hw2sol_fall03

# hw2sol_fall03 - AMS 310.01 Fall 2003 Homework 2 Solution...

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AMS 310.01 Fall 2003 Homework 2 Solution 3.28 (a) There are 52 possible outcomes. Two of those are red kings. Thus, the probability of red king is 2/52 = 1/26. (b) Again, there are 52 possible outcomes. There are 4 possible 3’s, 4’s, 5’s and 6’s in this event i.e. 16 outcomes. Thus, the probability is 16/52 = 4/13. 3.34 (a) N(A) = 2+8+54+20 = 84 (b) N(B) = 20+54+9+16 = 99 (c) N(C) = 8+54+9+14 = 85 (d) N( A B B ) = 20+54 = 74 (e) N ( A C B ) = 8+54 = 62 (f) N( A B C ) = 54 (g) N( A B B ) = 2+8+20+54+9+16 = 109 (h) N( B C B ) = 16+20+54+9+8+14 = 121 (i) N( A B C ) = 27+2+16+8+54+9+14 = 130 Another way to do this is N( A B C ) = 150-20 = 130. (j) N( ( ) B A C ) = 20+54+9 = 83 3.38 (a) 0.38 and 0.53 do not sum to 1. (b) Probability can not be negative. (c) The probability that the compressor or the fan motor is all right is 0.82+0.64-0.41=1.05 which is greater than 1. 3.42 (a) P(A) = N(A) / 150 = 84/150 = 0.56 (b) P(B) = N(B) / 150 = 99/150 = 0.66

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Unformatted text preview: (c) P( A C B ) = P( ( ) A C B ) = (16+27) / 150 = 43/150 = 0.287 (d) P( B C B ) = (54+9) / 150 = 63/150 = 0.42 3.60 (a) P(both awards) = 0.11 , and P(efficiency award) = 0.24. Thus, P(design award given efficiency award) = 0.11 / 0.24 = 0.458 (b) P( design award and no efficiency award) = 0.16 – 0.11 = 0.05 , and P(no efficiency award) = 1-0.24 = 0.76. Thus, P(design award given no efficiency award) = 0.05/0.76 = 0.066. i.e. Let’s denote D : design award , E : efficiency award P(D) = 0.16 , P(E) = 0.24 & P( D E B ) = 0.11 (a) P(D|E) = P( D E B ) / P(E) = 0.11 / 0.24 = 0.458 (b) P(D| E ) = P( D E B ) / P( E ) = [P(D) – P( D E B ) ] / [1-P(E)] = (0.16-0.11) / (1-0.24) = 0.05 / 0.76 = 0.066 3.66 The rolls are independent. (a) P(two 4’s) = (1/6)(1/6)=1/36 (b) P(a 4 and then a number lesser than 4 ) = (1/6)(3/6)=1/12...
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• Fall '03
• Mendell
• Initialisms, efficiency award

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