{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw6vqn - Vinh Nguyen Stat 120A(Fall 2007 HW5 Solutions Text...

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

Vinh Nguyen Stat 120A(Fall 2007): HW5 Solutions November 28, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 5.2.3, 5.2.4, 5.3.6, 5.3.7, 5.5.2, 5.4.1, 5.4.2, and Additional Problems 5.2.3 Let X be the number of heads in 10 tosses. We have X Bin (10 , 1 / 2). In order to get more heads than tails, we must have at least six heads. Thus, P ( X 6) = 1 - P ( X 5) = 1 - 5 X x =0 10 x (1 / 2) x (1 / 2) 10 - x = 0 . 3770 . 5.2.4 We have X Bin (15 , 0 . 4). Thus, P (6 X 9) = 9 X x =6 15 x (0 . 4) x (0 . 6) 15 - x = 0 . 5630 . 5.3.6 Suppose X 1 Bin ( n 1 , p ) and X 2 Bin ( n 2 , p ), where X 1 and X 2 are independent. Note that X 1 + X 2 Bin ( n 1 + n 2 , p ). We would like to compute the conditional distribution of X 1 given X 1 + X 2 = k , for k = 1 , . . . , n 1 + n 2 . In order to do so, we apply the definition of conditional probability: P ( X 1 = x | X 1 + X 2 = k ) = P ( X 1 = x, X 1 + X 2 = k ) P ( X 1 + X 2 = k ) (Definition) = P ( X 1 = x, X 2 = k - x ) P ( X 1 + X 2 = k ) = P ( X 1 = x ) P ( X 2 = k - x ) P ( X 1 + X 2 = k ) (Independence) = ( n 1 x ) p x (1 - p ) n 1 - x ( n 2 k - x ) p k - x (1 - p ) n 2 - k + x ( n 1 + n 2 k ) p k (1 - p ) n 1 + n 2 - k = ( n 1 x )( n 2 k - x ) ( n 1 + n 2 k ) Thus, X 1 | X 1 + X 2 = k follows a

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