~HW3_Sol

# ~HW3_Sol - ECEN 646 Homework 3 Solutions 1 C ¸inlar...

This preview shows pages 1–2. Sign up to view the full content.

ECEN 646 Homework 3 Solutions 1 C ¸inlar Exercise (5.1), Chapter 3 (a) ( X 1 ,...,X 8 ) = (1 , 0 , 0 , 0 , 1 , 1 , 0 , 1) (b) ( N 0 ,...,N 8 ) = (0 , 1 , 1 , 1 , 1 , 2 , 3 , 3 , 4) (c) ( T 0 ,...,T 4 ) = (0 , 1 , 5 , 6 , 8) 2 C ¸inlar Exercise (5.3), Chapter 3 (a) This is the probability that, out of the first four vehicles, only the 3rd vehicle turns left. It has probability (1 - 0 . 62)(1 - 0 . 62)(0 . 62)(1 - 0 . 62) = 0 . 0340. (b) This is the probability that the first vehicle turned left, exactly one of the 2nd or 3rd turned left, and the 4th turned right. There are two ways for this to happen, so the probability is 2(0 . 62) 2 (1 - 0 . 62) 2 = 0 . 1110. (c) This is the probability that 6 of the first 8 vehicles turned left, and then 12 - 6 = 6 of the next 7 turned left. Since these two are independent (the drivers act independently), the probability is, P { N 8 = 6 ,N 15 = 12 } = P { N 8 = 6 ,N 15 - N 8 = 6 } = P { N 8 = 6 ,N 7 = 6 } = parenleftbigg 8 6 parenrightbigg (0 . 62) 6 (1 - 0 . 62) 2 × parenleftbigg 7 6 parenrightbigg (0 . 62) 6 (1 - 0 . 62) 1 = 0 . 0347 3 C ¸inlar Exercise (5.5), Chapter 3 (a) E [ N 3 ] = E [ X 1 + X 2 + X 3 ] = 3 p = 3(0 . 8) = 2 . 4 E [ N 7 ] = 7(0 . 8) = 5 . 6 E [ N 3 + 4 N 7 ] = E [ N 3 ] + 4 E [ N 7 ] = 2 . 4 + 4(5 . 6) = 24 . 8 (b) Var ( N 3 ) = 3 pq = 3(0 . 8)(0 . 2) = 0 . 48 1

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern