EE351K_HW7_solution_FALL2007

EE351K_HW7_solution_FALL2007 - EE 351K Homework assignment...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 3

EE351K_HW7_solution_FALL2007 - EE 351K Homework assignment...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online