2. How many distinct arrangements are possible using the letters of the word MISSISSAUGA. =

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1 Counting Tool chest 1. Enumeration and counting by cases. 2. Addition and product rules. 3. Permutations and combinations. 4. Multinomial arrangements/permutations. 5. Inclusion/exclusion principle. 6. Pigeon-hole principle. Examples 1. The Quick Snack Bar offers 6 kinds of muffins, 8 kinds of sandwiches and five beverages (coffee, hot tea, iced tea, cola and orange juice). How many different snack combos are there if a customer orders either a muffin with a hot beverage, or a sandwich with a cold beverage? N=6*2+8*3=36 combos (using both addition and product rules) 2. How many distinct arrangements are possible using the letters of the word MISSISSAUGA. =!!!!!!! = from 11 letters (using multinomial arrangements.) 3. While in Edmonton, John needs to attend a meeting at a place 5 blocks east, and 3 blocks north of his office. How many distinct paths could he take on foot (meaning no traffic restrictions) to get to his destination? 8 letters : 5 E's and 3 N's giving 56 paths. (using multinomial arrangements) athletes participate in a race offering 5 trophies. In how many ways can the trophies be awarded? N=40P5= (using permutations) 5. (a) In how many ways can a student answer a 10 question true-false test? (b) As in (a), but student may skip answers (c) as in (b) where student skipped 3 questions (a) 2 =1024 (b) 3 = (c) 103 (2 )= (a) Six people are seated around a round table. How many arrangements are possible? (b) Sixteen people are to be seated around two tables with capacities ten and six respectively. How many arrangements are possible? Note: Arrangements that can be obtained by a simple rotation are considered identical. (a) 6!/6 = 5! = 120 (b) 16C6(10!/10)(6!/6)= In how many ways can a poker player draw five cards from a standard deck and get (a) a straight flush (b) a straight,(c) four aces (d) four of a kind (e) three aces and two jacks (f) three aces and a pair (g) a full house (h) three of a kind (i) two pairs (j) one pair (k) flush?

2 (a) straight flush: (4C1)(10)=40 (b) straight: 4^5(10)-straight flush= =10200 (c) four aces: 4(12)(4C4)=48 (d) four of a kind: 4(12)(13)(4C4) =624 (e) three aces and two jacks: (4C3)(4C2) =24 (f) three aces and a pair: (4C3)12(4C2)=288 (g) a full house: 13(4C3)12(4C2)=3744 (h) three of a kind = 13(4C3)(12C2)(4^2)=54912 (i) two pairs: (13C2)(4C2)(4C2)11*4= (j) one pair: 13(12C3)(4C2)(4^3)= (k) flush: (4C1)(13C5)- straight flush= = In how many ways can four oranges be distributed among up to four children, each receiving at least one orange? 1 child: (4) 2 children: (1,3),(2,2),(3,1) 3 children: (1,1,2),(1,2,1),(2,1,1) 4 children: (1,1,1,1) Total: =8 ways. 9. In a row of 6 seats, how many ways are there such that at least three women are sitting next to each other? Let M=man, W=woman, X=either WWWXXX=2 3 =8 ways MWWWXX=2 2 =4 ways XMWWWX=2 2 =4 ways XXMWWW=2 2 =4 ways Total = 20 ways 10. Ten objects are placed inside a square of side 6 cm. Show that at least two of the objects are within 2 2 cm of each other. Subdivide the square into 9 squares of side 2 cm. Place an object in each square. By the pigeon hole principle, the 10 th object must share a square with one of the previous ones. So the distance between these two objects cannot exceed 2 2 cm.

3 Elementary Probability Tool chest 1. Axioms: i. 0 P(E) 1 ii. P(S)=1 iii. P( )=( )+( )+ +( )= ( )=1 2.. Sets and Venn diagrams. 3. Tree diagrams. 4. Sample point method - enumerating sample points of equiprobability. 5. Addition and multiplication rule. 6. Conditional probability. 7. Law of total probability. 8. Bayes' rule. Examples (note: Ω = S = sample space, S = cardinality of S = size of sample space) 1. An airline offers 4 daily flights from Toronto to Vancouver, and 3 daily flights from Vancouver to Hong Kong. If flights are to be made on different days, how many flight arrangements can the airline offer from Toronto to Hong Kong? This is a two step problem and the multiplication rule applies. No. of outcomes for the first step = 4 no. of outcomes for the second step = 3 Total number of possible arrangements = 4*3 = McDougal burgers offers the following condiments: ketchup, mustard, mayonnaise, lettuce, tomato, onion, pickle, cheese, mushrooms, and hot pepper. If a customer can order as many or as few condiments as he wishes, how many different burgers are possible. There is a total of 10 condiments, meaning the set C={ketchup, mustard, mayonnaise, lettuce, tomato, onion, pickle, cheese, mushrooms, hot pepper} where C =10. The power set P(C) is a set of all possible subsets of C, including the null set. It can be shown that P(C) =2 C. Thus there are 2 10 =1024 different burgers possible. 3. Six fair dice are thrown and the upper faces are recorded. What is the probability that the recorded numbers are all distinct? (sample point method) No. of possible outcomes = 6*5*4*3*2*1 = 6! Size of sample space = S = 6^6 Probability of distinct upper faces = 6!/6^6 = 5/ A company wishes to place 3 orders among 5 suppliers. Each supplier has equal chance of getting any order, and there is no limit to the number of orders each supplier can get. Determine the probability that a. a given supplier gets exactly 2 orders. b. out of M supppliers, a supplier gets k orders out of n a. Size of sample space = SS = 5³ = 125 number of ways to give 2 orders out of three to one supplier = 3C2 = 3

4 number of ways to give 1 order to the remaining 4 suppliers = 4 1 = 4 probability of a certain supplier getting 2 orders = 3*4/125=12/125 b. Size of sample space = SS = M n number of ways to give k orders out of n to one suplier = nck number of ways to give remaining n-k orders to remaining M-1 suppliers = (M-1) n-k probability of giving k orders out of n to one supplier out of M = nck * (M-1) n-k / M n Conditional Probability & Bayes' Rule P(A B) = P(A B)/P(B) Two events are independent if any of the following holds: P(A B)=P(A) P(B A)=P(B) P(A B)=P(A)P(B) 5. If two events, A and B, are such that P(A)=0.5, P(B)=0.3, and P(A B)=0.1, find the following: a. P(A B) b. P(B A) c. P(A A B) d. Are events A and B independent a. P(A B)=P(A B)/P(B)=0.1/0.3=1/3 b. P(B A)=P(B A)/P(A)= P(A B)/P(A)=0.1/0.5=1/5 c. First establish P(A B) = P(A)+P(B)-P(A B)= =0.7 P(A A B)=P(A (A B))/P(A B)=P(A)/P(A B)=0.5/0.7=5/7 d. P(A)P(B)=0.5*0.3=0.15 P(A B), therefore A and B are not independent by definition of independence. 6. The following contingency table illustrates the outcome of a certain exam. Outcome Male (M) Female (F) Total Pass (A) Fail (A') Total a. Are the events A and M independent? b. Are the events A' and F independent? P(A)=0.60; P(A')=0.40; P(M)=0.40; P(F)=0.40; P(A M)=0.24; P(A' F)=0.24 a. P(A)P(M)=0.60*0.40=0.24 = P(A M), therefore A & M are independent. b. P(A')P(F)=0.40*0.60=0.24 = P(A' F), therefore A' and F are independent. 7. Cards are dealt from a standard deck. If the first 2 cards are spades, what is the probability that the next 3 cards are also spades? 50 cards are left to be dealt, of which 11 are spades. Probability that the next 3 cards are also spades = (11/50)(10/49)(9/48)=33/3920= In a survey of fifty students, 22 play tennis, 25 play football, 9 play both tennis and football, 17 play both tennis and baseball, 20 play both football and baseball, six play all three, and 4 play none. a. How many students play only tennis? b. How many students play only baseball? c. Given that a student plays football and baseball, what is the probability that he does not play tennis? Draw a Venn diagram and deduce that (a) 2 (b) 8 (c) P(T' B F)=P(T' (B F))/P(B F)=(14/50)/(20/50)=0.7

5 9. Diseases A and B are prevalent within a certain population. It has been estimated that 10% of the population will contract disease A, and 15% will contract disease B within their lifetime, while 3% will contract both. a. Find probability that a randomly chosen person will contract at least one disease. b. Find the conditional probability that a randomly chosen person from this population will contract both diseases, given that he or she has contracted at least one disease. a. P(A)=0.10; P(B)=0.15; P(A B)=0.03; P(A B)=P(A)+P(B)-P(A B)= =0.22 b. P(A B A B) = P(A B (A B))/P(A B) = P(A B)/P(A B)=0.03/0.22=3/22= A diagnostic test correctly detects a given disease 90% of the time when the person actually has the disease, and correctly reports negative 90% of the time when the person does not have the disease. Only 1% of the population is estimated to have the disease. What is the probability that a randomly chosen person actually has the disease if the test reports a positive result? Let D=event that the person actually has the disease P=event that the test reports positive. Method 1: use Bayes' theorem Then P(D)=0.01; P(D')=0.99; P(P D)=0.90; P(P' D)=1-0.90=0.10; P(P' D')=0.90; P(P D')=0.10 P(D P)=P(P D)P(D)/(P(P D)P(D)+P(P D')P(D')=0.9*0.01/(0.9* *(0.99)=0.009/( )=1/12= Method 2: use contingency table and the law of total probability Test result D: has disease D' : does not have disease Total P: tests positive 0.90*0.01= *0.99= P' : tests negative 0.10*0.01= *0.99= Total P(D P)=0.009/0.108= In a multiple choice examination, each question offers four possible answers. Suppose that a student knows the answer to 80% of the questions, and she makes guesses to the remaining questions with a success rate of If the answer to a particular question was correct, what is the probability that she actually knew the answer? K=event that she knows the answer; K'=event that she guesses C=event that the answer is correct; C'=event that the answer is incorrect Given P(K)=0.8; P(K')=0.2; P(C K)=1; P(C K')=0.25, then P(K C)=P(C K)*P(K)/(P(C K)*P(K) + P(C K')*P(K')=1*0.8/(1* *0.2)=0.8/0.85=16/17=0.9412