Suppose that joey arrived late once again this

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Suppose that Joey arrived late once again this morning and had to park in a no-parking zone. Can you find the probability that Joey will get a parking ticket? If so, do it. If you need additional information to find the probability, explain what is needed Problem 12: Two cards are dealt, one after the other, from a shuffled 52-card deck. Why is it wrong to say that the probability of getting two red cards is (1/2)(1/2) = 1/4? What is the correct probability of this event?
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Chapter 6: Probability and Simulation: The Study of Randomness Probability Review Disjoint or Mutually Exclusive Events do not have any outcomes in common. For example, if A is odd numbers on a six-sided dice and B is even numbers on a six-sided dice, then if we roll a die we can not have both event A and event B occur at the same time P(A and B) = 0 (no intersection in Venn). So the addition rule for disjoint events gives us P(A or B) = P(A) + P(B) Probability Rules for a Probabilistic Model: 1) Sum of all P(Events) = 1 2) All probabilities must be 0 < P(Events) < 1 3) P(Event) + P(Event’s Compliment) = 1 4) P(certainty) = 1 and P(impossibility) = 0 5) P(unusual event) < 0.05 or 5% number of ways E can occur Classical Method of Probability: P(E) = -------------------------------------------- total possible ways observations of F Empirical Method of Probability: P(F) = ------------------------------------------ total observations A B If events are not mutually exclusive or are not disjoint, then they have to have some part in common. As an example C is numbers < 5 on our six-sided die and D is numbers > 2. This common part would be double counted in the addition rule above; so the general addition rule takes this in account: P(C or D) = P(C) + P(D) – P(C and D) C D 2, 4, 6 1, 3, 5 3, 4 1, 2 5, 6 P(A) = 3/6 = 0.5 and P(B) = 3/6 = 0.5 so P(A or B) = P(A) + P(B) = 0.5 + 0.5 = 1 (not very common, but A and B are compliments!) P(C) = 4/6 = 0.67 , P(D) = 4/6 = 0.67 and P(C and D) = 2/6 = 0.33 so P(C or D) = P(C) + P(D) - P(C and D) = 0.67 + 0.67 – 0.33 = 1 (note: if we don’t remove the double counting then P(C or D) > 1!) Computing of At-Least Probabilities: P( at least one) = 1 – P(compliment of at least one) = 1 – P(none) At least one: x ≥ 1 so the complement is x = 0!
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Chapter 6: Probability and Simulation: The Study of Randomness Probability Review Two events are independent if the probability of them occurring is unaffected by the other occurring. Simplest example would be rolling two consecutive sixes on a die. The probability of rolling the second six is not dependent on rolling the first six. The multiplication rule for independent events is given by: P(E and F) = P(E) • P(F) (if E and F are independent!) P(rolling 2 consecutive sixes) = P(rolling 6) • P(rolling 6) = 1/6 • 1/6 = 1/36 = 0.0278 Two events are dependent if the probability of them occurring is affected by the other occurring. For example, if I drew from a deck of cards an Ace, then probability drawing another Ace from the deck (without replacing the first card) would be affected by drawing the first Ace.
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