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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: 103,681

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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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: 103,681

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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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: 103,681

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.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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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: 103,681

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.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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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: 103,681

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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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: 103,681

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?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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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: 103,681

Hi;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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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: 103,681

Hi;

Please, one question at a time.

2)

Any 2 different children? Or a specific 2 children?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**Doubtful****Guest**

Any 2 different children

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,681

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 )

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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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: 103,681

Hi Doubtful;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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