Hey All,
I posted a final practice exam. I created problems which seemed to match the description of the exam. I'm missing out on the end of the exam at the moment and wouldn't know how closely what I wrote up resembles the real exam. At any given time, you should make sure you have the most up-to-date version number.
Note that the copies in the Dropbox will get updated dynamically. While the copies in my Course Files folder will get updated periodically.
Current Version Numbers and Links to Files:
Final Practice Exam [v1.2.4.2]
Final Practice Exam Solutions [v1.2.4.2]
You can ask me questions via e-mail or posting a comment here. Cheers.
Hey Tim,
ReplyDeleteDo you know where the final exam is supposed to be? Wilson said in lecture that he didn't know what room.
Thanks!
Hi Emelynn, you can find the location on this exam schedule located on the department website. Happy studying! ExamScheduleFall2013.pdf
DeleteA student asked, is the book's answer to Question 55 in Section 12.4 wrong? I believe it is wrong. The correct answer would be
ReplyDelete$\frac{4!}{2!1!1!} \left( \frac{8}{18} \right)^2 \left( \frac{4}{18} \right)^1 \left( \frac{6}{18} \right)^1$
A student brought to my attention mistakes to several solutions which I have now fixed. Thanks! Do ask more questions and send in those corrections!
ReplyDeleteDue to a student's inquiry, I added a Monty Hall problem to the Practice Final Exam. Your questions help me help you! :D
ReplyDeleteA student asks, why is the number of straights for a five card hand
ReplyDelete${10 \choose 1} {4 \choose 1}^5$
and not
${10 \choose 1} {4 \choose 1}^4$.
Well, the way I solved the problem, I first only chose the value of the lowest card in the straight. To choose one out of ten of those possible values is ${10 \choose 1}$. The lowest value determines the values for the other four cards, because a straight is five cards in succession. Now we have five particular values and for each value we can choose one among four of them. Thus it is
${4 \choose 1} \cdot {4 \choose 1} \cdot {4 \choose 1} \cdot {4 \choose 1} \cdot {4 \choose 1}$ or more simply ${4 \choose 1}^5$.
A student asked about my solution to the Monty Hall problem and discovered that Monty doesn't open one of the doors you choose, but one of the doors that haven't been chosen.
ReplyDelete