Homework 2 Solution

# Homework 2 Solution - MSPE PROGRAM ECON 506 FALL 2010...

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

MSPE PROGRAM ECON 506- FALL 2010 SOLUTION TO HW2 (The problem numbers refer to numbers in 7 th edition) 3.11 There is a 1/3 chance a person has O + blood and 2/3 they do not. Similarly, there is a 1/15 chance a person has O blood and 14/15 chance they do not. Assuming the donors are randomly selected, if X = # of O + blood donors and Y = # of O blood donors, the probability distributions are 0 1 2 3 p ( x ) (2/3) 3 = 8/27 3(2/3) 2 (1/3) = 12/27 3(2/3)(1/3) 2 =6/27 (1/3) 3 = 1/27 p ( y ) 2744/3375 3*(196/3375) 3*(14/3375) 1/3375 Note that Z = X + Y = # will type O blood. The probability a donor will have type O blood is 1/3 + 1/15 = 6/15 = 2/5. The probability distribution for Z is 0 1 2 3 p ( z ) (2/5) 3 = 27/125 3(2/5) 2 (3/5) = 54/27 3(2/5)(3/5) 2 =36/125 (3/5) 3 = 27/125 3.12 E ( Y ) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2.0 E (1/ Y ) = 1(.4) + 1/2(.3) + 1/3(.2) + 1/4(.1) = 0.6417 E ( Y 2 – 1) = E ( Y 2 ) – 1 = [1(.4) + 2 2 (.3) + 3 2 (.2) + 4 2 (.1)] – 1 = 5 – 1 = 4. V ( Y ) = E ( Y 2 ) = [ E ( Y )] 2 = 5 – 22 = 1. 3.23 Define G to be the gain to a person in drawing one card. The possible values for G are \$15, \$5, or \$–4 with probabilities 3/13, 2/13, and 9/13 respectively. So,

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 ]}

### Page1 / 3

Homework 2 Solution - MSPE PROGRAM ECON 506 FALL 2010...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online