HW25Answers(9.1) - Math331, Spring 2008 Instructor: David...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 9.1 Homework Answers Homework: pgs. 383 - 384, #'s 1, 3, 5. 1. Let the number of hearts, diamonds, clubs, and spades be denoted by H, D, C, and S. We simply need to count the number of ways we can choose h hearts, d diamonds, c clubs, and s spades when h, d, c, s satisfy h + d + c + s = 13, and divide by the total number of 13 card combinations which is 52 . Simply by using the counting principle we get 13 P {H = h, D = d, C = c, S = s} = (13)(13)(13)(13) h d c s (52) 13 0 if h + d + c + s = 13 otherwise . 3. We have pX,Y (x, y) = pY,Z (y, z) = pX,Z (x, z) = 1 162 1 162 1 162 2 xyz = z=1 5 1 1 3xy = xy 162 54 1 1 9yz = yz, 162 18 1 1 6xz = xz, 162 27 for x = 4, 5, y = 1, 2, 3. for y = 1, 2, 3, z = 1, 2. for x = 4, 5, z = 1, 2. xyz = x=4 3 xyz = y=1 Using pY,Z calculated above: 3 2 E[Y Z] = y=1 z=1 yzpY,Z (y, z) = 1 18 3 1 18 3 2 y2z 2 y=1 z=1 = = 5y = y=1 5 70 14 = 18 18 35 . 9 5. Clearly X1 and X2 are independent. However, it is simple to see that once X3 is added to the mix, the three are no longer independent. Intuitively, this can be seen because knowing that X3 = 1 tells you that X1 + X2 = 1. More explicitly, P {X1 = 1, X2 = 1, X3 = 1} = 0 = and so they are not independent. 1 2 3 = P {X1 = 1}P {X2 = 1}P {X3 = 1}, 1 ...
View Full Document

This note was uploaded on 06/24/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.

Ask a homework question - tutors are online