EE351K_HW7_solution_FALL2007

# EE351K_HW7_solution_FALL2007 - 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

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## This note was uploaded on 01/20/2010 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas.

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EE351K_HW7_solution_FALL2007 - EE 351K Homework assignment...

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