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Chapter 6: Probability : The Study of Randomness “We figured the odds as best we could, and then we rolled the dice.” US President Jimmy Carter June 10, 1976

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6.1 Randomness (p. 310-316) Random phenomenon An outcome that cannot be predicted Has a regular distribution over many repetitions Probability Proportion of times that an event occurs in many repeated events of a random phenomenon Independent events (trials) Outcome of an event (trial) does not influence the outcome of any other event (trial)

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Examples: Independent event Rolling a single die twice Result on second roll is independent of the first Event that is not Independent Picking two cards from a deck (one at a time with no replacement) If you pick a card and do not replace it and reshuffle the deck, then the probability of a red card on the second pick is dependent upon what the first card happened to be.

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6.2 Probability Models (P. 317-337) It is often important and necessary to provide a mathematical description or model for randomness. Sample space The set of all possible outcomes of a random phenomenon

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Example of a Sample Space Consider the SUM obtained when two dice are rolled. Create a table to display the sums that can be obtained for this event. The sample space contains _______ outcomes.

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Probability Rules If A is an event, then the probability P(A) is a number between o and 1, inclusive. P(A does not happen) = 1 – P(A) The complement of A Two events are DISJOINT of they have no outcomes in common. If A and B are disjoint, then P(A or B) = P(A) + P(B) Sometimes the phrase MUTUALLY EXCLUSIVE is used to describe disjoint events

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Probability Rules: An Example In a roll of two dice Suppose A = rolling a sum of 7 and B = rolling a sum of 12, the P(A or B) = 6/36 + 1/36 = 7/36 (A and B are disjoint). Suppose C = rolling a sum of 7 and D = rolling an odd sum, the P(C or D) = P(C) + P(D) – P(C and D) = 6/36 + 18/36 – 6/36 = 18/36 = ½ (C and D are NOT DISJOINT since 7 is an odd number)

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Probability Rules Two events are INDEPENDENT if knowing that one occurs does not change the probability of the other. If events A and B are independent, then P(A and B) = P(A) X P(B)

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Examples: Suppose A = getting a head on a first toss and B = getting a head on a second toss. A and B are independent. P(A and B) = (1/2) X (1/2) = ¼ Suppose C = getting a red card by picking a card from a randomly shuffled deck of cards and D = getting a red card by picking a second card from the deck in which the first card WAS NOT REPLACED. C and D are NOT independent. P(C) = 26/52. If the first card is red, then P(C and D) = (26/52) X (25/51). If the first card had been placed back into the deck then P(C and D) = (26/52) X (26/52) because C and D ARE independent.

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Probability formulas are useful, however, sometimes they are not needed if sample spaces are small and you use some common sense. Consider a family of two children. There are four possible boy/girl combinations, ALL EQUALLY LIKELY. P(at least one child is a girl) = P(two girls) = P(two girls| one child is a girl) = P(two girls| oldest child is a girl) = P(exactly one boy | one child is a girl) = P(exactly one boy | youngest child is a girl) =

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Now suppose that a family unknown to you has two children. If one of the children is a boy, what is the probability that the other child is a boy? If the oldest child is a boy, what is the probability that the other child is a boy? Many intelligent people think that 50% is the answer to both of the above. You should be able to show that this is NOT the case.

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More about Probability (P. 341-355) There are basic laws that govern wise and efficient use of probability. FOR ANY TWO EVENTS A and B, P(A or B) = P(A) + P(B) – P(A and B) P(A and B) = 0 if A and B are disjoint P(B|A) = P(A and B)/P(A)

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Examples Using the Sum of Two Dice P(sum = 7 or sum = 11) = P(sum = 7 and sum = 11) = P(sum = 7 or at least one die shows a 5) = P(faces show same number or sum>9) = P(sum = 6 | one die show a 4) = P(sum is even | sum>9) =

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