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Chapter 6: Probability and Simulation, the study of randomness

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6.1 – Simulation

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Simulation: The imitation of chance behavior, based on a model that accurately reflects the experiment under consideration. Independent: One event doesnt influence the outcome of another event. (tossing a coin)

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5 steps of simulation: 1. State the problem or describe the random phenomenon 2. State the assumptions. 3. Assign digits to represent outcomes. 4. Simulate many repetitions 5. State your conclusions

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Example #1 Describe how you would simulate the following problems. a.Boy or girl? There is an equally likely chance of delivering a boy or a girl at birth. What is the likelihood of having a girl among their first four children? Flip a coin Roll Die: 1-3 girl, 4-6 boy Use # table, 0-4 girl and 5-9 boy

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Example #1 Describe how you would simulate the following problems. b. Choose one person at random from a group of which 50% are employed, 20% are unemployed, and 30% are not in the labor force. Use # table, 0-4 employed,5-6 unemployed,7-9 not in labor force

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Calculator Tip: Random Integers Math – Prob – RandInt (smallest #, largest #, n)

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Example #2 What is the likelihood of tossing a coin 5 times and getting at least 3 consecutive heads or 3 consecutive tails? Simulation Steps: 1. State the problem or describe the random phenomenon Find the likelihood of tossing a coin 5 times and getting 3 consecutive heads or tails 2. State the assumptions Probability of a heads or tails is 50% Coin tosses are independent

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Example #2 What is the likelihood of tossing a coin 5 times and getting at least 3 consecutive heads or 3 consecutive tails? Simulation Steps: 3. Assign digits to represent outcomes 0-4 heads, 5-9 tails 4. Simulate many repetitions Random digit table: Line 101

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First time: Second: Third: Fourth: H THHH T THHH H TTTT Fifth: Sixth: H TTHH T THHT H HTHH Seventh: Eighth: H HTHH T HTTH

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Example #2 What is the likelihood of tossing a coin 5 times and getting at least 3 consecutive heads or 3 consecutive tails? Simulation Steps: 5. State your conclusions P(3 consecutive heads or tails) = 3838 After 8 simulations, the probability of flipping a coin 5 times and getting at least 3 consecutive heads or tails is approximately 37.5%

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Example #4 A basketball player makes 70% of her free throws in a long season. In a tournament game she shoots 5 free throws late in the game and misses 3 of them. The fans think she was nervous, but the misses may simply be chance. You will shed some light by estimating the probability. a.Describe how to simulate a single shot if the probability of making each shot is 70%. Then describe how to simulate 5 independent shots. Assign numbers to represent baskets and misses. 0-6 basket, 7-9 miss Use Random Number table to choose 5 numbers

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Example #4 A basketball player makes 70% of her free throws in a long season. In a tournament game she shoots 5 free throws late in the game and misses 3 of them. The fans think she was nervous, but the misses may simply be chance. You will shed some light by estimating the probability. b. Use the calculator to simulate 20 repetitions of the 5 shots and record the number missed on each repetition. What is the approximate likelihood that the player will miss 3 or more of the 5 shots? randInt(0, 9, 5)

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14 6 # of times she didnt miss 3 out of 5: # of times she did miss 3 out of 5: P(missing 3 out of 5 shots) = 6 20 = 30%

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6.2 – Probability Models

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Probability: Chance behavior is unpredictable in the short run but has a regular and predictable pattern in the long run. Probability of an event equals the proportion of times it occurs in a long series of repetitions. P(Event) = # of successes Total #

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Random: An event is random if individual outcomes are uncertain, but there is a regular distribution of outcomes in a large number of repetitions.

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Probability Model: Mathematical description of a random phenomenon that has two parts: Sample Space Way of assigning probabilities to events

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Sample Space: Set of all possible outcomes S = { … } Event: Any outcome or set of outcomes. Subset of the sample space E = { … }

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Example #4 Fill in the answers below for each situation 1. Tossing a single die: a. List the sample space. S = {1, 2, 3, 4, 5, 6} b. Event A: an odd number E = {1, 3, 5}

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Example #4 Fill in the answers below for each situation 1. Tossing a single die: c. P(A) = probability of A occurring: P(A) = 3636 = 50%

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Example #4 Fill in the answers below for each situation 2. In a standard deck of playing cards there are 52 cards total. a. Drawing one card, P(Heart) P(Heart) = = 25%

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Example #4 Fill in the answers below for each situation 2. In a standard deck of playing cards there are 52 cards total. b. Drawing one card, P(King) P(King) = = 8%

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Example #4 Fill in the answers below for each situation 2. In a standard deck of playing cards there are 52 cards total. c. If I continue to draw cards from the deck, will the trials be independent? Why or why not? No,the first card influences the probability of the next card

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There are two ways to find the total number of outcomes in a sample space: 1. Tree Diagram:Branches off to represent each possibility 2. Multiplication Principle: If you can do one task in a number of ways and a second task in b number of ways, then both tasks can be done in a x b number of ways.

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Example #5 Draw a tree diagram for flipping a coin and then rolling die. Then determine the size of the sample space.

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H T # of outcomes: 12 Or Mult. Principle: a x b 2 x 6 12

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6.2 - continued

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Probability Rules: The probability P(A) of any event A satisfies 0 P(A) 1 The sum of the probabilities of all possible outcomes is 1. P(A) = 1 means the event will always occur P(A) = 0 means the event will never occur An event is empty if it has no outcomes in it.

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Venn Diagram: A picture representation of a sample space. S A B

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Example #1 Make a Venn diagram for the following situation: 45% of kids like Barney 25% of kids like Blue 55% of kids like Pooh 15% of kids like Blue and Pooh 25% of kids like Barney and Pooh 5% of kids like Blue but not Barney or Pooh 5% of kids like Barney, Blue, and Pooh

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Things kids like Barney Pooh Blue

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The probability that event A does NOT occur Written: A c P(A c ) = 1 – P(A) Complement Rule: S A

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Union: A B Either event A or event B occurs A B S

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Intersection: A B Event A and Event B occurs A B S

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Disjoint: (mutually exclusive) Two events A and B are disjoint if they have no outcomes in common, so never can occur simultaneously. A B S

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Addition Rule (for disjoint events only!!!!!): P(A or B) = P(A) + P(B) A B S Note: P(A and B) = 0

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: a. P(A) = Probability of drawing a spade P(A) = =25%

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: b. P(B) = Probability of drawing a King P(B) = =8%

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: c. P(C) = Probability of drawing an Ace P(C) = =8%

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: d. P(B c ) = P(B c ) = =92% 1 – Probability of not drawing an King

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: e. P(B C) = P(B C) = =15% Probability of Ace or King P(B) + P(C) = =

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Example #2 You have drawn one card from a standard deck of cards. Find the following probabilities: f. P(B C) = P(B C) = Probability of Ace and King 0

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What is the difference between disjoint and independent? Independence is when one event doesnt influence the probability of the next event. Disjoint means the two events have no outcomes in common, so never can occur simultaneously.

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Example #3 Determine if the given example is independent, disjoint, or neither. a.Drawing one card and getting a black King. Black King Deck of cards independent

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Example #3 Determine if the given example is independent, disjoint, or neither. b. Drawing two cards and getting a black King, without replacement. Black King Deck of cards neither

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c. Can you have a disjoint event be independent? Explain. A B Knowing event A happens means event B cannot

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Multiplication Rule (for independent events only!!!!!): P(A and B) = P(A) P(B)

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Example #4 The following table gives the gender and political party of the 100 delegates at a political convention. Suppose you encounter a delegate at random. The proportion of each person is given. WomenMen Republicans Democrats Independents a.Find the marginal percents

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b. Is it a valid probability distribution? How do you know? Example #4 The following table gives the gender and political party of the 100 delegates at a political convention. Suppose you encounter a delegate at random. The proportion of each person is given. WomenMen Republicans Democrats Independents Yes, the probabilities add to one

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c. What is the probability that the delegate will not be a man? Example #4 The following table gives the gender and political party of the 100 delegates at a political convention. Suppose you encounter a delegate at random. The proportion of each person is given. WomenMen Republicans Democrats Independents P(not man) =1 – P(man) =1 – 0.48 =0.52

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d. What is the probability that the delegate will not be an Independent? Example #4 The following table gives the gender and political party of the 100 delegates at a political convention. Suppose you encounter a delegate at random. The proportion of each person is given. WomenMen Republicans Democrats Independents P(not ind.) =1 – P(ind.) =1 – 0.10 =0.90

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e. What is the probability that the delegate will be a Republican woman? Example #4 The following table gives the gender and political party of the 100 delegates at a political convention. Suppose you encounter a delegate at random. The proportion of each person is given. WomenMen Republicans Democrats Independents P(Rep. and Woman) = 0.21 Note: not independent!

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Example #5 A gambler wants to know the probability of rolling one die twice and getting a 3 both times. P(3 3) = Note: Independent =3% 1616 = 1616 P(3) P(3) =

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Example #6 A string of Christmas lights contains 20 lights. The lights are wired in a series, so that if any light fails the whole string will go dark. Each light has probability 0.02 of failing during a 3 year period. The lights fail independently of each other. What is the probability the strings of lights will remain bright for 3 years? Note: Independent P(all 20 lights not fail for 3yrs) = (1 –.02) 20 =(.98)

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6.3 – General Probability Rules

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Problem: Addition Rule only works for disjoint events Solution: A B S Subtract P(A and B) from both probabilities P(A B) = P(A) + P(B) P(A B) = P(A) + P(B) – P(A B)

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Conditional Probability: The probability of an event occurring, given another event occurred P(A B) = P(A) happens given that B occurs P(B A) = P(B) happens given that A occurs

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A B S P(A B) = P(A and B) P(B)

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P(B A) = P(A and B) P(A) A B S

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P(A B) = P(A and B) P(B) P(A and B) = P(B) P(A B) General Multiplication Rule: If Independent: P(A and B) = P(A) P(B) In General:

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Determining Independence: Two events A and B are independent if P(A B) = P(A) P(B A) = P(B) OR:

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Example #1 The Adams Corporation stores spare parts at two warehouses, one in Baltimore and one in Plattsburgh. The number of defective and acceptable spare parts at each warehouse is given below: Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total If one of the 500 spare parts kept by the firm is chosen at random let D and A denote the events that the spare part is defective or acceptable, while B and P denote the events that it comes from Baltimore or Plattsburgh. Calculate the following probabilities:

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total a. P(B) = = 0.60

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total b. P(A B) = = Using Rule: P(A and B) = P(B) P(A B) P(A and B) = = 0.544

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total c. P(B c ) = 1 – P(B) =1 – = 0.40

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total d. P(A B) = – –.544 P(A) + P(B) – P(A and B) = = = 0.98

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total e. P(A B) = =

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P(A and B) P(B) Using Rule: P(A B) = = = = =

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total f. P(P D) = =

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P(P and D) P(D) Using Rule: P(P D) = = = = =

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Number of Spare Parts WarehouseDefectiveAcceptableTotal Baltimore28272 Plattsburgh10190 Total g. Determine if coming from Plattsburgh and being defective are independent. P(P D) = P(P) = No, they are not

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Tree Diagram: Some conditional probabilities can be solved by drawing a schematic way of looking at all possible outcomes.

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Example #2 Suppose a computer company has manufacturing plants in three states. 50% of their computers are manufactured in California and 85% of these are desktops, 30% of computers are manufactured in Washington, and 40% of these are laptops, and 20% of computers are manufactured in Oregon, and 40% of these are desktops. All computers are first shipped to a distribution site in Nebraska before being sent out to stores. If you picked a computer at random from the Nebraska distribution center, what is the probability that it is a laptop?

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CA OR State WA laptop desktop laptop desktop laptop desktop (.5.15) = (.5.85) = (.3.4) = 0.12 (.3.6) = 0.18 (.2.6) = 0.12 (.2.4) = 0.08 P(laptop) = = 0.315

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