Exam1_Answers_05a

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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) =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

This note was uploaded on 03/02/2011 for the course ESE 301 taught by Professor Keenan during the Fall '10 term at UPenn.

Ask a homework question - tutors are online