Errata
November 24th, 2007
| Date | Lesson | Description |
| 05/18/07 | E.2.3 | #13 – I solved for P[NSW | NC] rather than P[SW | NC]. The correct answer is C. |
| 06/08/07 | A.4.P | #1 – I solved for the probability of filing at least one claim, rather than more than one. The answer should be A. |
| 06/13/07 | B.1.P | #2 Derivatives – The final answer should be (2 ln 2) 2^(2x) (the 2 was missing in the exponent). |
| 06/13/07 | B.1.P | #4 Integrals – In the substitution, t=u-9 rather than 9-u. Because we have t^2 in the numerator, the sign cancels out so the original answer was correct. |
| 06/14/07 | A.4.P | #1 – New PDF with correct solution to #1 uploaded. |
| 06/14/07 | C.2.P | #3 – Answer choice D should be 28/15, and is also the correct answer. |
| 07/11/07 | D.1.P | #17 – The domain should be y < x, not x < y. |
| 07/11/07 | D.2.P | #13 – The correct answer is C. |
| #14 – The correct answer is D. | ||
| 11/24/07 | Practice Exam 4 | #9 – The correct answer is .11 |
| #10 – The correct answer is .035 |
EXAM P A.1 #2
I don’t see how you got .105 for P(one car and sports car).
Isn’t it .15 instead?
No, it should be .105. A different way to think about this problem is to imagine that you know how many total customers there are and to work in terms of that. Suppose that there are 1,000 total customers. 70% have more than one car, so 0.7*1,000=700 people have more than one car. Of those 700, 15% have a sports car, so we get .15*700=105 customers with more than one car that is also a sports car. P(more than one car and a sports car) is then 105/1,000=0.105.
On a side note, a better way to contact me is to send me e-mail as it is easier to respond to that when I am travelling and you will also get out of office messages when I am away.
Hello Dave. On B.1.P #35, one of the factors in the first part of the sum is (-2). Shouldn’t it be (-1)?
Thanks.
Yes, thanks for pointing this out.
Hi Dave. On Sample Test 2, number 26, shouldn’t “Method 2″ use the integral “1-e^-x/30″?
No. The cdf is 1-e^{-x/30}, but P[X>x] = 1-P[X<=x] = 1-(1-e^{-x/30}) = e^{-x/30}.