{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Midterm2009Answers-2

Midterm2009Answers-2 - Midterm 2009 Answers Question 1 A...

Info icon This preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Midterm 2009 Answers Question 1: Question 1: A business school graduate schedules 7 interviews in a given week. Suppose that the probability that an individual interview results in a job offer is 0.40. Assume that the outcomes of the interviews are independent of one another. Let X be the number of job offers that this person has at the end of the week. Write out one expression for P(X=x) where x may be any arbitrary positive integer between 1 and 7: 1 ≤ x ≤ 7 . Explain each part of the expression and why this makes sense.
Image of page 1

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

View Full Document Right Arrow Icon
Midterm 2009 Answers Question 1 Answer: The number of job offers at the end of the week is a Binomial Random variable with parameters p=0.40 and n=7. That is, we are looking at the sum of “successes” in n=7 independent trials.
Image of page 2
Midterm 2009 Answers Question 1 Answer: P(X=x) is given by the Probability Distribution Function for a Binomial Random Variable: 7 7 ( ) (1 ) for 0 , 1 , 2, ... , or, given that p=0.40, n=7 ( ) (0.4) (1 0.4) for 0 , 1 , 2, ... , n x x x n x x x P X x C p p x n P X x C x n - - = = × × - = = = × × - =
Image of page 3

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

View Full Document Right Arrow Icon
Midterm 2009 Answers 7 7 7 7 ( ) (0.4) (1 0.4) for 0 , 1 , 2, ... , Explanation: There are two parts of this equation: and (0.4) (1 0.4) We have 7 independent trials. We want to fin x x x x x x P X x C x n C - - = = × × - = × - 7 d the probability that x of these trials experience a success and n-x do not experience a success. The expression (0.4) (1 0.4) gives the probability of one particular sequence of outcomes that resu x x - × - 7 lts in x successes and 7-x failures. E.g., if x=2, then the probability of getting the sequence {1 0 0 0 1 0 0} is (0.4)(0.6)(0.6)(0.6)(0.4)(0.6)(0.6)=(0.4) (1 0.4) . However, this is only one x x - × - 7 very specific way that we can get two successes (successes on trials 1 and 5). The total number of ways that we can get x successes in 7 trials is given by , so the probability of observing any one x C 7 7 of these outcomes is given by the sum of the probabilities of all of these events, which is given by (0.4) (1 0.4) x x x C - × × -
Image of page 4
Midterm 2009 Answers Question 2: Consider the following estimates from the Sheboygan Fire Department: The probability that a house fire is due to arson is 0.10. The probability that the Fire Department correctly blames a house fire on arson is 0.82 (probability that they blame it on arson, given that the fire was actually caused by arson). The probability that the Fire Department incorrectly blames a house fire on arson is 0.05 (probability that they blame it on arson, given that it was not caused by arson) a) What is the probability that a house fire was actually caused by arson if the Fire Department blames the fire on arson? b) What is the unconditional probability that a house fire will be blamed on arson?
Image of page 5

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

View Full Document Right Arrow Icon
Midterm 2009 Answers Question 2 a) : This is a Question that can be answered using Bayes’ Theorem. First, define the events that we are working with: A: The event that a fire was due to arson B: The event that the Sheboygan Fire Department blamed the fire on Arson We want to find the probability that a fire was caused by arson if the fire was blamed on arson by the Fire Department. That is, we want to find P(A|B)
Image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern