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**Roosh****Member**- Registered: 2012-12-30
- Posts: 5

Hello everyone

am a beginner and have been trying to solve this one but couldn't :S

'In a population of 3000 children under 5 , those with malnutrition are 31%, if 5 children are selected randomly what is the probability of:

a)3 children with malnutrition ' ?

there's a second part to this question but I'll leave it for later ..

Thanks in advance

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi Roosh;

Welcome to the forum.

There are a couple of ways to do this:

1) By direct calculation

2) Or by using the hypergeometric distribution.

3) You could estimate the answer using a normal distribution.

4) You could use a binomial distribution to estimate it.

Which is quite close to the exact answer.

5) Computer simulation. Ran a simulation of 1000000 and got 142413 cases of 3 out of 5. For .1424 which is also close to the exact answer.

Bet the next question is a range say from 1 to 3 picked or something like that.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Roosh****Member**- Registered: 2012-12-30
- Posts: 5

Hello Bobbym,

Thank you for your quick reply ! i really appreciate it

the other part is

'at most 2 are malnourished' ?

there's another problem, it goes like

'The number of bacteria in culture growing by 16 each 4 days, in one day,

a) Find the probability of 2 growing bacterias

b) Find the probability of more than 3 growing bacteria '

i don't know what's up with me , but i can't seem to find my way around probabilities , & it's really devastating .

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Okay, let's do this one.

the other part is 'at most 2 are malnourished' ?

At most two, just means 0,1 or 2. We can handle each case one at a time to get the exact answer.

The heart of any combinatorics calculation is getting the same answer by a totally different method. Only then can you have faith in the answer.

I ran off a simulation of 1000000 trials and got .82355 which is very close to the exact answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Roosh****Member**- Registered: 2012-12-30
- Posts: 5

You are a genius ! Thanks a lot .

guess i need to go through the material again and again to let it settle in :S

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

You are a genius !Thanks a lot

My mother thought I was, I sure showed her!

Actually, I am a guy who has been doing these for a long, long, time. I still get them wrong though.

Try to do the other problem on your own and then come back.

What material are you studying.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Roosh****Member**- Registered: 2012-12-30
- Posts: 5

Moral of the story, never question what mothers think , because they just know! she definitely is right

*sigh* I'll give it another look then post whatever i come up with lol

It's a short Bio-statistics course , as a part of a postgraduate education , but math & statistics have always given me hard time hehe

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

It looks like a Poisson distribution problem.

We were just talking about a fellow who had an easy time with math. He was just about the smartest man that ever lived. The rest of us will have to struggle through it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Roosh****Member**- Registered: 2012-12-30
- Posts: 5

wish i was gifted =_=

Thanks A LOT for your time ! & I'll post my answers ASAP

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Me too! But I never had much luck wishing for things.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Doubtful****Guest**

Hi,

Could you please help me with this question:

In your early post you mentioned one solution was:

1) By direct calculation

10 * (930*929*928*2070*2069) / (3000 * 2999 * 2998 * 2997 * 2996)

based on the above, would the below be right ?

1. Is the probability of choosing one child out of the total = 1 / 3000

2. Is the probability of choosing 2 children out of the total = 2 / 3000 or is it (1 * 2) / (3000 * 2999)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi Doubtful;

1. Is the probability of choosing one child out of the total = 1 / 3000

Child with malnutrition, 310 / 3000. Any particular child 1 / 3000.

Do you understand up to here?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Doubtful****Guest**

Yes i understand any one child is 1/3000

But for malnutrition children it should be 930/3000, am i right here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Yikes, of course you are right! A simple error on my part.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Doubtful****Guest**

No problem at all... I appreciate your quick response.

About my other question:

2. Is the probability of choosing 2 children out of the total = 2 / 3000 or is it (1 * 2) / (3000 * 2999)

I have one more question:

3. What is the probability of choosing 2 malnutrition children out of all the children: (2 * 930) / 3000

**Doubtful****Guest**

One more doubt:

4. You mentioned the probability of choosing one child is 1/3000 and the probability of choosing one malnutrition child is 930/3000

How is it possible that the probability of choosing one child is less than the probability of choosing a malnutrition child?

should'nt the probability of choosing one child be more than choosing one malnutrition child?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Please, one question at a time.

2)

Any 2 different children? Or a specific 2 children?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Doubtful****Guest**

Any 2 different children

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Since you said any two children,

(3000 / 3000 ) for the first child and (2999 / 2999) for the second child which equals 1. Not a very interesting pick.

A specific two kids,

(2 / 3000 ) ( 1 / 2999 )

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Doubtful****Guest**

Thanks lot for your reply.

Can you pls help me with my other question:

What is the probability of choosing 2 malnutrition children out of all the children is it: (930/3000) * (929/2999)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi Doubtful;

Looks like you do not need any help with that one you are correct! Very good.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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