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## #1 Re: Exercises » Expected Claim Count » 2011-04-18 18:10:56

Got it - so use baye's theorem to get the probability for high risk and low risk drivers in 1998 given that he had one claim and then use the logic I did

thanks

## #2 Re: Exercises » Exponential Distribution Verbiage » 2011-04-18 17:39:35

haha - I see the joke! thanks bobbym

## #3 Exercises » First time licensees! » 2011-04-18 17:38:28

getback0
Replies: 2

A large pool of adults earning their first drivers license includes 50% low-risk drivers,
30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior
driving record, an insurance company considers each driver to be randomly selected from
the pool. This month, the insurance company writes 4 new policies for adults earning their
first drivers license. What is the probability that these 4 will contain at least two more
high-risk drivers than low-risk drivers?
(A) 0.006 (B) 0.012 (C) 0.018 (D) 0.049 (E) 0.073

I figured -

P(L) = 0.5
P(H) = 0.2
P(M) = 0.3

Now, probability (atleast 2 more high risk drivers than low risk drivers) =

p( 0 L, 2 H, 2 M) + p(0L, 3 H, 1M) + p(0L, 4H, 0M) + p(1L, 3H)

= (0.2^2)*(0.3^2) + (0.2^3*0.3) + (0.2^4) + (0.5* 0.2^3)

= .0116

I am not getting any answers right!!!!

I dont know whats going to be my plight in the exams!

## #4 Exercises » Expected Claim Count » 2011-04-18 17:31:45

getback0
Replies: 3

An insurance company designates 10% of its customers as high risk and 90% as low risk.
The number of claims made by a customer in a calendar year is Poisson distributed with
mean θ and is independent of the number of claims made by a customer in the previous
calendar year. For high risk customers, θ = 0.6, while for low risk customers θ = 0.1.
Calculate the expected number of claims made in calendar year 1998 by a customer who
made one claim in calendar year 1997.

My solution:

Given that claims made by a customer in a year are independent of those made in the previous year, I disregarded the 1997 claim filing.

For any year, the probability that a customer makes one claim would be -

f(k) = [0.1 * (e ^ -0.6) * (0.6 ^ k) / k! ] + [0.9 * (e ^ -0.1) * (0.1 ^ k) / k! ]

E(x) = ∑ xf(x)

works out to 0.1*0.6 + 0.9*0.1 = 0.15

doesnt match the key

I am way off here, i guess!

## #5 Re: Exercises » Exponential Distribution Verbiage » 2011-04-18 17:18:59

Got it  - makes sense -

1. I made a mistake assuming density function is 0 beyond x = 250- it is clearly 250

2. Clearly, if the median for the exponential density function works out to be less than the benefit limit, the limit doesnt change the median

Thanks bobbym! I got 2 more problems for you - Stay tuned!

## #6 Re: Exercises » Exponential Distribution Verbiage » 2011-04-18 16:54:18

if I assume, the density function goes all the way to infinity, then yes - I get c = 0.004 and the median is the solution of the equation

integral (0.004 e ^ -0.004*x ) over [0, x ] = 0.5  -works out to 173.4

However, since it mentioned maximum benefit of 250, to determine c, I used -

integral (pdf) over [0, 250] = 1 - this gives c of 0.00633

and then solving the same equation above for x gives 94.97

## #7 Exercises » Exponential Distribution Verbiage » 2011-04-18 10:36:26

getback0
Replies: 6

I am not for sure if it is that I am misunderstanding the problem statement here or that the problem is just poorly stated -

================================================================================
An insurance policy reimburses dental expense, X, up to a maximum benefit of 250. The
probability density function for X is

f(x) =   ce^−0.004x for x >= 0,
0 otherwise,

where c is a constant. Calculate the median benefit for this policy.
(A) 161 (B) 165 (C) 173 (D) 182 (E) 250

================================================================================

My question is -

When the question states the maximum benefit is 250 units, shouldnt the benefit density function be the value for f(x) only until 250 and 0 beyond?

I ask as this changes the answer. With benefit density limited to 250 (as the question states), my median calculation is around 94. Withouth the 250 unit limitation, the median works out to be 173 (Option - C) which matches the solution in my textbook.

Why is the solution not accounting for 250 unit limit? Any ideas?

- G

## #8 Re: Help Me ! » real huge numbers » 2011-03-23 07:04:14

sorry - i didnt realize your response changed - was responding to your older post -

hey AND I HAVE NEVER FOUND YOU grouchy! To me, you are an extremely kind-hearted person who takes all the effort to help out random people (who may not always be nice in return).

THANKS SIR!

## #9 Re: Help Me ! » real huge numbers » 2011-03-23 06:59:09

not sure if it sounds like binomial to me (not disagreeing - just not sure) - for the following reasons -

experiment following binomial distribution should have -

1. n identical trials
2. each should have only 2 outcomes - success and failure
3. probability of success on each trial should stay the same
4. all trials should be independent of each other

IT COULD WELL BE JUST ME - as with these problems, it just seems one needs to have the 'eye' to fit a problem to an established distribution!

## #10 Re: Help Me ! » real huge numbers » 2011-03-23 05:04:57

Hi bobbym! Hope this finds you well - THANKS AS always for getting back!

However, I disagree with the solution given the verbiage - if the probability in question was for two or more professors to pick the same course, the question is misleading since it says "two professors picking the same course"

Said another way, what would be your take if the question was "only 2 professors pick the same course"?

## #11 Help Me ! » real huge numbers » 2011-03-23 04:32:34

getback0
Replies: 6

I am having trouble matching my answer to the key provided -

Question:

Suppose each of 100 professors in a large mathematics department picks at random one of 200 courses. What is the probability that two professors pick the same course?

My Solution:

Sample space outcomes = 200 to the power 100

Assuming professor-1 and professor-2 of the 100 professors were to pick the same course of the 200 available, this could happen in

n(E1) = C(200, 1) * P(199, 98)  .... (1)

Since there are C(100, 2) ways of picking professor-1 and professor-2 , we have -

n(E) = C(100, 2) * n(E1)           .... (2)

Hence, probability is

P(E) = n(E)/n(S)

P(E) = [ C(200, 1) * P(199, 98) * C(100, 2) ] / (200 ^ 100 )

Microsoft Excel calculates this to be -

P(E) = 3.26715E-12 --> in other words, highly unlikely!

Key provided

## #12 Re: Help Me ! » More probability! » 2010-11-07 15:11:15

Sorry about the HTML error! Here's the corrected response -

Awesome solution with Q4! Although, I have to ask on the Two-pair hand problem, because I am having trouble unwinding out of the thought -

Why

and not

P(13,3) ?

Is the 5-card hand drawn as -

4-cards at the first go and

1 card on the second go?

That would explain the solution. If not, I am having trouble there.

Leon

## #13 Re: Help Me ! » More probability! » 2010-11-07 14:28:16

Yep - they make sense - thanks for all the guidance and encouragement! I will keep trying.

I will let you know if I get stuck again.

- getback0

## #14 Re: Help Me ! » More probability! » 2010-11-06 14:58:37

I will take your word! Will start working on the material in that link you sent.

THANKS again,

getback0

## #15 Re: Help Me ! » More probability! » 2010-11-06 13:17:24

First up! THANKS A LOT - for the prompt response! I need to be more regular at this stuff.

I have a few questions though -

1.  For Q #4:
=========

In the solution when we substitute

; we are assuming A and B to be independent events - data that the problem has not provided, correct?

2.  For Q #3:
=========

In the solution again, we have used -

which suggests we use -

for the probability of a certain number of claims

Do we?? If so, this is the probability of a number of injury claims, is it not??

3.  For Q #2:
=========

When they say the proportion of people choosing coverage A is

, how does that suggest
? You clearly got it right - why am I not getting it even when someone's laying the solution out for me??????????

4.  For Q #1:
==========

Ok ... let me just say .... YOU ARE AN AUTHORITY on this subject!!!

Could you help me understand what's wrong with my thought process? I totally see your solution. I like to verbalize the experiment and then translate that into a "counting" problem.

From your solution, it seems you translated this to -

X: Pick 2 different ranks, and pick any 2 cards from a total of 4 for each rank ->

Y: Pick another rank, and pick any 1 card from a total of 4 for each rank        ->

n(E) = X * Y

What is wrong with my line of reasoning?

## #16 Re: Help Me ! » More probability! » 2010-10-28 17:50:03

Just 2 more items -

Q - #3:
=====

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 that month?

Key: 2/5
=======

Confusion:
========

Honestly, the problem seems a little off to me - especially around the bold part. If I was to assume, the task is to find the probability of at least 1 injury claim, given at most 4 claims, I believe I would also need to probability of a certain number of claims, wouldnt I?  I cannot think much of this one!

Q - #4:
======

Given P(A U B) = 0.7

P(A U B') = 0.9

Determine P(A)

Key - 0.6
========

Confusion
========

I got to say the problem sounds incomplete as it is - I could convert these into equations using set theory, but no matter what I do, I end up with 2 equations and 3 unknowns (that's can't be solved???)

I have seen this problem somewhere with the clause, A and B are independent - and that would give P(A intersec. B) = P(A) * P(B) and that would allow for the above situation to simplify to 2 equations and 2 unknowns and I do get the 0.6 answer.

However, I am not sure of the problem as-is.

THANKS again!

- Leon

## #17 Help Me ! » More probability! » 2010-10-28 17:14:58

getback0
Replies: 11

Hello again! I have a bunch of problems this time that I cannot match my answers on (to the solution in Appendix). I have listed below -

the problems,
the key in the Appendix (that does not match ),
my solution/confusion

Any pointers would be truly appreciated!

Q - #1.
=====

A 5-card poker hand is dealt from a standard deck of cards. What is the probability that you get 2 pairs ( QQ993 )?

Key: 0.0475
=========

Confusion
=========

n(S) = C(52,5)

Assuming by 2 pairs is meant 2 different pairs, n(E) = C(13,3) * C(4,2) * C(4,2) * C(4,1) => P(E) = 0.0158

Assuming by 2 pairs is meant 2 same/different pairs,

n(E) = ( C(13,3) * C(4,2) * C(4,2) * C(4,1)    .. as above) + ( C(13,2) *  C(4,4) * C(4,1) ) )       => P(E) = 0.016

Q - #2.
=====

An insurer offers a health plan to the employees for a large company. As part of this plan, the individual employees may choose exactly two of the supplementary coverages A, B and C or they may choose none. The proportions of the company's employees that choose coverages A, B and C are 1/4, 1/3, and 5/12 respectively. Determine the probability that a randomly chosen employee will choose no supplementary coverage?

Key: 1/2
=========

Confusion
=========

P(A) = 1/4
P(B) = 1/3
P(C) = 5/12

Choosing any coverage seems to be independent of choosing another. So,

P(A intersec B) = P(A).P(B) = 1/12
P(B intersec C) = P(B).P(C) = 5/36
P(C intersec A) = P(C).P(A) = 5/48

Hence, P(A U B U C) = P(A) + P(B) + P(C) - P( A int. B ) - P( B int. C) - P( C int. A) + P(A int. B int. C)

P(A int. B int. C) = 0 as choosing all 3 coverages is not permitted

Hence P(A U B U C) = 1/4 + 1/3 + 5/12 - 1/12 - 5/36 - 5/48
= 97/144

Hence, P (no coverage is chosen) = 47/144
= 0.3264

Of course, this is a trifle different from the key

HELP!!!! I am so going to flunk!!!!

Leon

## #18 Re: Help Me ! » Event with repitition » 2010-10-17 13:53:49

Thanks for the encouragment, bobby! Much appreciated!

I agree - splitting a compound/complex problem down to simpler events which involve counting through combinations/permutations presents a very efficient way of solving problems. And the elaborate explanation you put together for n(E2) helped me get through my exercises quickly - so thanks again!

Caution: If only you are not tired of me talking about my thought process on this 1 problem, I would suggest reading on ... by the way, I am almost through with thinking about this one problem, so if I post again, very likely, it will be a different problem

I just wanted to find out what was leading me to get double the correct answer when I did [ C(4, 2)*C(2, 2) ] * P( 10,2 ) to arrive at n(E2) - because in my mind, the combinations-product was not accounting for any particular arrangement of numbers, while the permutations were - so I could not have been double counting.

But then in going through your explanation, I realized, that although I was using a combinations-product - the total partition of 4 people itself was being counted twice (for example, P-1 > [ (1,2), (3,4) ] and P-2 > [ (3, 4), (1, 2) ] - and the permutation (10,2) on top of this was not going to help.

Besides, I started thinking - why do combinations seem like they care for a particular order? And after some deliberation, it occured that combinations worked exactly the same as they are defined to - it was what I was looking to do with them - come up with unique-group of combinations and not combinations themselves (e.g. for my approach of using P(10,2), I could have counted either P-1 OR P-2 as they are the same partitions and I would have got the 270 answer - but they are different if you want to use C(10,2) for a final answer).

I guess a know a little more if I run into this situation again.

Leon

## #19 Re: Help Me ! » Event with repitition » 2010-10-17 10:26:56

Alright! Alright! Alright!

I think I got it!!! until someone proves this wrong ...

The number of unique partitions (k) of n objects into p groups of m objects each (where n = m*p) is -

Hence, for the example, number of uniqe partitions of a group of 4 people into 2 groups of 2 each is -

C(4,2) * C(2,2) / P(2,2) = 6*1/2 = 3

This is verified in the example on the original post.

Here is why I think "k" should have the value from (1.1) -

1. The relation in (1.1) without the P(p,p) or p! is the familiar result for the outcome of an experiment that deals with picking m objects of n successively for p times when n = m*p

2. However, in this result, the same combination of objects appear multiple times - once for each of the p groups. In our example, [ (1,2), (3,4) ] is a partition.

This is different from [ (3,4), (1,2) ] if you define Groups - A and B - so, the C(4,2) * C(2,2) is trying to count both partitions. But if you consider them to be the same partition, just arranged in a different order, then this is just a rearrangement of the 2 partitions - in the general case, the p partitions would be rearranged p! times - inflating the results by a factor p!

3. Consequently, the final result needs to be (1/p!) times the outcome from C(n, m) * C(n - p, m) * C(n - 2p, m) * ... * C(m, m)

I tried verifying this with dividing 6 objects into groups of 2 objects -

conventionally this should have been C(6, 2) * C(4, 2)  * C(2, 2) = 15 * 6 * 1 = 90

However, if you were to actually list the combinations down, you will see we start repeating combinations around in a different orde - rightly so, if you are looking for number of ways to group them in Groups A, B and C. However, if you are simply looking to split them in groups of 2, you will need to weed out 3! - 1 redundant counts - for each combination. If you do that, you should end with 15 unique partitions - which is the answer from 1.1

Thoughts bobby?

Leon

## #20 Re: Help Me ! » Event with repitition » 2010-10-15 07:08:46

You sound a lot like me!!! Never mind - I think you set me on the right path - I will get back

## #21 Re: Help Me ! » Event with repitition » 2010-10-15 05:17:31

may be not - I dont know any more

## #22 Re: Help Me ! » Event with repitition » 2010-10-15 05:07:50

OH NO - CORRECTION!!

The total number of unique partitions (k) of n objects when it is divided in p groups of m objects each for a given partition is -

k = C(n,m) / C(  C(n,m) , p )

Does that seem right?

Leon

## #23 Re: Help Me ! » Event with repitition » 2010-10-14 18:03:34

Hello Bobbym,

I meant to get back sooner - first up, thanks for the detailed explanation! It makes sense to me - I was just struggling conceptually in that for E2, I felt I could -

1.  Determine number of ways to split 4 people in groups of 2 IN ANY ORDER
2.  Determine number of ways to pick 2 numbers from 1 - 10 and assign them in a specific order to all the combinations in #1

Since #1 above had no sense of an order, I used C(4,2)*C(2,2).

Also, since #2 above sounded like it had order involved, I used P(10,2).

... giving me C(4,2)*C(2,2)*P(10,2) = 540.

However, i now realize that using C(4,2) [number of combinations of 4 objects taken 2 at a time] would cause repititions in the splits I make. For example,

P-1 P-2         P-3 P-4
P-1 P-3         P-2 P-4
P-1 P-4         P-2 P-3
================ (=3) this right here is all possible partitions of 4 people in groups of 2. However, C(4,2) continues ...
P-2 P-3         P-1 P-4
P-2 P-4         P-1 P-3
P-3 P-4         P-1 P-2
================ (=6) all these partitions in the second group here are dups of the one above

I realize now that #1 above was really the unique number of partitions I could make (each partition is a group of combinations). So, in this case (1/2)*C(4,2) would the number of outcomes of #1 and hence -

(1/2)*C(4,2)*P(10,2) = 270

Now that sounds like a desperate attempt to match the right answer - however, I would feel better if you could weigh in on the following hypothesis -

The total number of unique partitions (k) of n objects when it is divided in p groups of m objects each for a given partition is -

k = (1/p)*C(n,m)

As always, thanks a lot for the guidance!

Leon

## #24 Help Me ! » Event with repitition » 2010-10-10 18:00:32

getback0
Replies: 11

Hello,

I am sure this is a piece of cake for you guys - but for some reason - i am not able to get it right (lets not focus on the reason for now! ).

The Problem:

Four people pick a number at random, each, between 1 and 10. In how many ways can at least 2 people pick the same number?

Deliberation:

n(S) = n(Sample space) = 10 * 10 * 10 * 10 = 10,000

E = at least 2 people pick the same number

~E = no two people pick the same number
n(~E) = 10 * 9 * 8 * 7 = 5,040

n(E) = n(S) - n(~E)
= 10,000 - 5,040
n(E) = 4,960                           ..... (A)

Now that's one solution, and I get that.

My dilemma:

However, shouldnt the more direct method of defining possible outcomes for E yield this same answer? I tried but could not get it to match.

Here's what I did (clearly erroneous) and I am seeking your help to point out my error. Any help would be truly appreciated!

E = E1 U E2 U E3 U E4
where
E1 = two people pick the same number and the other two pick different numbers altogether
E2 = two people pick the same number and the other two pick the same number but different from the first one
E3 = three people pick the same number
E4 = all 4 pick the same number

AND

all these events are mutually exclusive.

Using the notation,

P(n, r) for permutations of n objects taken r at a time, and
C(n, r) for combinations of n objects taken r at a time

I feel,

n(E1) = ( P(10, 1)*C(4, 2) ) * ( P(9, 2)*C(2, 2) ) = 4,320

n(E2) = ( P(10, 1)*C(4, 2) ) * ( P(9, 1)*C(2, 2) ) = 540

n(E3) = ( P(10, 1)*C(4, 3)) *  ( P(9, 1)*C(1, 1) ) = 360

n(E4) = ( P(10, 1)*C(4, 4))                                = 10

However, with this, n(E) = 4,320 + 540 + 360 + 10
n(E) = 5,230                ..... (B)

Clearly, (A) and (B) do not agree.

What am I doing wrong? I dont want to grow any older thinking about this, but cant help it!!!!

Again, thanks for the guidance!

Leon