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#21
12-09-2006, 03:41 PM
 Nonpareil Note Contributor Join Date: Nov 2006 Location: Rocket City Studying for Exam C Posts: 833

Quote:
 Originally Posted by bjengs Yes, and I was surprised at what was emphasized, at least in my batch of questions. All of the discrete distributions were tested, and I had 2 instances of Poisson used in interesting scenarios. But as for continuous dist's, one only need have learned, inside and out, the distributions for the uniform, and exponential distributions, and you could make a case for the Pareto (but solving the problem I saw wasn't dependent on knowing it was a Pareto or any of its particular characteristics). Case in point on the exponential: there was a q. that you posted (one of the ones the moderator deleted) that involved finding E(XgivenY) and dividing the given f(x,y) by the derived f(y) yielded (1/y)e^{-x/y}. Now you could go ahead and do the math or you could spot that, since you are working in terms of a fixed Y, that is the pdf for an exponential dist. with mean y and that is your answer.
I only had one Poisson, and nothing on Pareto or gamma, at least that I could recognize as such.

bjengs, didn't you also take Exam FM in November as well? How did the difficulty of the two exams compare, in your opinion?
#22
12-09-2006, 04:12 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by KingWithoutACrown Did anyone get the problem, that looked something like F(Y) - F(X) = (1/X^3) - (1/Y^2) and you had to find, f(2)/F(2) I couldn't figure that out during the exam, or even after, Also I had no idea how to solve the M.A.D. problem either
If that's the form of the question then it's either ambiguous or contradictory. First, if we are to assume that X and Y are not identically distributed, then which distribution is intended when computing f(2)/F(2), the distribution for X, or for Y? Second, if we are to assume that X and Y are identically distributed, then the LHS equals zero when X = Y, but the RHS is not zero except when X = Y = 1. That makes no sense.
__________________
Spoiler:
"No, Moslems don't believe Jesus was the messiah.

Think of it like a movie. The Torah is the first one, and the New Testament is the sequel. Then the Qu'ran comes out, and it retcons the last one like it never happened. There's still Jesus, but he's not the main character anymore, and the messiah hasn't shown up yet.

Jews like the first movie but ignored the sequels, Christians think you need to watch the first two, but the third movie doesn't count, Moslems think the third one was the best, and Mormons liked the second one so much they started writing fanfiction that doesn't fit with ANY of the series canon."

-RandomFerret
#23
12-09-2006, 04:23 PM
 ba553th0unD Member Join Date: Sep 2006 Posts: 39

I agree with you, atomic. VERY ambiguous question. That's exactly what I thought when I first came across it, and realized it could be interpreted in different ways.
#24
12-09-2006, 04:54 PM
 karend26 Member Join Date: May 2006 Posts: 85

Did anyone have this

A=1,2,3,4,5 with prob of .3
B=1,2,3,4 with prob of .1
C=6 with prob .7

What is AorB + AorC + BorC?

I can't remember if it asked for expected value or what but all the answers were >1. I know two of the answer choices were 2.10 and 2.17.

Any ideas?
#25
12-09-2006, 05:23 PM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by karend26 Did anyone have this A=1,2,3,4,5 with prob of .3 B=1,2,3,4 with prob of .1 C=6 with prob .7 What is AorB + AorC + BorC? I can't remember if it asked for expected value or what but all the answers were >1. I know two of the answer choices were 2.10 and 2.17. Any ideas?
I can't even figure out what you're trying to say in the question. It's not clear what the distribution of A, B, and C are, and I assume you mean $\Pr[A \cup B] + \Pr[A \cup C] + \Pr[B \cup C]$.
__________________
Spoiler:
"No, Moslems don't believe Jesus was the messiah.

Think of it like a movie. The Torah is the first one, and the New Testament is the sequel. Then the Qu'ran comes out, and it retcons the last one like it never happened. There's still Jesus, but he's not the main character anymore, and the messiah hasn't shown up yet.

Jews like the first movie but ignored the sequels, Christians think you need to watch the first two, but the third movie doesn't count, Moslems think the third one was the best, and Mormons liked the second one so much they started writing fanfiction that doesn't fit with ANY of the series canon."

-RandomFerret
#26
12-09-2006, 07:37 PM
 scx43 Member SOA AAA Join Date: Jul 2006 Posts: 66

I got that question on my exam. How I approached it was like a roll of a dice. You see that B is a subset of A, and that C is disjoint from both A and B and then use that knowledge to get the answer. I believe I got 2.1.

On my exam I had a significant amount of insurance based questions. I was surprised to have multiple questions testing the deductible and policy limits. I know someone before had previously commented on not having a lot of continuous distributions tested, however on my test I would say that the majority of the distributions were continuous, however nothing exotic like lognormal or weibull (thankfully). One thing that I found being very useful was being able to recognize a distribution as a beta or a gamma and using that knowledge to simplify the integral, or using their expectation/variance formulae. All in all I thought the test was moderately difficult... now we just have to twiddle our thumbs until January.

Last edited by scx43; 12-09-2006 at 07:54 PM..
#27
12-09-2006, 08:46 PM
 neb7777 Join Date: Sep 2006 Posts: 2
another question from the exam

You are given 3 dice.

die1 has sides numbered 1,2,3,4,5,6.
die2: 2,2,4,4,6,6
die3: 6,6,6,6,6,6.

You choose two at random and roll them. What is the probability that the sum is 11?

Here is how I approached the problem.

The probabilty is (1/3) that you choose either die1 and die2, die1 and die3, or die2 and die3.

For each case, the probability of rolling an 11 is (1/18), (1/6), and 0, repsectively.

Therefore, the desired probability should be

(1/3)(1/18) + (1/3)(1/6) +(1/3)(0) = .074.

I don't remember if this was one of the choices, though. Does anyone have any other approach for the problem?
#28
12-10-2006, 12:32 AM
 Gator Cane Member CAS Join Date: Jun 2006 Location: FL Favorite beer: Sam Adams Boston Ale Posts: 377

Quote:
 Originally Posted by christian I only had one Poisson, and nothing on Pareto or gamma, at least that I could recognize as such. bjengs, didn't you also take Exam FM in November as well? How did the difficulty of the two exams compare, in your opinion?
Actually, the Pareto may have been on a practice exam I completed the morning of the exam, can't remember now.

I found P more difficult - this is my second sitting - but especially since studying for and taking FM I realize that I have yet to find a manual for P that even comes close to how effective Harold Cherry's ASM manual for FM is. It just seemed to anticipate all my questions as I went. It also helps that there are more than 3 times as many released problems pertinent to FM than there are for P, and the ASM manual has nearly all of them with solutions.

Then again, I'm not a math major. YMMV.
__________________
"After this exam, Im seriously considering the SOA path to General Insurance, maybe they are less retarded at making exams, I dont understand how [the CAS] can make such inconsistent exams and ask questions about consistency." -euler said it here

Last edited by Gator Cane; 01-20-2007 at 03:34 PM.. Reason: the wisdom that comes with the passing of time
#29
12-10-2006, 12:43 AM
 Gator Cane Member CAS Join Date: Jun 2006 Location: FL Favorite beer: Sam Adams Boston Ale Posts: 377

Quote:
 Originally Posted by neb7777 You are given 3 dice. die1 has sides numbered 1,2,3,4,5,6. die2: 2,2,4,4,6,6 die3: 6,6,6,6,6,6. You choose two at random and roll them. What is the probability that the sum is 11? Here is how I approached the problem. The probabilty is (1/3) that you choose either die1 and die2, die1 and die3, or die2 and die3. For each case, the probability of rolling an 11 is (1/18), (1/6), and 0, repsectively. Therefore, the desired probability should be (1/3)(1/18) + (1/3)(1/6) +(1/3)(0) = .074. I don't remember if this was one of the choices, though. Does anyone have any other approach for the problem?
I believe you are correct. To just do it by counting, die 1 * die 2 allows for 18 different rolls, die 1 * die 3 allows for 6 different rolls, and die 2 * die 3 allows for 3 different rolls. There are only two rolls that create 11 (5 on die1 plus 6 on either of the other 2). 2/27 = .074
__________________
"After this exam, Im seriously considering the SOA path to General Insurance, maybe they are less retarded at making exams, I dont understand how [the CAS] can make such inconsistent exams and ask questions about consistency." -euler said it here

Last edited by Gator Cane; 01-20-2007 at 03:35 PM.. Reason: the usual
#30
12-10-2006, 01:00 PM
 ba553th0unD Member Join Date: Sep 2006 Posts: 39

There maybe 3, even 4 questions that were really different from anything that I've ever seen before (even expected). Here's to hoping that they were all pilot questions!

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