EE351K_HW7_solution_Fall2009

EE351K_HW7_solution_Fall2009 - EE 351K Homework assignment...

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EE 351K Homework assignment No.7 Problem 1. Binomial distribution ! ( ) ( ) 0,1,2, , !( )! kn k X n Pk PX k pq f o r k n knk !! ! ! In this problem n = 5, so k = 0,1,2,3,4,5 (a) P H = 0.6, P T = 0.4 35 3 32 5! (5)(4) ( 3) (0.6) (0.4) (0.6) (0.4) (10)(0.216)(0.16) 3!(5 3)! (1)(2) X Pk ! ! =0.3456 (b) Probability of exactly two tails = Probability of exactly three heads = 0.3456 (c) Probability of 1,2, or 3 heads = (1 ) (2 ) (3 XX X ) !" ! " ! 15 1 4 5! 5! ( 1) (0.6) (0.4) (0.6)(0.4) 5(0.6)(0.0256) 0.0768 1!(5 1)! 4! X ! ! ! 25 2 23 5! (3)(4)(5) ( 2) (0.6) (0.4) (0.6) (0.4) 10(0.36)(0.064) 0.2304 2!(5 2)! (1)(2)(3) X ! ! ! X PX ! ) ) =0.3456 from (a) ) ) X ! =0.0768+0.2304+0.3456=0.6528 (d) Probability of at least 1 tail = P(1T)+ P(2T)+ P(3T)+ P(4T) +P(5T)=1-P(0T) 55 5 5 5! (0 ) (5 ) (0.6) (0.4) (1)(0.6) (1) 0.07776 5!(5 5)! PT PH ! ! Probability of at least 1T = 1-0.07776 = 0.92224 (e) Probability of less than 3H = P(0H)+ P(1H)+ P(2H)=P(0H)+ 0.0768+0.2304 (from part (c)) 05 0 5! (0 ) (0.6) (0.4) (1)(1)(0.01024) 0.01024 0!(5 0)! PH ! Probability of less than 3H = 0.01024+0.0768+0.2304 = 0.31744 1/3
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Problem 2. Binomial distribution problem n=10, p=0.4, q=0.6 (a) E=np=10(0.4)=4 hits (b) V=npq=(10)(0.4)(0.6)=2.4 != 2.4 V ! =1.549 (c) k=4 46 10! (7)(8)(9)(10) ( 4) (0.4) (0.6) (0.4) (0.6) (10 4)!4! (1)(2)(3)(4) X Pk !! ! (210)(0.0256)(0.046656) ! =0.2508 (d) k=0 01 0 10! ( 0) (0.4) (0.6) (0.6) 0)!0! X ! 10 =0.00605 (e) k=10 10 0 10 10! ( 10) (0.4) (0.6) (0.4) 0!10! X ! =0.00010 (f) P(k=1,2,3, ˎ ,10)=1-P(k=0)=1-0.00605=0.99395 (g) P(k=2,3, ,10)=1-P(k=0)- P(k=1) 19 10! ( 1) (0.4) (0.6) (10)(0.4)(0.01008) 0.04031 9!1! X ! ! P(k=at least 2)=1-0.00605-0.04031=0.95364
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EE351K_HW7_solution_Fall2009 - EE 351K Homework assignment...

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