Exam P Question

[Mikel_85]

I would like to kindly know if there's anyone out here who can help me solve this problem..

COURSE 1 Nov 2001 # 32

The number of injury claims per month is modeled by a random variable N with
P[N = n] = 1/(n + 1)(n + 2)
, where n ≥ 0 .
Determine the probability of at least one claim during a particular month, given
that there have been at most four claims during that month.
(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 5/6

Please explain your solutions as well as i dun know from where 1/6 +1/12+1/20+1/30(based on solution) come from...thank you very much...

[edt966]

It's discrete. It asks for the prob. of at least 1 / prob. at most 4. This makes a finite amount for n. n={1,2,3,4}/ n= {0,1,2,3,4}. Sub plug chug. [1 to 4]/[0 to 4] of the p(x) function.

[xpkx]

I would like to kindly know if there's anyone out here who can help me solve this problem..

COURSE 1 Nov 2001 # 32

The number of injury claims per month is modeled by a random variable N with
P[N = n] = 1/(n + 1)(n + 2)
, where n ≥ 0 .
Determine the probability of at least one claim during a particular month, given
that there have been at most four claims during that month.
(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 5/6

Please explain your solutions as well as i dun know from where 1/6 +1/12+1/20+1/30(based on solution) come from...thank you very much...

the answer of this question is B.

the question asks you to find P(n>=1|n<=4)
By the conditional probability, P(1<=n<=4)/p(n<=4). Also, this is a discrete case.


the P(n<=4) =1/2+1/6 +1/12+1/20+1/30, sub n=0,1,2,3,4

and P(1<=n<=4)=1/6 +1/12+1/20+1/30

then put them back to the equation, and you get 2/5

Hope this help

[edwinjaxfl]

xpkx got it right, again!

We need to find P(N ≥ 1|N ≤ 4), which is the same as P(1 ≤ N ≤ 4)/P(N ≤ 4).

P(1 ≤ N ≤ 4) = Σ(n=1,4) P(N=n)
= 1/6 + 1/12 + 1/20 + 1/30
= 1/3

P(N ≤ 4) = Σ(n=0,4) P(N=n)
= 1/2 + 1/6 + 1/12 + 1/20 + 1/30
= 5/6

So, P(1≤N≤4)/P(N≤4) = (1/3)/(5/6) = 2/5

[Mikel_85]

thanks a lot...thanks 4 d explanation as well..

the discrete probability thing isn't given in GUO...any idea where i can look it up?

[edwinjaxfl]

Look in the SOA website for a list of references. Here's the link to their syllabus:

http://www.soa.org/files/pdf/edu-2008-fall-exam-p.pdf

You can also purchase the Actex or ASM study guides, which have everything you need to know for the exam. :wink:

[Mikel_85]

Guo isn't as good as those two?

[Mikel_85]

Another question, can you guys help please?

An auto insurance company has 10,000 policyholders. Each policyholder is classified as
(i) young or old;
(ii) male or female; and
(iii) married or single.
Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The
policyholders can also be classified as 1320 young males, 3010 married males, and 1400
young married persons. Finally, 600 of the policyholders are young married males.
How many of the company’s policyholders are young, female, and single?
(A) 280
(B) 423
(C) 486
(D) 880
(E) 896

my main problem, was i couldnt understand the formula used in the solution

[Puoya]

You are given that 3000 are young

1320 are young males then 1680 are young females

We are given that

1400 are young and married

600 young married males

then we must have

800 young married females

If 800 are young married females, it must be that the remainder of

1680 - 800 are young single females. 880.

I think they just threw a bunch of info just to add confusion.

[leerobb1966]

Hi,
I couldn't figure out the problem myself for a while.
They I came across a book titled "Elementary Probability".
The writer drew a big circle with one smaller circle inside, both of the same center.
Then the write drew two diameters to make the big circle into 8 different parts.
you can label them by pairs of mutually exclusive groups.

It's much easier for visual learner like me.
Hope this is helpful.