hw7_418_10 - 48. Let X be the number of defective disks in...

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48. Let X be the number of defective disks in one package. X B (10 , 0 . 01). Then P ( X 1) = 0 . 99 10 + ( 10 1 ) 0 . 01(1 - 0 . 01) 9 = 0 . 09135. i.e. p ( X > 1) = 0 . 004248 . Let Y denotes the number of packages returned out of 3. Then Y B (3 , 0 . 004248). Then p ( Y = 1) = ( 3 1 ) 0 . 004248(1 - 0 . 004248) 2 = 0 . 01264 . 49. Let X1 denotes the number of times coin 1 lands on heads. X2 denotes the number of times coin 2 lands on heads. a) X 1 B (10 , 0 . 4), X 2 B (10 , 0 . 7). p ( X 1 = 7) = ( 10 7 ) 0 . 4 7 (1 - 0 . 4) 0 . 3 = 0 . 4247 , p ( X 2 = 7) = ( 10 7 ) 0 . 7 7 (1 - 0 . 7) 0 . 3 = 0 . 26683 . p(Coin lands on head exactly 7 of the 10 times)=p(X1=7)p(coin 1)+p(X2=7)p(coin 2)=0.0424(1/2)+0.26683*(1/2)=0.15465. b) A=coin 1 is selected. B=heads results, A C =coin 2 is selected. p ( A | B ) = p ( B | A ) p ( A ) p ( B | A ) p ( A )+ p ( B | A C ) p ( A C ) = 0 . 4 * (1 / 2) 0 . 4 * 1 / 2+0 . 7 * 1 / 2 . Given that the first flip lands on heads, there are 9 tosses left and exactly 6 more need to land on heads. p
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This note was uploaded on 01/23/2011 for the course STAT 418 at Pennsylvania State University, University Park.

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hw7_418_10 - 48. Let X be the number of defective disks in...

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