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 Probability Old Exam P Forum

#21
04-03-2006, 10:20 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,099

I believe the solution interprets the wording to mean:

The family car will be driven some known distance per year, e.g. 40,000 km. (For an alternative distance, everything changes proportionately.)

The mother will drive 20,000 km; the father will drive 16,000; the son will drive 4,000.

If the mother were to drive all 40,000 and the others 0, or the father to drive all 40,000 and the others 0, the probability of an accident is 1%. If the son were to drive all 40,000 and the others 0, the probability of an accident is 5%.

That's a reasonable interpretation, but the words could also mean:

If the mother were to drive her 20,000 and the others 0, or the father to drive his 16,000 and the others 0, the probability of an accident is 1%. If the son were to drive his 4,000 and the others 0, the probability of an accident is 5%.

Edited to add: the intended meaning is probably the better choice, since the alternative could be stated much more simply as "the probability that the mother in the year is in an accident is 1%; that the father is, is 1%; that the son is, is 5%"

Last edited by Gandalf; 04-03-2006 at 10:23 AM..
#22
04-03-2006, 02:18 PM
 sam_broverman Member Join Date: Feb 2005 Posts: 639

I agree that the wording makes the problem somewhat
ambiguous. I will try to clean up the wording to make it
more clear.

Sam Broverman
#23
04-10-2006, 08:37 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639
Aprl 10 Question

You can find the question at:
http://www.sambroverman.com/apr10p.pdf

Sam Broverman
2brove@rogers.com
#24
04-17-2006, 08:41 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639
April 17 Question

You can find the question at:
http://www.sambroverman.com/apr17p.pdf

A corrected solution has been posted, April 21, 2006.

Sam Broverman
2brove@rogers.com

Last edited by sam_broverman; 04-21-2006 at 05:44 PM..
#25
04-18-2006, 10:56 AM
 thekind78 Member Join Date: Dec 2005 Posts: 116
Re: Apr 17th problem

Hello,
The question says there is a max of 8 on the payment the insurer will pay, however the solution states the insurer will pay 10 if X+Y> 10. I was wondering if I was misreading this?
#26
04-21-2006, 07:42 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639

Thanks for pointing that out. The solution should have a maximum
payment of 8, not 10, so the integral with the max
payment of 10 should have a max of 8. I'll post a correction
shortly.

Sam Broverman
#27
04-24-2006, 07:51 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639
April 24 Question

You can find the question at:
http://www.sambroverman.com/apr24p.pdf

Sam Broverman
2brove@rogers.com
#28
05-01-2006, 08:26 AM
 sam_broverman Member Join Date: Feb 2005 Posts: 639
May 1 Question

You can find the question at:
http://www.sambroverman.com/may1p.pdf

Sam Broverman
2brove@rogers.com
#29
05-02-2006, 08:29 PM
 mallkins Member CAS Join Date: Feb 2006 Favorite beer: Guinness Posts: 348

Hi

Would you think that something as convoluted as this would come up on the exam, or is this an example of an extended problem to help extend the material

http://www.sambroverman.com/dec26p.pdf

Cheers,
#30
05-02-2006, 09:59 PM
 sam_broverman Member Join Date: Feb 2005 Posts: 639

Mallkins,

It is quite convoluted, and I wouldn't expect a question
like this one on the exam. It really is a stretch for the
Exam P material. I must have been in a bad
mood when I made it up (or hungover, it was Boxing Day after all) .

Sam Broverman