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**katerinaaa****Member**- Registered: 2010-01-27
- Posts: 4

Hi all, I have a problem with probabilities but I can't solve it.

The problem is :

"I have 3 folders. A folder of them has a letter while the others are empty.

The probablity to find the letter in the folder, if we touch the top of the folder is p_j , if there is letter in this folder.

We check folder 1 and we did not touch the letter.

Which is the possibility the letter to be inside the folder 1 ??"

I don't understand, the possibility is 1/3 again I think.

Could you help me please ??

thanks a lot

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi katerinaaa;

Are asking what is the probability of choosing a folder with the letter in it if there are 3 folders and only one folder has a letter in it?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Three things could have happened:

1) The letter is in folder 1, and you missed it when you checked earlier.

This has probability 1/3 * (1-p_j).

2) The letter is in folder 2. This has probability 1/3.

3) The letter is in folder 3. This also has probability 1/3.

And for completeness:

[4) The letter is in folder 1, and you found it earlier. Probability 1/3 * p_j.]

We already know the 4th option didn't happen, so the probability of the letter being in folder 1 is

or more simply,

This means that the probability of the letter being in folder 1 is:

1/3, if p_j = 0.

0, if p_j = 1.

1/5, if p_j = 1/2.

etc.

Why did the vector cross the road?

It wanted to be normal.

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**katerinaaa****Member**- Registered: 2010-01-27
- Posts: 4

thanks a lot for your help!

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**leealden06****Member**- Registered: 2010-09-06
- Posts: 4

In probability, the problem of birthday, or accession to the birthday paradox probability that a number of people chosen at random for a couple of them have the same problem birthday.The is to calculate approximately the probability that a room of people, have at least two the same birthday.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,072

hi leealden06

I've met this before. Are you asking how to calculate this?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**frankMwarimbo****Member**- Registered: 2011-02-26
- Posts: 1

whats is the difference between independent events and dependent event?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi frankMwarimbo;

Two events are independent if the occurrence of one has no effect on the probability of the other.

Please take a look here:

http://www.mathsisfun.com/definitions/i … event.html

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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