This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EL630: Homework4 Solutions 1) {First 2 appears at n th roll} = } 2 , 2 , . . . . , 2 , 2 , 2 { . That is, we get 1 n rolls in which 2 does not appear, and then the 2 appears on the n th roll. Since the rolls are independent, the probability of this happening is = 6 1 6 5 } 2 { 1 n th roll n at appears First P . Thus, = = 6 1 6 5 ) ( 1 n n X P . Therefore, = + < < = = = ... , 3 , 2 , 1 ; 1 6 1 6 5 1 ) ( ) ( 1 1 m m x m x x X P x F m n n X 2) This problem looks like an application of the DemoivreLaplace theorem with 5000 = n , 2 / 1 = p , 2 / 1 = q , except that the lower limit of 2,225 is not within the range of the theorems assumptions. That is, 35 ) 2 / 1 )( 2 / 1 )( 5000 ( = = npq in this case, but the lower limit is more than 35 away from the np point of 2,500. But we can use Bernoulis theorem from the end of lecture 2 here. we can use Bernoulis theorem from the end of lecture 2 here....
View
Full
Document
 Spring '11
 voltz

Click to edit the document details