## Errata

November 24th, 2007
1. September 7th, 2009 at 20:57 | #1

EXAM P A.1 #2
I don’t see how you got .105 for P(one car and sports car).

2. September 14th, 2009 at 07:56 | #2

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.

3. February 29th, 2012 at 20:54 | #3

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.

4. March 5th, 2012 at 10:57 | #4

Yes, thanks for pointing this out.

5. September 23rd, 2012 at 07:42 | #5

Hi Dave. On Sample Test 2, number 26, shouldn’t “Method 2″ use the integral “1-e^-x/30″?

6. September 24th, 2012 at 10:29 | #6

No. The cdf is 1-e^{-x/30}, but P[X>x] = 1-P[X<=x] = 1-(1-e^{-x/30}) = e^{-x/30}.