MGMT650-HW2-G01 Prob Questions 2017 - 1_AwadK

MGMT650-HW2-G01 Prob Questions 2017 - 1_AwadK - Probability...

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Unformatted text preview: Probability is the relative frequency with which an event occurs. These measures help managers make educ their companies. Learning Objectives: Describe the types of probability: 1) Empirical Probability is determined by observation. For example, the number of defects in 10,000 widgets be 25. So, P(defective) = 25/10,000. 2) Subjective Probability - Subjective probability measures the likelihood of an event subjectively ● Rationally derived from knowledge and experience, ● A "personal belief" allowing substantial variation in estimates. Based on my experience and conditions of that the probability of candidate A winning the election is about 55%. 3) Classical Probability - Classical probability involves enumerating all possible outcomes, counting the num of interest, and computing probability of the outcomes of interest = number of the outcomes of interest / th of outcomes. If there are 40 even numbers and 60 odd numbers randomly distributed in a box, the probabil even number, P(even) = 0.40. Learning Objective: What are the probability rules? ● Probability of an outcome can take any value from 0(0%, impossible) to 1(100%, certainty). ● The probability of all possible outcomes is 100%. ● An event consists of 1 or more outcomes. ● The probability of an event occurring is 1 minus the probability that it doesn't occur. Question # 1 Highlight the values which are NOT VALID for probabilities ● 0.50 ● -0.10 ● 1.10 The Law of Large Numbers means that as the number of observations of an event becomes larger, the runn relative frequency of the event appears to settle down to a constant value, the probability of the event. You have just flipped a fair coin 7 times getting all heads. Does the coin owe you a tail? Although the mythic Averages predicts you are due for a tail, no, the probability of a head on the next toss hasn't changed. 2 Is a good hitter in baseball who has struck out the last six times due for a hit his next time up? ● Yes the hitter is due for a hit. This is an example of the Law of Averages ● This is an example of the so-called Law of Averages. No, the hitter is not due for a hit because the Law ● Yes, the hitter is due for a hit. This is an example of the Law of Large Numbers ● The Law of Large Numbers predicts the hitter is not due for a hit. 3 The running mean of the relative frequency of the event appears to settle down to a constant value. ● This is an example of the Law of Large Numbers. ● This is an example of the Law of Averages. Counting Outcomes Outcomes with Replacement Draw a card from a deck, replace it, and shuffle the deck. There are 52 equally likely outcomes on each draw ● P(draw ace of hearts) = 1/52 is the same from draw to draw. ● For each of the 52 ways in which the 1st draw occurs., there are 52 ways the 2nd draw occurs. So, there a two draws can occur. P(acefollowed by ace) = 4/52 x 4/52 Outcomes without Replacement Draw a card from a deck and not replace it. P(heart) = 13/52. P(heart followed by spade) = 13/52 x 13/51 Outcomes with Replacement Draw a card from a deck, replace it, and shuffle the deck. There are 52 equally likely outcomes on each draw ● P(draw ace of hearts) = 1/52 is the same from draw to draw. ● For each of the 52 ways in which the 1st draw occurs., there are 52 ways the 2nd draw occurs. So, there a two draws can occur. P(acefollowed by ace) = 4/52 x 4/52 Outcomes without Replacement Draw a card from a deck and not replace it. P(heart) = 13/52. P(heart followed by spade) = 13/52 x 13/51 Permutations A permutation is a sequence of outcomes where outcomes are drawn without replacement and order is imp For a sequence of n things drawn from N things without replacement, there are N ways the 1st draw occurs. For each way of the 1st trial, there are N-1 ways the 2nd trial can occur. For 2 draws, there are N x (N-1) poss Continuing, there are N x (N-1) x (N-2) triples for 3 draws. In general, it's N!/(N - n)! , or =PERMUT( N, n) Combinations A combination is a group of outcomes where outcomes are drawn without replacement and order is not imp Example: How many groups of 4 can be formed from 7 people? We know there are 840 sequences possible. 4 things can arrange themselves 4 x 3 x 2 x 1 = 4! (read 4 factorial) ways. so, there are 840 / 4! groups. In ge of combinations = the number of permutations/the number of sequences of n things =N! / (n! x (N - n)!), or =COMBIN(N,n) Sequences of 4 of 7 things 840 Groups of 4 of 7 things 35 # of poker hands Click on the cells to see the formulas 2,598,960 4 The number of sequences (order is important) formed by drawing without replacement of 5 The number of sequences (order is important) formed by drawing with replacement of 6 The number of committees of size 4 formed from 8 Mutually Exclusvie Events If event A = Bill is a Democrat, and event B = Bill is a Republican, events A and B cannot occur jointly (simult and B are called mutually exclusive. P(A and B) = 0. P(A or B) = P(A) + P(B), which is the Simple Addition Rul mutually exclusive events = 1. In the following, mark the correct answer. 7 P(Democrat and Republican) > 0 ● Correct ● Incorrect 8 P(Democrat, Republican, or Independent) = 0, or impossible ● Correct ● Incorrect 9 P(Democrat, Republican, or Independent) = 1, or certain ● Correct ● Incorrect 10 P(Democrat and Republican) = 0, or impossible ● Correct ● Incorrect Example: Toss 2 dice, and let the event be the sum of the top face values. The possible events are computed in the following table . Example: Toss 2 dice, and let the event be the sum of the top face values. The possible events are computed in the following table . toss of 1st die 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 toss of 2nd die 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 The probabilities of the events are Event Probability 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 Only one of the events 2 through 12 can occur in a toss of the dice. These are mutually exclusive event Simple Addition Rule applies. P(X not more than 7) or P(X at most 7) = P(X ≤ 7) = P(2) +P(3) +P(4) +P(5) +P(6) +P(7) P(X not less than 7) or P(X at least 7) = P(X ≥ 7) = P(7) +P(8) +P(9) +P(10) +P(11) +P(12) P(X less than 7) or P(X < 7) = P(2) +P(3) +P(4) +P(5) +P(6) P(X greater than 7) or P(X > 7) = P(8) +P(9) +P(10) +P(11) +P(12) P(X is not 9) = P(X≠9) = 1 – P(X=9) P(no event) = 0 P(any or some event) = 1 11 Assume the mutually exclusive events {1,2,3,4,5} occur with the following probabilities: X P(X = x) 1 0.047 2 0.136 3 0.553 4 0.198 5 0.067 Total 1.000 Which applies? ● Simple Multiplication Rule: P(A and B) = P(A)*P(B) ● Simple Addition Rule: P(A or B) = P(A) + P(B) ● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) 12 13 P(X not more than 4) = P(X at least 3) = 0.933 0.817 Let Excel compute for you by first typing = followed by the computatio with addresses to values in the blue cells. This will avoid typing errors. 14 15 16 P(X = 2) = P(X > 1) = P(X < 6) = 0.136 0.953 1.000 Let Excel compute for you by first typing = followed by the computatio with addresses to values in the blue cells. This will avoid typing errors. Non-Mutually Exclusive Events Non-mutually exclusive events can occur jointly. Example: In a deck of 52 cards, there are 4 suites (heart, di spade), each suite has 4 face cards (ace, king, queen, jack) and 9 number cards (2 – 10). If event A is the car heart and event B is ace, these events can occur jointly. Adding P(ace) + P(heart), we get = 4/52 + 13/52 = 1 know there are only 16 cards (13 hearts and 3 more aces). So that P(A and B) is not double counted, P(A or P(B) – P(A and B), which is the General Addition Rule. So, P(A or B) = 13/52 + 4/52 – 1/52 = 16/32 ∩ means "and". If you add P(A) and P(B), an extra P(A∩B) is added 17 P(A) = P(B) = P(C) = P(A and B)= 0.418 0.242 0.524 0.184 Note: If P(A and B) = 0, A and B are mutually exclusive. Otherwise, A and B can 18 Which applies? ● Simple Multiplication Rule: P(A and B) = P(A)*P(B) ● Simple Addition Rule: P(A or B) = P(A) + P(B) ● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) 19 P(A or B) = 20 P(A or B or C) = 0.559 1.000 Independent Sequential or Simultaneous Events An event which is independent does not affect the probability of occurrence of the next event. Suppose that the probability of winning the lottery, event A, is 1 in 10 million, and that the probability of ex airplane crash, event B, is 2.5 in 1 million (I looked this up). These events are independent. You have been g trillion lifetimes to determine the probability of winning the lottery and crashing on the way to a resort to e winnings. On average, of 10 trillion lifetimes, you won 10T x 1/10M = 1M lottery tickets. On the accompanyi your plane crashes 2.5 times. So, the probability that you will win and crash is 2.5/10T. In general, P(A) of the time A occurs. P(B) of the time B occurs. When A and B share the same time, the eve coincide P(A) x P(B) of the time, or P(A and B) = P(A) x P(B), which is the Simple Multiplication Rule. So, P(AB 2.5/M. You should fly, but you may consider buying something other than the lottery ticket. The assumption of independence lead to P(A and B) = P(A) x P(B). Conversely, if P(A and B) = P(A) x P(B), A a independent. P(A) = P(B) = P(A and B) = 21 0.433 0.348 0.150 Are A and B independent events? ● Yes ● No 22 How do you test independence? Testing to see if the occurance of one event doesn't make it less probable for the other event to occur. Two events are independent if any of the following are true: 1.P(A|B)=P(A) 2. P(B|A)=P(B) 3. P(A and B) = P(A Two customers enter a store. Independently, they make decisions to purchase or not to purchase. The follow shows how the events can occur and combine with red bolding of sequence Customer 1 does not purchase a 2 does not purchase The possible outcomes are shown on the right. If the event is the number of purchases events: 0 purchases, 1 purchase, and 2 purchases 23 If each of the branches on the right of the figure are equally likely, the P(1 purchase) = 24 Which Rule applies to the sequence of independent events? ● Simple Multiplication Rule: P(A and B) = P(A)*P(B) ● Simple Addition Rule: P(A or B) = P(A) + P(B) ● General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) Assume the following probabilities: P(customer makes purchase) = P(customer does not purchase) = 25 26 27 28 Event Purchases 0 2 1 (0,1,or 2) neither purchases both purchase one xor other purchases 0.887 0.113 Probability 0 1/3 2/3 1.000 Hint: 1= P(Purchases = 0, 1, or 2) = P(Purchases = 0 or 2) + P(Pu P(Purchases = 1) = 1 - P(Purc Explain why P(Purchases = 0, 1, or 2) = 1 There are only three options, 0 1 and 2 and if purchases =0,1,or 2 then it’s a 100% chance one of the number Contingency Tables, Joint, Marginal, and Conditional Probabilities In a relative frequency contingency tables, joint probabilities are located inside the margins. In the margin of of a relative frequency continency table, you sum joint probabilites across a row or sum acr Contingency Tables, Joint, Marginal, and Conditional Probabilities In a relative frequency contingency tables, joint probabilities are located inside the margins. In the margin of of a relative frequency continency table, you sum joint probabilites across a row or sum acr It is called a marginal probability. The sum of all marginal probabilities in the bottom row is 1. The sum of all probabilities in the right column is 1. When the probability of event B is affected by the occurrence of event A, the events are not independent. Le the probability of B given the condition that A has occurred. This is called a conditional probability. P(B | A) ≠ P(A | B) ≠ P(A). P(A and B) = P(A) x P(B | A). This is the General Multiplication Rule. ● Type | by holding down Shift and type \ 29 Fill in the blanks in the following relative frequency contingency table for events A, B, C, D, E D E A 0.25 0.14 B 0.13 0.33 C 0.17 0.11 0.28 0.62 0.80 0.11 means P(C and E) = 0.11 0.33 means P(B) = 0.33 Suppose we have the probabilities: P(Democrat and Yes) is called a joint probability because it involves joint events of Democrat and Yes. It can directly from the table. P(Democrat and Yes) = 0.15. P(Democrat) is a marginal probability located in the bottom margin and column Democrat. It is computed by joint probabilities P(Democrat and Yes) + P(Democrat and No) = 0.15 + 0.25 = 0.40. P(Yes) is a marginal probability located in the right margin and row Yes. It is computed by adding the joint pr P(Democrat and Yes) + P(Republican and Yes) = 0.15 + 0.20 = 0.35. Conditional probability is simply a probability given a condition that some event has occurred. This reduces the row or column of the event that occurred. P(Yes | Democrat) is the probability of event Yes given the co the event Democrat has occurred. The table reduces to the Democrat column. In condition Democrat, Yes occurs at a rate of 0.15 in 0.40. So P(Yes | Democrat) = 0.15/0.40 = 0.37 30 Compute the marginal probabilities for following summary of poll results, Male Female Republican 0.611 0.126 Democrat Independent 0.003 0.048 0.125 0.088 31 P(Male and Republican) is type ● Joint ● Marginal ● Conditional 32 P(Male and Republican)= 33 P(Male) is type ● Joint ● Marginal ● Conditional 34 P(Male) = 35 P(Male | Republican) is type ● Joint ● Marginal ● Conditional 36 P(Republican | Male)= (Compute using the values in the Male row) 37 P(Male | Republican) = (Compute using the values in the Republican column) Additional Reading ures help managers make educated decisions for ber of defects in 10,000 widgets was observed to n event subjectively y experience and conditions of the pesent, I feel e outcomes, counting the number of outcomes of the outcomes of interest / the total number stributed in a box, the probability of drawing an https://www.youtube.com/watch?v=AY3O_qsSnbE 100%, certainty). sn't occur. event becomes larger, the running mean of the he probability of the event. you a tail? Although the mythical Law of next toss hasn't changed. s next time up? due for a hit because the Law of Averages is non-existent and doesn't predict anything. to a constant value. y likely outcomes on each draw. he 2nd draw occurs. So, there are 52 x 52 ways http://davidmlane.com/hyperstat/probability.html y likely outcomes on each draw. he 2nd draw occurs. So, there are 52 x 52 ways https://www.boundless.com/algebra/textbooks/boundless-algebra-textbook/com t replacement and order is important. are N ways the 1st draw occurs. draws, there are N x (N-1) possible ways. N - n)! , or =PERMUT( N, n) placement and order is not important. re are 840 sequences possible. But here are 840 / 4! groups. In general, the number n things =N! / (n! x (N - n)!), people = 4 things from 8 things= 1680.000 4 things from 8 things= 70.000 d B cannot occur jointly (simultaneously). A which is the Simple Addition Rule. P(All The possible events are 1/2 https://www.boundless.com/statistics/textbooks/boundless-statistics-textbook/p The possible events are se are mutually exclusive events. The ) +P(6) +P(7) +P(11) +P(12) xcel compute for you by first g = followed by the computation addresses to values in the blue This will avoid typing errors. xcel compute for you by first g = followed by the computation addresses to values in the blue This will avoid typing errors. ds, there are 4 suites (heart, diamond, club, ds (2 – 10). If event A is the card drawn is eart), we get = 4/52 + 13/52 = 17/52, but we ) is not double counted, P(A or B) = P(A) + 4/52 – 1/52 = 16/32 clusive. Otherwise, A and B can occur jointly of the next event. n, and that the probability of experiencing an independent. You have been granted 10 hing on the way to a resort to enjoy your ery tickets. On the accompanying flights, s 2.5/10T. B share the same time, the events A and B ple Multiplication Rule. So, P(AB) = 1/10M x e lottery ticket. y, if P(A and B) = P(A) x P(B), A and B are I don't understand what I am supposed to do here for #17. it says P(A and B that would mean that .418x.242 is actually 0.1013077 http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Statistical_independe https://www.boundless.com/statistics/textbooks/boundless-statistics-textbook/p http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_ProbabilityMultipli he other event to occur. (B|A)=P(B) 3. P(A and B) = P(A) * P(B) e or not to purchase. The following sketch Customer 1 does not purchase and Customer ent is the number of purchases, there are 3 Sketches make probability intuitive 2/3 Hint: 1= P(Purchases = 0, 1, or 2) = P(Purchases = 0 or 2) + P(Purchases = 1). So, P(Purchases = 1) = 1 - P(Purchases = 0 or 2) 00% chance one of the numbers are pulled de the margins. abilites across a row or sum across a column. de the margins. abilites across a row or sum across a column. bottom row is 1. The sum of all marginal events are not independent. Let P(B | A) be onditional probability. P(B | A) ≠ P(B). Also, Rule. A, B, C, D, E and E) = 0.11 s: nts of Democrat and Yes. It can be read mn Democrat. It is computed by adding the 0.40. omputed by adding the joint probabilities ent has occurred. This reduces the table to ability of event Yes given the condition that n. | Democrat) = 0.15/0.40 = 0.375. publican column) https://www.youtube.com/watch?v=H02B3aMNKzEhttps://people.richland.edu/ s/boundless-algebra-textbook/combinatorics-and-probability-343/combinatorics-57/permutations-238-5523/ ks/boundless-statistics-textbook/probability-8/probability-rules-34/the-addition-rule-170-4444/ o here for #17. it says P(A and B)=.184 but P(Aand B) = P(A) x P(B) ??? ki100k/docs/Statistical_independence.html ks/boundless-statistics-textbook/probability-8/probability-rules-34/the-multiplication-rule-171-4445/ ?file=Algebra_ProbabilityMultiplicationRule.xml NKzEhttps://people.richland.edu/james/lecture/m170/ch05-rul.html ...
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