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- bobbym
- 2013-01-15 05:12:50
Hi Doubtful;
Looks like you do not need any help with that one you are correct! Very good.
- Doubtful
- 2013-01-15 03:09:33
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
- 2013-01-10 03:55:09
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 )
- Doubtful
- 2013-01-10 00:56:36
Any 2 different children
- bobbym
- 2013-01-09 03:37:51
Hi;
Please, one question at a time.
2)
Any 2 different children? Or a specific 2 children?
- Doubtful
- 2013-01-09 03:36:19
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?
- Doubtful
- 2013-01-09 03:28:48
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
- bobbym
- 2013-01-09 03:05:39
Hi;
Yikes, of course you are right! A simple error on my part.
- Doubtful
- 2013-01-09 01:52:58
Yes i understand any one child is 1/3000
But for malnutrition children it should be 930/3000, am i right here?
- bobbym
- 2013-01-08 03:20:53
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?
- Doubtful
- 2013-01-08 03:18:13
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
- 2012-12-31 03:06:52
Me too! But I never had much luck wishing for things.
- Roosh
- 2012-12-31 02:59:27
wish i was gifted =_=
Thanks A LOT for your time ! & I'll post my answers ASAP 
- bobbym
- 2012-12-31 02:48:43
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.
- Roosh
- 2012-12-31 02:45:58
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