If he does not enter he is certain to break even but

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Unformatted text preview: spades. The probabilities come from the table, because the two game outcomes ARE statistically independent. P(G1 ­0 ∩ G2 ­0) = P(G1 ­0) P(G2 ­0) = (0.370) (0.370) = 0.137 P(G1 ­1 ∩ G2 ­0) = P(G1 ­1) P(G2 ­0) = (0.427) (0.370) = 0.158 P(G1 ­0 ∩ G2 ­1) = P(G1 ­0) P(G2 ­1) = (0.370) (0.427) = 0.158 These are mutually exclusive, so we can add them. SUM = 0.453 That’s P(LoseMoney) P(BreakEvenOrWin) = 1 – P(LoseMoney) = 0.546 Note that this is NOT P(BreakEvenOrWin)2. Why? You could break even overall by making up a loss in one game with a win in the other game. This is looking forward to when we combine distributions… 3/7 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 5 Expected Values, More Distributions The student will enter twice if he rolls a five or six on the dice. If he does not enter, he is certain to break even (but of course he will not win money). Two events are possible with the die. P(1234) = P(NotPlay) = 0.667; P(56) = P(Play) = 0.333. P(BreakEvenOrWin) = P(BreakEvenOrWin | Play) P(Play) + P(BreakEvenOrWin | NotPlay) P(NotPlay) = 0.546 (0.333) + 1 (0.667) = 0.848 This student’s pocketbook is not in danger. 2) For a discrete probability distribution, show that E[(X ­µ)2] = (Σ x2 P(x)) – µ2 E[(X ­µ)2] = Σ(x ­µ)2 P(x) = Σ (x2  ­2µx + µ2 )P(x) = Σ x2 P(x)  ­ Σ 2µx P(x) + Σ µ2 P(x) = Σ x2 P(x)  ­ 2µ Σ x P(x) + µ2 Σ P(x) Summation in middle term is E(X) = µ, and summation in last te...
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