Math Is Fun Forum

lindah

Member

Registered: 2010-01-25

Posts: 121

Random Variables question

Hi,

I am experiencing a brain freeze on how to approach the following question:

A group of five components contains two defective components. If components are taken from the five at a time and tested, let X be the trial at which the second defective is found.

a) What is the probability distribution of X?

b) Find the mean of X

My initial approach was to list the possible ordered scenarios {DDNNN, NNNDD, NDNDN} which would have 5! possible sequences, but then I am not sure how to derive a probability distribution.

Can someone point me in the right direction?

Thank you in advance for your help

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Random Variables question

hi lindah

My first thought was that I ought to recognise this as a text book distribution and I could not see which one.

But your suggestion of listing the possibilities is a good one.  Sometimes the number of possibilities is just too big to list them all but, for this question quite small.  You should have done it!  If you want the satisfaction of solving this yourself, then read as little or as much as you need from what follows.

It's not 5! however.  Because one D is just like the other and similarly, the  nonDs are the same,  you have to divide by 2! and 3!

so

Whoops, I uploaded the images in the order I wanted but the software has reversed them.

I've listed them below in image 2.  (D = defective,   G = good)

And image 1 is the distribution, showing X, Frequency of X, and Probability of X.

You can get the mean by either

or by

Hope that helps

Bob

Last edited by Bob (2010-08-26 08:23:36)

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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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Random Variables question

Hi lindah;

Welcome to the forum. The mean of X in this case is called the expected number and is written E(X). In case you do get one with a sample space bigger than 10 there is a distribution to help you.

You can compute it like this.

So the mean of X is E(x) = 4. You can expect 4 tries to get both defectives. You can verify this answer by using Bob's table and manually adding up the number of tries and taking the average.

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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.

lindah

Member

Registered: 2010-01-25

Posts: 121

Re: Random Variables question

Hi Bob and bobby,

I see I've overlooked repeating combinations.

Thank you very much for the clear explanations!!

Regards
Linda

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Random Variables question

Hi lindah;

Your welcome. Bob and bobby?! Sort of reminds me of Starsky and Hutch, not that awful movie, you know the old tv show!

---

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.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,623

Re: Random Variables question

Hi lindah,

Glad to be of help.

Hi bobbym,

Bags I get to drive the Gran Torino!

Bob

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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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob