More probability!

getback0 · 2010-10-28 17:14:58

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

Neither answer matches the key!!!!!

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

getback0 · 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

bobbym · 2010-10-28 18:24:42

Hi Leon;

For Q - #3:

This is a discrete distribution so we have a sum.

But you are not done yet.

Q - #4: Is fine. You can do it in two ways one by expanding the two expressions or by using a Venn diagram.
(You are right though, I find the problem to be a little loose)

bobbym · 2010-10-28 20:11:09

Hi Leon;

Q - #1.

From the ranks 1 to 13 you pick 2. They will automatically be different. Do you see why?
Now you give each the 6 combinations. For a pair of deuces for instance (2h 2s) ( 2h 2c) ( 2h 2d) (2s 2c) (2s 2d) (2d 2c).
You have 11 ranks left pick one and any suit.

Please go here to get a feel for all of these card problems.

http://en.wikipedia.org/wiki/Poker_probability\

Q - #2.

You have the 3 probabilites. This is a little tricky, draw a Venn diagram to see the probabilites.
The diagram suggests a 3x3 simultaneous set of linear equations to solve. You will need to stop and think right here.

You get:

1/12 +1/6 + 1/4 = 1/2 that choose two plans so 1/2 do not.

getback0 · 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?

bobbym · 2010-11-06 14:17:24

Hi;

I agree about Q4 sometimes you just have to surmise what the question wants, especially with book problems being presented by teachers.

For Q3 yes I believe that is correct. You notice that if you sum that series from 0 to infinity the sum = 1 which is what you would expect.

For Q2, you are not looking at the Venn diagram with the right idea. The problem says.

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?

If someone chooses A he has to choose B or C with it. He cannot just have A!
So there is really only A intersection B and A intersection C.
This has to be the 1 / 4 they are talking about. Same for the others.

getback0 wrote:

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

Now, you have said something that does not hold up.
Everyone and I mean everyone makes mistakes in combinatorics and probability.

Around these parts there is a guy who is the best.
Works at a large secret installation and frequently is called in when math guys and physicists mess up.
Can solve mostly any problem, he says, "bobbym, Yer know why them fellas do all that there topology and logic?"
"No, I do not, I says.( mocking his hillbilly accent )" "It's because they's a scared of combinatorics and probability."

Here is what Gomer Pyle had to say about me. "Yer know bobbym, you is the only fella I knows that can solve three problems
using three different methods and make four mistakes."

If you do a lot of reasoning about probability - combinatorics your head is going to explode. Seriously your only chance is to see
and use the methods that work. Memorize them and train your mind to think in the same fashion as the good problem solvers.
That is why I directed you to that page. Look at what he has done. Do it the same way, ask questions later.

There is no substitute for doing it the right way in these two fields.

Now as regard to your question. How are you thinking about it, that is different from mine?

getback0 · 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

bobbym · 2010-11-06 15:02:54

Hi getback0;

Are the other answers understandable to you?

As regarding my lecturing... You are on your way as a matter of fact you are good.
Do not let errors or slight misconceptions deter you. I make them too.

Here is a better proof for Q4, it uses the fact that:

is true whether A and B are independent or not. It looks the same because of cancellations.
It is also kind of experimental.

getback0 · 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

bobbym · 2010-11-07 14:55:44

Glad to help! Pop in to let me know what you are doing.

getback0 · 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

bobbym · 2010-11-07 17:56:33

Hi Leon;

It is important to understand that you can do these using permutations rather than combinations just as long as you are consistent.
That is how I used to do them. Got about 50 - 50 right! Want to know why?
Because know one else did them that way so I could never ask for help or look in a book.

Let me explain what he did up there. Maybe it will clear up some problems.
First of all he picked 2 different pairs.
From the set { 13,12,11,10,9,8,7,6,5,4,3,2,1} He chose two.
They were automatically different, because there is only one of each.
This is important because they way he picks means there is no chance of getting 3 of a kind, 4 of a kind, or a full house. That would cause overcounting.

Now there are 13 * 12 ways to pick them when order counts but 22 and 44 is the same hand as 44 and 22.
So you (13 * 12) / 2 which is 13 choose 2.
Each pair has six combinations as I showed you that is where the 4 choose 2 squared comes from.
Four cards are done but only 2 ranks. For instance with the 22 and 44 out the set now looks like this:

{1,3,5,6,7,8,9,10,11,12,13} exactly 11 of them are left, he chooses 1, 11 choose 1 and there 4 of them.
We are done!

Math Is Fun Forum

#1 2010-10-28 17:14:58

More probability!

#2 2010-10-28 17:50:03

Re: More probability!

#3 2010-10-28 18:24:42

Re: More probability!

#4 2010-10-28 20:11:09

Re: More probability!

#5 2010-11-06 13:17:24

Re: More probability!

#6 2010-11-06 14:17:24

Re: More probability!

#7 2010-11-06 14:58:37

Re: More probability!

#8 2010-11-06 15:02:54

Re: More probability!

#9 2010-11-07 14:28:16

Re: More probability!

#10 2010-11-07 14:55:44

Re: More probability!

#11 2010-11-07 15:11:15

Re: More probability!

#12 2010-11-07 17:56:33

Re: More probability!

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