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Exam1_Answers_05a

Exam1_Answers_05a - 0.6^4(4 choose 3(0.4^2(0.6)^3(4 choose...

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ESE 301 Engineering Probability First Exam Answers, February 17, 2005 1. n is at least 250 but this uses Chebyshev’s Inequality which we haven’t come to yet. 2. 171 3. From Bayes Theorem, the desired probability is (.99*.005)/(.99*.005 + .01*.995) = 0.332. 4. Binomial summation of sum(k=0 to 3)(1000 choose k) 0.01^k(0.99)^(1000-k); (.99^4)*(.01). 5. (2 choose 1)*(3 choose 1)*(5 choose 1)*(2 choose 1)/(12 choose 4). 24/(12 choose 4) 6. 2/3 7. Requires an analytical derivation. 8. Desired probability is P(win in 4 )+ P(win in 5) + P(win in 6 )+ P(win in 7) =
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Unformatted text preview: 0.6^4 + (4 choose 3 )*(0.4^2)*(0.6)^3 + (4 choose 2 )*(0.4^4)*(0.6)^2 + (4 choose 3 )*(0.4^2)*(0.6)^4 + (4 choose 1 )*(0.4^6)*(0.6)^1 + 2*(4 choose 2 ) *(0.4^4)*(0.6)^3+ (4 choose 3 )*(0.4^2)*(0.6)^5 9. Desired probability = (.1*.45)+(.12*.35)+(.15*.2) . 10. Because (12 choose 1)* (11 choose 1)* (10 choose 1) orders the denominations of the 3 single cards. Since the denominator does not consider order we have to be consistent and not order the numerator and so it should be (12 choose 3)....
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