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Nossi ch 10
 

Nossi ch 10

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Contemporary Math ch 10

Contemporary Math ch 10

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    Nossi ch 10 Nossi ch 10 Presentation Transcript

    • Chapter 10 Probability
    • Section 10.1 Simple Experiments
      • Goals
        • Study probability
          • Experimental probability
          • Theoretical probability
        • Study probability properties
          • Mutually exclusive events
          • Unions and intersections of events
          • Complements of events
    • Interpreting Probability
      • Probability is the mathematics of chance.
      • For example, the statement “The chances of winning the lottery game are 1 in 150,000” means that only 1 of every 150,000 lottery tickets printed is a winning ticket.
    • Probability Terminology
      • Making an observation or taking a measurement is called an experiment .
      • An outcome is one of the possible results of an experiment.
      • The set of all possible outcomes is called the sample space .
      • An event is any collection of possible outcomes.
    • Example 1
      • An experiment consists of rolling a standard six-sided die and recording the number of dots showing on the top face.
        • List the sample space.
        • List one possible event.
    • Example 1
      • The sample space contains 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
      • One possible event: The event of rolling an even number: {2,4,6}
    • Example 2
      • An experiment consists of tossing a coin 3 times and recording the results in order.
        • List the sample space.
        • List one possible event.
    • Example 2, cont’d
      • Solution: The sample space contains 8 possible outcomes and can be written {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
      • One possible event is {HTH, HTT, TTH, TTT}, which is the event of getting a tail on the second coin toss.
    • Example 3
      • An experiment consists of spinning a spinner twice and recording the colors it lands on.
        • List the sample space.
        • List one possible event.
    • Example 3
      • Solution: The sample space contains 16 possible outcomes and can be written {RR, RY, RG, RB, YR, YY, YG, YB, GR, GY, GG, GB, BR, BY, BG, BB}.
      • One possible event is {RR, YY, GG, BB}, which is the event of getting the same color on both spins.
    • Example 4
      • The experiment consists of rolling 2 standard dice and recording the number appearing on each die.
        • List the sample space.
        • List one possible event.
    • Example 4
      • Solution: The sample space contains 36 possible outcomes and can be written {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.
    • Example 4
      • Solution, cont’d: One possible event is {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)}, which is the event of getting a total of 7 dots on the two dice.
    • Probability, cont’d
      • The probability of an event is a number from 0 to 1, and can be written as a fraction, decimal, or percent.
        • The greater the probability, the more likely the event is to occur.
        • An impossible event has probability 0.
        • A certain event has probability 1.
    • Experimental Probability
      • One way to find the probability of an event is to conduct a series of experiments.
    • Example 5
      • An experiment consisted of tossing 2 coins 500 times and recording the results
      • Let E be the event of getting a head on the first coin Find the experimental probability of E .
    • Example 5
      • The even of getting a head on the first coin:
      • {HH, HT}.
      • occurred a total of
      • 137 + 115 = 252 times out of 500.
        • The experimental probability of E is
    • Theoretical Probability
      • Another way to find the probability of an event is to use the theory of what “should” happen rather than conducting experiments.
    • Example 6
      • An experiment consists of tossing 2 fair coins.
      • Find the theoretical probability of:
        • Each outcome in the sample space.
        • The event E of getting a head on the first coin.
        • The event of getting at least one head.
    • Example 6, cont’d
      • Solution:
        • There are 4 outcomes in the sample space: {HH, HT, TH, TT}. Each outcome is equally likely to occur.
    • Example 6
      • The event E is {HH, HT} and the theoretical probability of E is the number of outcomes in E divided by the number of outcomes in the sample space.
    • Example 6
      • c) The event of getting at least one head is E = {HH, HT, TH}.
    • Example 7
      • An experiment consists of rolling 2 fair dice.
      • Find the theoretical probability of:
        • Event A : getting 7 dots.
        • Event B : getting 8 dots.
        • Event C : getting at least 4 dots.
    • Example 7
      • Solution: There are 36 outcomes in the sample space. (see pg 635 for a list of the sample space.)
        • Event A : getting 7 dots contains 6 outcomes:
        • {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}.
        • Each outcome is equally likely to occur.
    • Example 7
      • Solution, cont’d:
        • Event B : getting 8 dots contains 5 equally likely outcomes:
        • {(2,6), (3,5), (4,4), (5,3), (6,2)}.
    • Example 7
      • Solution, cont’d:
        • Event C : getting at least 4 dots (or a total of 4 or more) contains 33 equally likely outcomes:
        • { (1,3), (1,4), (1,5), (1,6), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }.
    • Example 8
      • A jar contains four marbles:
      • 1 red, 1 green, 1 yellow, and 1 white.
    • Example 8
      • If we draw 2 marbles in a row, without replacing the first one, find the probability of:
        • Event A : One of the marbles is red.
        • Event B : The first marble is red or yellow.
        • Event C : The marbles are the same color.
        • Event D : The first marble is not white.
        • Event E : Neither marble is blue.
    • Example 8
      • If we draw 2 marbles in a row, without replacing the first one, find the probability of:
      • Event A : One of the marbles is red.
        • A = {RG, RY, RW, GR, YR, WR}.
    • Example 8
      • If we draw 2 marbles in a row, without replacing the first one, find the probability of:
        • Event B : The first marble is red or yellow.
          • B = {RG, RY, RW, YR, YG, YW}.
        • Event C : The marbles are the same color.
    • Example 8
      • If we draw 2 marbles in a row, without replacing the first one, find the probability of:
        • Event D : The first marble is not white.
        • D = {RG, RY, RW, GR, GY, GW, YR, YG, YW}.
          • e) Event E : Neither marble is blue.
    • Union and Intersection
      • The union of two events, A U B , refers to all outcomes that are in one, the other, or both events.
      • The intersection of two events, A ∩ B, refers to outcomes that are in both events.
    • Mutually Exclusive Events
      • Events that have no outcomes in common are said to be mutually exclusive .
      • If A and B are mutually exclusive events, then
    • Example 9
      • A card is drawn from a standard deck of cards.
        • Let A be the event the card is a face card.
        • Let B be the event the card is a black 5.
      • Find and interpret P( A U B ).
    • Example 9
      • Solution: The sample space contains 52 equally likely outcomes. (this chart is on pg 639)
    • Example 9
      • Event A has 12 outcomes, one for each of the 3 face cards in each of the 4 suits.
        • P( A ) = 12/52.
      • Event B has 2 outcomes, because there are 2 black fives.
        • P( B ) = 2/52.
      • Events A and B are mutually exclusive because it is impossible for a 5 to be a face card.
        • P( A U B ) = 12/52 + 2/52 = 14/52 = 7/26.
        • This is the probability of drawing either a face card or a black 5.
    • Complement of an Event
      • The set of outcomes in a sample space S , but not in an event E , is called the complement of the event E .
        • The complement of E is written Ē .
        • For example: The weather man says there is a 45% chance of rain. The complement of this event is the chance that it will not rain which is 55%.
    • Complement of an Event, cont’d
      • The relationship between the probability of an event E and the probability of its complement Ē is given by:
    • Example 10
      • In a number matching game,
        • First Carolan chooses a whole number from 1 to 4.
        • Then Mary guesses a number from 1 to 4.
      • What is the probability the numbers are equal?
      • What is the probability the numbers are unequal?
    • Example 10
      • Solution: The sample space contains 16 outcomes: { (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4) }.
      • Let E be the event the numbers are equal.
        • P( E ) = ¼
      • Then Ē (complement of E) is the event the numbers are unequal.
        • P( Ē ) = 1 – ¼ = ¾
    • Example 12
      • An experiment consists of spinning the spinner once and recording the number on which it lands.
    • Example 12
      • Define 4 events:
        • A: an even number
        • B: a number greater than 5
        • C: a number less than 3
        • D: a number other than 2
      • Find P( A ), P( B ), P( C ), and P( D ).
      • Find and interpret P( A U B ) and P( A ∩ B ).
      • Find and interpret P( B U C ) and P( B ∩ C ).
    • Example 12
      • Solution: The sample space has 8 equally likely outcomes: {1, 2, 3, 4, 5, 6, 7, 8}.
      • Find P( A ),P( B ), P( C ), and P( D ).
        • A = {2, 4, 6, 8}, so P( A ) = 4/8 = 1/2
        • B = {6, 7, 8}, so P( B ) = 3/8
        • C = {1, 2}, so P( C ) = 2/8 = 1/4
        • D = {1, 3, 4, 5, 6, 7, 8}, so P( D ) = 7/8.
    • Example12
      • Solution, cont’d:
      • Find and interpret P( A U B ) and
      • P( A ∩ B ).
        • A and B are not mutually exclusive, so
        • A ∩ B = {6, 8}, so P( A ∩ B ) = 2/8
    • Example 12
      • Solution, cont’d:
      • Find and interpret P( B U C ) and
      • P( B ∩ C ).
        • B and C are mutually exclusive, so
        • and
    • Section 10.2 Multistage Experiments
      • Goals
        • Study tree diagrams
          • Study the fundamental counting principle
        • Study probability tree diagrams
          • Study the additive property
          • Study the multiplicative property
    • 10.2 Initial Problem
      • A friend who likes to gamble wagers that if you toss a coin repeatedly you will get 2 tails before you get 3 heads.
      • Should you take the bet?
        • The solution will be given at the end of the section.
    • Example 1
      • A tree diagram is drawn for the experiment of drawing 1 ball from a box containing 1 red ball, 1 white ball, and 1 blue ball.
    • Example 2
      • There are 12 possible outcomes.
      • 12 is the product of 4 branches times 3 branches.
    • Fundamental Counting Principle
      • The number of outcomes in an experiment can also be determined using the fundamental counting principle :
        • If an event or action A can occur in r ways, and , for each of these r ways, an event or action B can occur in s ways, then the number of ways events or actions A and B can occur, in succession, is r times s .
        • The principle can be extended to more than two events or actions.
    • Example 3
      • The options on a pizza are:
        • Small, medium, or large
        • White or wheat crust
        • Sausage, pepperoni, bacon, onion, mushrooms
      • How many different one-topping pizzas are possible?
    • Example 3
      • Solution:
        • The first action is choosing 1 of 3 sizes.
        • The second action is choosing 1 of 2 crusts.
        • The third action is choosing 1 of 5 toppings.
      • There are 3(2)(5) = 30 different one-topping pizzas possible.
    • Example 4
      • Find the probability of getting a sum of 11 when tossing a pair of fair dice.
    • Example 4
      • Solution: There are 36 equally likely outcomes in the sample space.
        • 6 possible outcomes on the first roll
        • 6 possible outcomes on the second roll
        • 6(6) = 36
      • There are 2 ways of rolling a sum of 11: (5,6) and (6,5).
        • The probability is 2/36 = 1/18.
    • Probability Tree Diagrams
      • Tree diagrams can also be used to determine probabilities in multistage experiments.
      • Tree diagrams that are labeled with the probabilities of events are called probability tree diagrams .
    • Probability Tree Diagrams
    • Example 6
      • Draw a probability tree diagram to represent the experiment of drawing one ball from a container holding 2 red balls and 3 white balls.
    • Example 6, cont’d
      • Solution: The first tree has one branch for each ball.
      • The second tree was simplified by combining branches.
    • Example 7
      • Draw a probability tree diagram to represent the experiment of spinning the spinner once.
      • Find the probability of landing on white or on green.
    • Example 7
      • Solution: There are 4 outcomes in the sample space.
        • They are not all equally likely.
        • They are all mutually exclusive. (no overlapping)
        • Use the central angles to find the probability of each outcome and draw the probability tree diagram.
    • Example 7
      • The probability of white or green is:
    • Example 8
      • A jar contains 3 marbles, 2 black and 1 red.
      • A marble is draw and replaced, and then a second marble is drawn. What is the probability both marbles are black?
    • Example 8
      • Solution: Draw a probability tree diagram to represent the experiment.
    • Example 8
      • Assign a probability to the end of each secondary branch. (multiply fractions as you move down branches, add factions as you move across the end of branches.)
    • Example 8
      • Either tree diagram can be used to find that P( BB ) = 4/9.
    • Example 9
      • A jar contains 3 red balls and 2 green balls.
      • First a coin is tossed.
        • If the coin lands heads, a red ball is added to the jar.
        • If the coin lands tails, a green ball is added to the jar.
      • Second a ball is selected from the jar.
      • What is the probability a red ball is chosen?
    • Example 9
      • Solution: A probability tree diagram is created.
    • Example 9
      • The probability of choosing a red ball is found by adding the probabilities at the end of the branches labeled R .
    • Example 10
      • A jar contains 3 marbles, 2 black and 1 red.
      • A marble is drawn and not replaced before a second marble is drawn.
      • What is the probability that both marbles were black?
    • Example 10
      • Solution: Create a probability tree diagram to represent the experiment.
    • Example 10
      • The probability of choosing 2 black marbles is:
    • Example 11
      • A jar contains 2 red gumballs and 2 green gumballs.
      • An experiment consists of drawing gumballs one at a time from a jar until a red one is chosen.
      • Find the probability of:
        • A: only 1 draw is needed
        • B: exactly 2 draws are needed
        • C: exactly 3 draws are needed
    • Example 11
      • Solution: Create a probability tree diagram.
      • The probabilities are:
    • 10.2 Initial Problem Solution
      • A friend who likes to gamble wagers that if you toss a coin repeatedly you will get 2 tails before you get 3 heads. Should you take the bet?
      • You can figure out what you should do by creating a probability tree diagram.
    • Initial Problem Solution, cont’d
    • Initial Problem Solution
      • Add the probabilities of the outcomes that result in a win for you .
        • You are much more likely to lose than to win, so you probably should not take the bet.
    • Section 10.3 Conditional Probability, Expected Value, and Odds
      • Goals
        • Study conditional probability
          • Study independent events
        • Study odds
    • 10.3 Initial Problem
      • If you bet $100 on one number on the roulette wheel, what is your expected gain or loss?
        • The solution will be given at the end of the section.
    • Conditional Probability
      • In the experiment of tossing 3 fair coins suppose you know the first coin came up heads.
        • The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
        • Then the conditional sample space is {HHH, HHT, HTH, HTT} because the first coin was Head.
    • Conditional Probability
      • The probability of A given B is called a conditional probability .
      • The probability of A given B means the probability of event A occurring within the conditional sample space of event B .
        • The probability of A given B is written P( A | B ),
    • Conditional Probability
      • Suppose A and B are events in a sample space S and that the probability of B is not zero.
      • The formula for conditional probability is
    • Example 1
      • One jar of marbles contains 2 white marbles and 1 black marble.
      • Another jar of marbles contains 1 white marble and 2 black marbles.
    • Example 1
      • A coin is tossed.
        • Heads: a marble is selected from the first jar.
        • Tails: a marble is selected from the second jar.
      • Find the probability that the coin landed heads up, given that a black marble was drawn.
    • Example 1
      • A coin is tossed.
        • Heads: a marble is selected from the first jar.
        • Tails: a marble is selected from the second jar.
      • Find the probability that the coin landed heads up, given that a black marble was drawn.
      • Solution: Create a probability tree diagram.
        • The sample space is {HW, HB, TW, TB}.
        • The probabilities are labeled on the diagram.
    • Example 1
      • Solution, cont’d:
    • Example 2
      • Suppose a test for a viral infection is not 100% accurate.
        • Of the population, ¼ is infected and ¾ is not.
        • Of those infected, 90% test positive.
        • Of those not infected, 80% test negative.
      • What is the probability the test is correct?
      • Given that a person’s test is positive, what is the probability the person is infected?
    • Example 2
      • Solution: Create a probability tree diagram.
        • Of the population, ¼ is infected and ¾ is not.
        • Of those infected, 90% test positive.
        • Of those not infected, 80% test negative.
    • Example 2
      • The probability of a correct test is the probability of a positive test for an infected person or a negative test for an uninfected person.
    • Example 2
      • Given that a person’s test is positive, what is the probability the person is infected? Find the conditional probability:
        • P(positive) =
    • Example 2
      • Solution, cont’d:
        • P(infected and positive) =
        • P(infected|positive)
        • =
    • Example 3
      • Results from an inspection of a candy company’s 2 production lines are shown in the table.
      • If a customer find a sub-standard piece of candy, what is the probability it came from the Bay City factory?
    • Example 3
      • Solution: Use the numbers in the table to solve:
        • P(Bay City and sub-standard) =
    • Example 3
      • Solution, cont’d:
        • P(sub-standard) =
    • Example 3
      • Solution, cont’d:
        • P(Bay City|sub-standard) =
    • Independent Events
      • Two events are called independent if one event does not influence the other.
      • When 2 events are independent, their probabilities follow the rule given below:
    • Example 5
      • A student’s name is chosen at random from the college enrollment list and the student is interviewed.
        • Let A be the event the student regularly eats breakfast.
        • Let B be the event the student has a 10:00 AM class.
      • Explain in words what is meant by:
    • Example 5,
      • Solution:
        • : the probability the student regularly eats breakfast and has a 10:00 AM class.
        • : the probability the student regularly eats breakfast given that the student has a 10:00 AM class
        • : the probability the student does not regularly eat breakfast.
    • Expected Value
      • The average numerical outcome for many repetitions of an experiment is called the expected value .
        • If E = 0, the game is said to be fair .
    • Example 6
      • An experiment consists of rolling a fair die and noting the number on top of the die.
      • Compute the expected value of one roll of the die.
    • Example 6
      • Solution: The calculations are shown below: Multiply 1(1/6) and 2(1/6) and 3(1/6) etc.
    • Example 7
      • How much should an insurance company charge as its average premium in order to break even?
    • Example 7
      • Solution: The calculations are shown above. The product row is the product of the 2 numbers directly above each value in the product row.
        • The average premium should cost $760.
    • Odds
      • The odds in favor of an event compare the number of favorable outcomes to the number of unfavorable outcomes.
        • If the odds in favor are a : b , then
    • Odds
      • The odds against an event compare the number of unfavorable outcomes to the number of favorable outcomes.
        • The odds in favor of an event E are
        • The odds against an event E are
    • Example 8
      • Suppose a card is randomly drawn from a standard deck.
      • What are the odds in favor of drawing a face card?
    • Example 8
      • Solution: There are 12 face cards in the deck, and there are 40 other cards.
      • The odds in favor are 12:40, which simplifies to 3:10.
    • Example 9
      • Find P( E ) given the following odds:
        • The odds in favor of E are 3:7.
        • The odds against E are 5:13.
    • Example 9, cont’d
      • Solution:
        • The odds in favor of E are 3:7.
        • The odds against E are 5:13.
    • Example 10
      • Find the odds in favor of event E , given the following probabilities:
        • P( E ) = 1/4.
        • P( E ) = 3/5.
    • Example 10, cont’d
      • Solution:
        • The odds in favor of E are 1:3.
        • The odds in favor of E are 3:2.
    • 10.3 Initial Problem Solution
      • A roulette wheel has 38 slots numbered 00, 0 and 1 through 36. You place a bet on a number or combination of numbers. If you bet on the winning number, you win your bet plus 35 times your bet. If you lose, you lose the money you bet.
      • If you bet $100 on one number, what is your expected gain or loss?
    • Initial Problem Solution
      • Solution:
        • The probability of winning is 1/38 and the amount won would be $3500.
        • The probability of losing is 37/38 and the amount lost would be $100.
    • Initial Problem Solution
      • The expected value is approximately -$5.26. You should expect to lose this amount, on average, for every $100 you bet.
    • Chapter 10 Assignments
      • Section 10.1 pg 645 (7,11,15,17,31,35)
      • Section 10.2 pg 664 (1,3,11,13,19,25)
      • Section 10.3 pg 686 (5a,6a,15,27,3339,41)