4-1 BASIC CONCEPTS OF PROBABILITY

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1 CHAPTER BASIC CONCEPTS OF PROBABILITY Identify probabilities as values between 0 and 1, and interpret those values as expressions of likelihood of events Develop the ability to calculate probabilities of events Define the complement of an event and calculate the probability of that complement. 4-2 ADDITION RULE AND MULTIPLICATION RULE 1. Develop the ability to calculate the probability that in a single trial, some event A occurs or some event B occurs or they both occur. Apply the addition rule by correctly adjusting for events that are not disjoint (or overlapping). 2. Develop the ability to calculate the probability of an event A occurring in a first trial and an event B occurring in a second trial. Apply the multiplication rule for events that are not independent 3. Distinguish between independent events and dependent events. 4-3 COMPLEMENTS, CONDITIONAL PROBABILITY AND BAYES THEOREM Compute the probability of at least one occurrence of an event A Apply the multiplication rule by computing the probability of some event, given that some other event has already occurred. 4-4 COUNTING Develop the ability to apply the fundamental counting rule, factorial rule, permutations rule, and combinations rule. Distinguish between circumstances requiring the permutations rule and those requiring the combinations rule

2 4-1 BASIC CONCEPTS OF PROBABILITY DEFINITIONS An event is any collection of results or outcomes of a procedure A simple event is an outcome or an event that cannot be further broken down into simpler components. The sample space for a procedure consists of all possible simple events. That is the sample space consists of all outcomes that cannot be broken down any further

3 NOTATION FOR PROBABILITIES P denotes a probability A, B, and C denote specific events. P(A) denotes the probability of event A occurring. Probabilities are always values between 0 and 1 0 P(A) 1 Solution: 0, 3/5, 1, In exercises express the indicated degree of likelihood as a probability value between 0 and If you make a random guess for the answer to a true/false test question, there is a chance of being correct. Solution: P(correct) = 0.5 or ½ 14. When making a random guess for an answer to a multiple choice question on a SAT test, the possible answers are a, b, c, d, e, so there is 1 chance in 5 of being correct. Solution: P(correct) = 0.2 or 1/5 15. Based on a Harris poll, if you randomly select a traveler, there is a 43% chance that his or her luggage is black. Solution: P(black)=.43 or 43/ Based on a report in Neurology magazine, 29.2% of survey respondents have sleepwalked. Solution: P(sleepwalked)=.292 or 292/ Sydney Smith wrote in On the Conduct of the Understanding that it is impossible to fit a square peg in a round hole. Solution: P(square peg in round hole) = Benjamin Franklin said that death and taxes is a certainty of life. Solution: P(death and taxes) = 1

4 Three Common Approaches to Finding the Probability of and Event. 1. Relative Frequency Approximation of Probability Conduct (or observe) a procedure and count the number of times that event A occurs. P(A) is then approximated as follows: number of times A occurred P(A) = number of times the procedure was repeated 2. Classical Approach to Probability (Requires equally Likely Outcomes) If a procedure has n different simple events that are equally likely, and if event A can occur in s different ways, then number of ways A occurs s P(SA) = = number of different simple events n 3. Subjective Probabilities P(A), the probability of event A is estimated by using knowledge of the relevant circumstances. Sometimes none of the preceding three approaches can be used. A simulation of a procedure is a process that behaves in the same ways as the procedure itself so that similar results are produced.

5 Caution: Use fractions or decimals for probabilities NOT percents. LAW OF LARGE NUMBERS As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability. Caution: 1. The law of large numbers applies to behavior over a large number of trials, and it does not apply to any one individual outcome. Gamblers sometimes foolishly lose large sums of money by incorrectly thinking that a string of losses increases the chances of a win on the next bet, or that a string of wins is likely to continue 2. If we know nothing about the likelihood of different possible outcomes, we should not assume that they are equally likely. For example, we should not think that the probability of passing the next statistics test is ½ or 0.5 (because we either pass or do not pass). The actual probability depends on factors such as the amount of preparation and the difficulty of the test.

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7 Example: Relative frequency probability Solution: P(Green) = 428/580 or.738; Mendel claimed. (because = 580) So this is close to ¾ as Example: Solution: 21. 5/555 or The employer would suffer because it would be at risk by hiring someone who uses drugs.

8 Solution: /555 or The person tested would suffer because he or she would be suspected of using drugs when in reality he or she does not use drugs. Example: Solution: Classical probability. a. P(correct date)= 1/365 Since there are 365 equally likely days to choose from. b. Yes. 1/365 <.05 c. He already knew d. 0 (Triola humor) COMPLEMENTARY EVENTS DEFINITION The complement of event A, denoted which event A does not occur., consists of all outcomes in Solution: A = uses social networking sites P(A) = 3732/( ) = 3732/5112 =.730 So A = does not use social networking sites P( A ) = 1380/5112 =.270 It is not unlikely for someone to not use social networking sites.

9 Relationship between P(A) and P( A ) is that they sum to one. P(A) + P( A ) = 1 or put another way P( A ) = 1- P(A).

10 ODDS Expressions of likelihood are often given as odds, such as 50:1 (or 50 to 1). Advantages of probabilities and odds: Odds make it easier to deal with money transfers associated with gambling. Probabilities make calculations easier, so they tend to be used by statisticians, mathematicians, scientists, and researchers in all fields.

11 Example:

12 Solution: a. Odds against winning = P( A )/P(A) = = 9999 :1 b. Find the payoff odds: (net profit): (amount bet) = 4999:1 c. The description is not accurate. The odds against winning are 9999:1 and the odds in favor of winning are 1:9999, not 1:10,000

13 4-2 ADDITION RULE AND MULTIPLICATION RULE Use addition rule for finding P(A or B) Use multiplication rule for finding P(A and B) DEFINITION A compound event is any event combining two or more simple events. ADDITION RULE INTUITIVE ADDITION RULE: To find P(A or B), add the number of ways event A can occur and the number of ways event B can occur, but add in such a way that every outcome is counted only once. P(A or B) is equal to that sum, divided by the total number of outcomes in the sample space. FORMAL ADDITION RULE: P(A OR B) = P(A) + P(B) P(A and B) Where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure. Example: Solution: P(McD or Accurate) = P(McD) + P(accurate) P(McD and accurate) = 362/ / /1118 = 1020/1118 or.912

14 These are not disjoint events (that is they CAN occur simultaneously: Both McD and accurate) DEFINITION Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap) Example: Disjoint events: Event A: Adopting a pet that is male. Event B: Adopting a pet that is female. Not Disjoint events: Event A: Adopting a pet that is a dog. Event B: Adopting a pet that is male. Solution: P(not accurate or Wendy s) = P(not accurate) + P(Wendy s) P(Both not accurate and Wendy s) 131/ / /1118 = 380/1118 or.340 These events are not disjoint because it is possible to be BOTH inaccurate and Wendy s.

15 COMPLEMENTARY EVENTS AND THE ADDITION RULE P (A or A ) = P(A) + P( A ) = 1 Note that these events must be disjoint! Solution: A = event of arriving on time A = event of not arriving on time So P( A ) = 1 P(A) = =.197 (Remember that probabilities are between 0 and 1 and NOT a percent) * Try # 5, 7,8 in HW section MULTIPLICATION RULE Caution: The notation P(A and B) has two meanings, depending on its context. For the multiplication rule, P(A and B) denotes that event A occurs in one trial and event B occurs in another trial; for the addition rule we use P(A and B) to denote that the events A and B both occur in the same trial. Intuitive Multiplication Rule: To find the probability that event A occurs in one trial and event B occurs in another trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B is found by assuming that event A has already occurred.

16 INDEPENDENCE AND THE MULTIPLICATION RULE DEFINITION Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are independent if the occurrence of any does not affect the probabilities of the occurrence of the others.) If A and B are not independent, they are said to be dependent. Example: Solution: Let T1 = event first is from Taco Bell Let T2 = event that second is from Taco Bell a. P(T1 and T2) = P(T1) * P(T2/T1) = P(T1) * P(T2) = i = These are independent because it is with replacement. b. P(T1 and T2) = P(T1) * P(T2/T1) = i = These events are NOT independent because it is without replacement so the sample space has changed. *Try # in the HW section Note: Sampling with replacement VS sampling without replacement Sampling with replacement: Selections are independent events. Sampling without replacement: Selections are dependent events.

17 However, there is a caveat. When the sample is large, then sampling without replacement is almost the same a sampling with replacement. Below is a guideline: Example: Solution: a. P(Transported) = b. Since 5/ = or.003% < 5% so we can use the guideline to treat this as if it was sampled with replacement P( all 5 transported) = REDUNDANCY: IMPORTANT APPLICATION OF MULTIPLICATION RULE The principle of redundancy is used to increase the reliability of many systems. Our eyes have passive redundancy in the sense that if one of them fails, we continue to see. An important finding of modern biology is that genes in an organism can often work in place of each other. Engineers often design redundant components so that the whole system will not fail because of the failure of a single component.

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19 Example: Solution: Let F1 be the event that the first generator fails. Let F2 be the event that the second generator fails. a. P(F1 and F2) = P(F1)*P(F2) = (.22)*(.22) =.0484 b. P(at least one works) = 1 P(both fail) = =.952 Rationale for the Multiplication Rule Consider two independent events. You are given a quiz and the first question is true/false. The second question is multiple choice with possible answers a,b,c,d,e. So the possible scenarios are Ta Tb Tc Td Te Fa Fb Fc Fd Fe (10 possible outcomes) Hence the probability of getting both correct is 1/10 or 0.1 because there are 10 possible outcomes and only one is right. Or P(1 st is correct and 2 nd is correct) = P(1 st correct)*p(2 nd correct) = 1 2 i 1 5 = 1 10

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21 4-3 COMPLEMENTS, CONDITIONAL PROBABILITY, AND BAYES THEOREM COMPLEMENTS: The Probability Of At Least One At least one has the same meaning as one or more The complement of getting at least one particular event is that you get no occurrences of the event. Example: Solution: Let A = event of getting at least one girl. Then A = event of getting no girl Thus P(A)= 1 P( A ) = 1 - (.5) 3 =.875 Solution: It s the same as P(Boy) since each birth is independent of the rest. Hence it is 0.5 or ½ Solution: As in question #5

22 Thus P(A)= 1 P( A ) = 1 - (.545) This could not continue indefinitely at this rate as women would become extinct. L * continue to try HW #9 - #12 CONDITIONAL PROBABILITY DEFINITION A conditional probability of an event is a probability obtained with the additional information that some other event has already occurred. Notation: P(B\A) denotes the conditional probability of event B occurring, given that event A has already occurred. INTUITIVE APPROACH FOR FINDING P(B\A) The conditional probability of B occurring given that A has occurred can be found by assuming that event A has occurred and then calculating the probability that event B will occur. FORMAL APPROACH FOR FINDING P(B\A) The probability P(B\A) can be found by dividing the bprobability of events A and B both occurring by the probability of event A: P(A and B) P(B\A) = P(A)

23 For the following 2 examples let G = purchased gum, M = kept money, Q = given quarters, and B = given $1 bill Solution: a. P(M\B) Intuitive approach:. P(G\B) = 12/46 or.261 Formal approach: P(G\B) = = =.261 b. P(M\B) = 34/46 or.739 P(G and B) P(B) = P(Bought gum and was given $1 bill) P(Given $1 bill) c. It appears that when students are given a $1 bill, they are more likely to keep the money than spend it.

24 Solution: a. P(M\Q) = 16/43 =.372 b. P(M\B) = 34/46 =.739 c. It appears that students are more likely to keep the money when given a $1 bill than when given four quarters. Other examples: HW #17-20 Confusion of the Inverse: Note that P(B\A) is not necessarily the same as P(A\B)

25 BAYES THEOREM Sometimes in the medical profession doctor s give misleading information when they experience confusion of the inverse. They tend to confuse P(positive test result\cancer) with P(cancer\positive test result). About 95% of physicians estimated P(cancer\positive test result) to be about 10 times too high. For Example 4 use the following table:

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27 The most general form of Bayes Theorem is DEFINITION: A prior probability is an initial probability value original obtained before any additional information is obtained. A posterior probability is a probability value that has been revised by using additional information that is later obtained.

28 4-4 COUNTING 1. Multiplication Counting Rule the multiplication counting rule is used to find the total number of possibilities from some sequence of events. MULTIPLICATION COUNTING RULE For a sequence of events in which the first event can occur ways, the second event can occur ways, the third event can occur ways, and so on, the total number of outcomes is HW NOTE: in exercises 5 36 express all probabilities as fractions. Example Solution: There are 10 different possibilities for each digit, so the total number of different possible guesses of the first 5 digits is n 1 i n 2 i n 3 i n 4 i n 5 = 10 i10 i10 i10 i10 = So the P(correct guess) = 1/100, Factorial Rule

29 The factorial rule is used to find the total number of ways that n different items can be rearranged (order of items matters). The factorial rule uses the following notation: Notation: The factorial symbol! denotes the product of decreasing positive numbers. For example 5!= 5 i 4 i 3i 2 i1 = 120 Also note that 0! = 1 FACTORIAL RULE The number of different arrangements (order matters) of n different items when all n of them are selected is n! Example Solution: P(graded in alphabetical order) = 1/8! = 1/40320 Permutations and Combinations

30 When using different counting methods, it is essential to know whether different arrangements of the same item are counted only once or are counted separately. DEFINITIONS Permutations of items are arrangements in which different sequences of the same items are counted separately. (the letter arrangements of abc, acb, bac, bca, cab, and cba are all counted separately as six different permutations.) Combinations of items are arrangements in which different sequences of the same items are counted as being the same. (the letter arrangements of abc, acb, bac, bca, cab and cba are all considered to be the same combination) 3. Permutations Rule (When All of the Items are Different) The permutations rule is used when there are n different items available for selection, we must select r of them without replacement and the sequence of the items matters. The result is the total number of arrangements (or permutations) that are possible. (Remember rearrangements of the same items are counted as different permutations.) PERMUTATION RULE When n different items are available and r of them are selected without replacement, the number of different permutations (order counts) is given by Example: Solution:

31 Since she is selecting 5 (r) out of 50 (n) and order does matter use permutations. The number of ways she can visit 5 capitals is 50 P 5 = 50! (50 5)! = So the probability that it is in the specific order given in #11 is Permutations Rule (When Some Items are Identical to Others) When n items a2e all selected without replacement, but some items are identical, the number of possible permutations (order matters) is found by using the following rule. PERMUTATIONS RULE (WHEN SOME ITEMS ARE IDENTICAL TO OTHERS) The number of different permutations (order counts) when n items are available and all n of them are selected without replacement, but some of the items are identical to others, is found as follows: where Example: How many unique arrangements of the word MISSISSIPPI are there. Solution: Notice there are 4 S, 4 I, 2 P and 1 M. 11! So there are = arrangements. 4!i 4!i 2! 5. Combinations Rule The combinations rule is used when there are n different items available for selection, only r of them are selected without replacement, and order does not matter. The result is the total number of combinations that are possible. (Remember: Rearrangements of the same items are considered to be the same combination.) COMBINATION RULE: When n different items are available, but only r of them are selected without replacement, the number of different combinations (order does not matter), is found as follows:

32 Example: Solution: The winner must guess 5 out of 75 numbers in any order AND 1 out of 15. So we are going to use the multiplication rule here as these are independent events. The number of ways to select 5 out of 75 numbers without regard to order is 75 C 5 = 75! (75 5)!5! = The number of ways to select 1 out of 15 = 15 So the number of ways to BOTH select 5 out of 75 AND 1 out of 15 is i15 =

33 Hence the probability of winning is 1/258,890,850. It is more likely to be struck by lightning Note: Try other HW questions to practice detecting which formula applies to the solution. Examples Solution. a. How many ways can 3 players be selected from 11 players? Since order does not matter use combinations. 11C 3 = 165 b. How many ways can 3 players be designated as first, second, and third? This would be 3! = 6 Solution: There are two cases. First with 2 letters and second with 3 letters. For 2 letters: The first spot can be either K or W so there are 2 choices. The second spot can be any of 26 letters. The third spot can be 26 letters. So there are 2 x 26 x 26 = 1352 possibilities. For 3 letters: The first spot can be either K or W so there are 2 choices. The second spot can be any of 26 letters. The third spot can be any of 26 letters. The fourth spot can be 26 letters. So there are 2 x 26 x 26 x 26 =35152 possibilities. Total number of possibilities is = 36,504 Solution: We are choosing 2 out of 5 and since order does not matter we use combinations: 5 C 2 =10

34 Solution: The probability is 1 # ways = 1 5! = Solution: The number of way since order matters is 4! = 24 So the probability is 1/24. Solution: We are selecting 6 out of 9 locations (order matters) so use permutations. Hence the number of different routes is 9 P 6 = Solution: a. To choose 4 out of 10 with them going into designated slots we use permutation. So 10 P 4 = 5040 b. To choose 4 out of 10 with no regard to order use combinations: So 10 C 4 = 210 c. P(choosing 4 youngest) = 1/210