 xy 0 = y + x = (x + y)1 . Suppose it is true for n = l, i.e. l (x + y)l = k=0 l k xk y l-k . Then for n = l + 1, l (x + y)l+1 = (x + y)(x + y)l = (x + y) k=0 l l k l k xk y l-k xk y l+1-k = k=0 l+1 l k l xk+1 y l-k + k=0 l = i=1 l i-1 l xi y l+1-1 + i=0 l i xi y l+1-i xi y l+1-i + y l+1 = xl+1 + i=1 l l i-1 l+1 i + l i = xl+1 + i=1 l+1 xi y l+1-i + y l+1 = i=0 l+1 i xi y l+1-i . Thus it is also true for n = l+1. This proves the binomial thereom by induction. 5 Homework 1. Problems 1, 7, 9, 13, 17, 18, 20, 24
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Event # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 34A a) b)type of event effect on total assets asset source increases Equ asset source increases Lia asset use decrease Equ NA NA Equ asset exchange unchanged asset use decrease Equ NA NA asset source incr
Drexel - CIVE - 310
Peak Streamflow Histogram16141210 Frequency86420 frequency0-9,9990,00-19,999 1 20,000-29,999 30,000-39,999 40,000-49,999 50,000-59,999 60,000-69,999 70,000-79,999 80,000-89,999 90,000-99,999 100000-109,999 110000-119,999 0 7 12 15
SUNY Oswego - ACCT - 201
ACC 201 Lab 5February 28, 2008Name_BAL 1 4 7Cash 26,200 1,800 15,000 3,600 5,400 7,000 28,000 62,200 -2 3 8BALOffice Equipment 27,000 27,000 -Accounts Payable 1,000 1,0005Service Revenue 31,000 5,400 36,4006 10Acccounts Receiva
SUNY Oswego - ACCT - 201
Lab 7Cash Flow Cash 50,000 OA (1,200) Inventory 6,000 32,000 1,200 (1,000) (620) Land 10,000 = Accounts Payable 32,000 (1,000) (620) (30,380) 42,000 (30,000) (1,900) (8,000) 1,200 33,300 Sales Rev Cost of goods shipping out selling exp gain on sale
SUNY Oswego - ACCT - 201
Lab 7Cash Flow Cash 50,000 OA (1,200) Inventory 6,000 32,000 1,200 (1,000) (620) Land 10,000 = Accounts Payable 32,000 (1,000) (620) (30,380) 42,000 (30,000) (1,900) (8,000) 1,200 33,300 Sales Rev Cost of goods shipping out selling exp gain on sale
SUNY Oswego - ACCT - 201
Jason Rozenberg Problem A 1) \$ 2) \$ 1,050 3) 4) 5) \$ 91,050 \$ 93,150 \$ 2,100Problem B A. \$ B. C. D. E. F. G. H. I. J. K. L. M. N. O. P. Q. R. S. T. U. V. W. X. Y. Z.150 = Rev - Exp\$ 8,000 =Beginning CS - Ending CS \$ 150 =Net Income \$ 125 =Net i
SUNY Oswego - ACCT - 201
Sheet1 Problem 29A Pg 208 Date 2008 1 Cash Comm. Stk 2 Ppd. Rent Cash 3 Cash Unearned Revenue 4 A/R Service Revenue 5 Operating Expense A/P Cash A/R 7 Salaries Exp. Cash 8 Adjusting Entries 9 Rent Expense Prepaid Rent 10 Unearned Revenue Service Reve
SUNY Oswego - ACCT - 201
Accounting Title Cash Certificate of Deposit Accrued Interest Accounts Receivable Supplies Prepaid Rent Office Equipment Accounts Payable Unearned Revenue Interest Revenue Common Stock Retained Earnings Salaries Payable Service Revenue Salaries Expen
Western Michigan - ACTY - 2100
ACCOUNTS RECEIVABLE AND BAD DEBT Accounts Receivable represents money owed to the business from customers for the sale of products or services. The Allowance for Bad Debts (a contra asset) represents what portion of the accounts receivable that the b