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Preparation for Test 1 -solutions - Preparation to test 1...

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Preparation to test 1 Solutions #1 Binomial Distribution A – stick with tour group P(A) = 0,27 B – own experience P(B) = 0,73 a) B – success N=5 P=0,73 F(x≤3) - ? F(x≤3) = 1- F(x>3) = 1 – F(x=4) – F(x=5) = 1 – 0,383 – 0,207 = 0,41 k n k k n p p C k x f - - = = ) 1 ( ) ( )! ( ! ! k n k n C k n - = 5! F(x=4) = 4!1! * 4 ) 73 , 0 ( * 1 ) 27 , 0 ( = 0,383 5! F(x=5) = 5!0! * 5 ) 73 , 0 ( * 0 ) 27 , 0 ( = 0,207 b) N=4 P=0,27 A – success F(x 2) -? F(x 2) = 1-F(x<2) = 1- F(x=0)- f(x=1)= 1-0,283-0,420= 0,297 k n k k n p p C k x f - - = = ) 1 ( ) ( )! ( ! ! k n k n C k n - = 4! F(x=0)= 0! 4! * 0 ) 27 , 0 ( * 4 ) 73 , 0 ( =0,283 4! F(x=1)= 1!3! * 1 ) 27 , 0 ( * 3 ) 73 , 0 ( = 0,42
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#2 Binomial Distribution A – system will detect attack P(A)= 0,85 a) n=4 (four system operates independently) we are asked about probability that at least two of the systems will not detect the attack therefore P=1-0,85 = 0,15 F(x 2) - ? F(x 2) = 1 – f(x<2) = 1- f(x=0) – f(x=1) = 1-0,522-0,368 = 0,11 F(x=0) = k n k k n p p C - - ) 1 ( = 0 4 C 4 0 ) 85 , 0 ( ) 15 , 0 ( = ! 4 ! 0 ! 4 4 ) 85 , 0 ( 1 =0,522 F(x=1) = k n k k n p p C - - ) 1 ( = 1 4 C 3 1 ) 85 , 0 ( ) 15 , 0 ( = ! 3 ! 1 ! 4 3 ) 85 , 0 ( 15 , 0 = 0,368 b) n = 5 (because it is written that five systems installed) since we are asked about probability that at most three of the systems will detect the attack P=0,85
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