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Summary of Chapter 5 Probability Models Statistics Section 5.1 A sample space is a complete list or description of possible outcomes of a chance process. The possible outcomes must be disjoint (mutually exclusive), meaning that one outcome cannot be another outcome. An event is a subset of a sample space. For example, if we roll a six-sided dice, the sample space is the numbers 1, 2, 3, 4, 5, and 6 (complete list of possible outcomes). Each element of the sample space is disjoint (rolling a 2 is not the same as rolling a 3). An example of an event for this sample space might be rolling a 2. Another example of an event for this sample space might be rolling an even number (2, 4, or 6). The event rolling a 2 and the event rolling an even number are not disjoint (mutually exclusive) because rolling a 2 is also rolling an even number. However, the event rolling a 2 and the event rolling an odd number are mutually exclusive because rolling a 2 is not rolling an odd number. Probability is a numerical expression of how likely it is for an event to occur. Numerically, it is equal to Number of outcomes of event Number of equally likely outcomes The probability of an event, A, occurring, P(A), is always 0 ≤ P(A) ≤ 1. A probability distribution gives a list a complete set of possible events from a sample space and the probability of each event. Since the list is complete, the sum of all the probabilities is equal to 1. For example, a two-way table for the sums of rolling two four-sided die and the associated probability distribution are shown below: + 1 2 3 4 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 If a sampling is random, the Law of Large Numbers says as the sample size becomes large, the probability of Event A for the sample approaches what the probability of Event A is for the population. For example, if a fair 6 sided dice is tossed 10 times, the proportion of 5’s might be quite a bit higher or lower than 1/6. However, as the number of times the dice is tossed increases to a very large value, the proportion of 5’s would approach 1/6. The total number of possible combinations for a number of separate possibilities (stages) can be computed by the Fundamental Principle of Counting, which says that “if there are k stages, with ni possible outcomes for stage i, then the number of possible outcomes for all k stages taken together is n1n2n3 . . . nk. For example, say for a meal we have a choice of 2 main dishes, 3 drinks, and 2 desserts. Each type of choice (main dish, drink, dessert) would be a stage. The total number of possible combinations is 2 x 3 x 2 = 12. These 12 possible combinations can be shown with a tree diagram: B1 M1 B2 B3 B1 B2 M2 B3 D1 M1B1D1 D2 M1B1D2 D1 M1B2D1 D2 M1B2D2 D1 M1B3D1 D2 M1B3D2 D1 M2B1D1 D2 M2B1D2 D1 M2B2D1 D2 M2B2D2 D1 M2B3D1 D2 M2B3D2 Section 5.2 Simulations A simulation is a procedure where a model of a chance process is used to copy (simulate) a real situation. Simulations are useful to determine the probability of something occurring or, if something has occurred already, to determine how likely that event would be if it was controlled by chance alone. Digits are used to represent situations. The steps involved in running a simulation are: 1.) State the assumptions being made about the real-life situation. 2.) Describe the model to be used to represent the real-life situation. The model should include instructions for how to perform one run. 3.) Repetition. Repeat the model run many times, recording the summary statistic (what you are recording from each run) in a frequency table. 4.) Write a conclusion in the context of the situation. Part of the conclusion should include an estimated probability. For example, if a team has a record of winning only 10% of its games, you could assume that the team’s chance of winning any game at random is 10%. Say you want to determine the probability that this team will win 2 out of every five games played. To do this you could let the digit 0 represent a win and the digits 1 to 9 represent a loss. Using a random digit table or a random digit generator (such as RandInt on the calculator) a run would consist of 5 digits, each digit representing one of the five games. If 2 0’s are present, the run is a “success” (meaning that the team won 2 of the 5 games). If there are not 2 0’s, the run is a “failure” (meaning the team did not win 2 of the 5 games). Repeat this run many times, keeping track of the number of “successes” and “failures”. The simulation runs can be stopped when the proportion value stays essentially constant. Once the runs are completed, you can write a conclusion, which in this case would include your estimated probability of the team winning 2 out of 5 games. Section 5.3 The Addition Rule and Disjoint Events In statistics, the phrase “A or B” can mean A or B or Both. If 2 events are disjoint (mutually exclusive) they cannot occur at the same time. A necessary condition for disjoint events, then, is that P(A and B) = 0 (case for disjoint events) If 2 events are disjoint, the probability of 1 event or the other happening is just the sum of the two probabilities: P(A or B) = P(A) + P(B) (addition rule for disjoint events) (in more general terms, if there are more than 2 events the probability of one of them happening would be the sum of the probabilities for each). For example, in rolling a six sided die, the events rolling a 1, 2, 3, 4, 5, 6 are all mutually exclusive. Each has a probability of 1/6. The probability of rolling a 1, 4, or 5 is (1/6 + 1/6 + 1/6) = 3/6 = ½. If events are not disjoint (for example, rolling a 2 or rolling an even number) when computing the probability for P(A or B) the portions of the two events that overlap will be accounted for twice, so it is necessary to subtract out the portion that is treated twice. The addition rule for events that are not disjoint, then, is: P(A or B) = P(A) + P(B) – P(A and B) (Addition Rule) Note that this rule is very general, as it will also work for disjoint events, because for disjoint events P(A and B) = 0. As presented in a two-way table form we can calculate the P(A or B) in two ways. B Yes No TOTAL Yes a b a+b A No c d c+d TOTAL a + c b + d a + b + c + d Using P(A or B) = P(A) + P(B) – P(A and B) we have: ab ac a abc abcd abcd abcd abcd Using P(A or B) as the sum of three inner cells we have: a b c abc . abcd abcd abcd abcd Section 5.4 Conditional Probability Probabilities that involve the terms “given that” are called conditional probabilities. In conditional probabilities, the probability is potentially affected by some additional restriction or information about the situation. One situation where conditional probabilities are involved is when one deals with draws in a small population without replacement. For example, if we have 3 black socks and 4 white socks, the color of sock we draw for the first draw has an effect on the probability of what is drawn on the second. At the first draw, we have a 3/7 probability of drawing a black sock. If a black sock is drawn (and not replaced) we have a 2/6 = 1/3 probability of drawing a black sock on the second draw. The probability of the second draw is conditioned by what has happened on the first draw. In this situation, we might say “The probability of drawing a black sock on the second draw given that the first draw was black is 2/6”. In probability notation, this is written as P(black 2nd draw | black 1st draw) = 2/6. When you see the symbol “|” think of the words “given that”. Using the two-way table from above, B Yes No TOTAL Yes a b a+b A No c d c+d TOTAL a + c b + d a + b + c + d P(A) = ab a , but P(A | B) = (the probability A given that B is abcd ac yes). Using a tree diagram, such as the titanic example on page 327 of the text, it can be shown that: P(A and B) = P(A) P(B | A) or P(A and B) = P(B) P(A | B) This is called the Multiplication Rule. Rearranging the Multiplication Rule gives the Definition of Conditional Probability: P A and B . P A | B P B Section 5.5 Independent Events Two events are independent if the result of one event does not have an effect on the outcome of the other event. Using the notation for conditional events of the last section (where what happens in one event can have an effect on the outcome of another event) we have the Definition of Independent Events: Assume P(A) > 0 and P(B) > 0. Events A and B are independent events if and only if: P(A | B) = P(A) P(B | A) = P(B) (given that B occurred has no effect on A) or (given that B occurred has no effect on A). For independent events, the multiplication rule, P(A and B) = P(B) P(A | B), is simplified to: P(A and B) = P(B) P(A) because for independent events P(A | B) = P(A). For more than 2 independent events, the multiplication rule becomes: P(A1 and A2 and A3 and . . . and An) = P(A1) P(A1) P(A1) . . . P(An) A type of problem often encountered with independent events is the situation of at least one. This might involve a large number of possible situations, but the only situation it will not involve is that of none. To find the probability for at least one scenario, simply compute its equivalent, 1 – P(none).