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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) = 10.24 = 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(DE) = 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 ) ] / [1P(E)] = (0.160.11) / (10.24) = 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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This homework help was uploaded on 01/31/2008 for the course AMS 310.01 taught by Professor Mendell during the Fall '03 term at SUNY Stony Brook.
 Fall '03
 Mendell

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