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Unformatted text preview: the total probability, or by applying Bayes’ theorem to these two events. Instead of using (check the underlined part) P(D) = P(D | M ∩~W) P(M ∩~W) + P(D |~M ∩W) P(~M ∩W) + ... P(D |M ∩W) P(M ∩W) + P(D |~M ∩ ~W) P(~M ∩ ~W) Many students used only: P(D) = P(D | M ∩~W) P(M) + P(D |~M ∩W) P(W) + P(D |M ∩W) P(M ∩W) + ... P(D |~M ∩ ~W) P(~M ∩ ~W) 5/9 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) € Homework 3 Bayes’ Theorem and Probability Dist. 5) Three balls are drawn (without replacement) from a box containing 6 black balls and 3 green balls. Sketch the probability distribution for the number of green balls drawn. (Note: “without replacement” means that after you draw the first ball, you don’t put it back) Ans: 6 black balls and 3 green balls Method A: ȹ 9ȹ N = ȹ ȹ = 84 ȹ 3Ⱥ i) 3 greens = 1 way Probability (3 greens) = 1/84 = 0.012 ȹ 3ȹ ii) 2 greens = ȹ ȹ = 3; ȹ 2Ⱥ The other ball (black) can be filled in 6 ways; 3x6 = 18; Probability (2 greens and 1 black) = 18/84 = 0.214 iii) 1 green = 3 ways; € ȹ 6ȹ The other 2 balls (black) can be filled in ȹ ȹ ways = 15; ȹ 2Ⱥ...
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