binomial and hypergeometric probability distributions

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April 19th 2012, 02:53 PM
hollymxox

binomial and hypergeometric probability distributions

binomial and hypergeometric probability distributions:

from a box of 12 flares, four are selected at random. If the box contains four flares that do no work:

a. what is the probability that all four will not work?
b. what is the probability that, at most, two will not work?
c. of the four flares selected, how many would you expect to not work?

probability distribution for discrete random variables:

Jason has designed a game where if you roll a prime number with two six-sided dice, you receive twice the amount of your roll, but if you roll a composite number (not prime), you must must pay the value of the roll.

a. what is the expected value [E(x)] per roll?
b. is this game fair? explain.

Thanks for the help everyone :)
April 19th 2012, 03:09 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

binomial and hypergeometric probability distributions:
from a box of 12 flares, four are selected at random. If the box contains four flares that do no work:
a. what is the probability that all four will not work?
b. what is the probability that, at most, two will not work?
c. of the four flares selected, how many would you expect to not work?

You simply must show some effort on these questions.
Suppose that $X$ is the numbers of duds in the four tested.
Then $\mathcal{P}(X=k)=\dfrac{\binom{4}{k}\binom{8}{4-k}}{\binom{12}{4}}$ .

Now you show us some work.
April 19th 2012, 03:31 PM
hollymxox

Re: binomial and hypergeometric probability distributions

I thought it would be

= C(12,4) x C(12,8)
C(12,4)

which would equal to 495.

Or would it be better to use indirect way so it would be:

1-C(12,4) x C(12,8)
C(12,4)
April 19th 2012, 03:53 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

I thought it would be

= C(12,4) x C(12,8)
C(12,4)

which would equal to 495.

Or would it be better to use indirect way so it would be:

1-C(12,4) x C(12,8)C(12,4)

None that is correct.
Study what I posted. If still do not understand then please ask for a sit-down conference with your instructor.
You appear to be really confused.
May 2nd 2012, 12:39 PM
hollymxox

Re: binomial and hypergeometric probability distributions

Hey,

could you please review my solution for the second question I asked?

Jason has designed a game where if you roll a prime number with two six-sided dice, you receive twice the amount of your roll, but if you roll a composite number (not prime), you must must pay the value of the roll.

a. what is the expected value [E(x)] per roll?
2, 3, 5, 7, and 11 are prime, 4, 6, 8, 9, 10, and 12 are composite.
Let X be the variable of the net value of each roll. So if, for instance, you roll a 5, X is 10, but if you roll a 6, X is -6.

is it correct to do it by making this table?

. X ... p(X) .... X*p(X)
-12 ... 1/36 ... -12/36
-10 ... 3/36 ... -30/36
-9 ..... 4/36 ... -36/36
-8 ..... 5/36 ... -40/36
-6 ..... 5/36 ... -30/36
-4 ..... 3/36 ... -12/36
4 ...... 1/36 ... 4/36
6 ...... 2/36 ... 12/36
10 .... 4/36 ... 40/36
14 .... 6/36 ... 84/36
22 .... 2/36 ... 44/36

Add up the third column to get E(X) = 24/36 = 2/3

b. The game is not fair since E(X) is not equal to zero.
May 2nd 2012, 01:58 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

Hey,

could you please review my solution for the second question I asked?

Jason has designed a game where if you roll a prime number with two six-sided dice, you receive twice the amount of your roll, but if you roll a composite number (not prime), you must must pay the value of the roll.

a. what is the expected value [E(x)] per roll?
2, 3, 5, 7, and 11 are prime, 4, 6, 8, 9, 10, and 12 are composite.
Let X be the variable of the net value of each roll. So if, for instance, you roll a 5, X is 10, but if you roll a 6, X is -6.

is it correct to do it by making this table?

. X ... p(X) .... X*p(X)
-12 ... 1/36 ... -12/36
-10 ... 3/36 ... -30/36
-9 ..... 4/36 ... -36/36
-8 ..... 5/36 ... -40/36
-6 ..... 5/36 ... -30/36
-4 ..... 3/36 ... -12/36
4 ...... 1/36 ... 4/36
6 ...... 2/36 ... 12/36
10 .... 4/36 ... 40/36
14 .... 6/36 ... 84/36
22 .... 2/36 ... 44/36

Add up the third column to get E(X) = 24/36 = 2/3

b. The game is not fair since E(X) is not equal to zero.

That seems to be correct.
May 2nd 2012, 02:00 PM
hollymxox

Re: binomial and hypergeometric probability distributions

Thank you :)

I am still working on the first question, thank you for helping out
May 2nd 2012, 02:05 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

I am still working on the first question, thank you for helping out

What don't you understand about my reply to #1?
$\dbinom{12}{4}=\dfrac{(12)(11)(10)(9)}{(4)(3)(2)(1 )}$
May 3rd 2012, 03:17 PM
hollymxox

Re: binomial and hypergeometric probability distributions

I g ave the flare question another shot let's see if I got any of the parts right lol, sorry I'm doing this course all alone so I dont really have anyone to turn to for more help...

a. C(12,4) = 12 x 11 x 10 x 9
4 x 3 x 2 x 1
= 11880 = 495
24
b. 12 – 4 = 8 will work
p(x>2) = 1 – p(x=0) – p(x=1)
p(x>2) = 1 – C(4,0) x C(8,4) – C(4,1) x C(8,3)
C(12,4) C(12,4)
= 1 - .1414141414 - .4525252525
= .4060606061 = 41%

c. 4 x .4060606061
= 1.6 = approximately 2 out of the 4 will not work
May 14th 2012, 11:32 AM
hollymxox

1 Attachment(s)

Re: binomial and hypergeometric probability distributions

Hey Plato, and others,

I am on this unit (that's attached below) and I am really struggling to understand it, in the first example they find the line of best fit and when I tried to find it was different... I am just pretty confused about this whole unit.

It would be great if you could help me out..

Thanks,

Holly
May 14th 2012, 03:01 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

I am on this unit (that's attached below) and I am really struggling to understand it, in the first example they find the line of best fit and when I tried to find it was different... I am just pretty confused about this whole unit.

Comment 1. This is a new question so it should be in a new thread.

2. Why do you expect any of us to down-load a file and read it. It is not our job. If you are too lazy to type out your questions, you have no reasonable expectation of help.
May 14th 2012, 03:47 PM
hollymxox

Re: binomial and hypergeometric probability distributions

That is in no way what I meant to do, it was not a lazy move. I wrote that I am having trouble with the whole line of best fit equation and such I just wanted to show the work I was working with and that is why I attached the file, it was not meant to be lazy in any way...

Would you like me to post the file in a new thread?
May 14th 2012, 05:05 PM
Plato

Re: binomial and hypergeometric probability distributions

Quote:

Originally Posted by hollymxox

That is in no way what I meant to do, it was not a lazy move. I wrote that I am having trouble with the whole line of best fit equation and such I just wanted to show the work I was working with and that is why I attached the file, it was not meant to be lazy in any way...Would you like me to post the file in a new thread?

It seems to me that is up to you.
But don't post a link. Write up the actual question.
May 14th 2012, 06:02 PM
hollymxox

Re: binomial and hypergeometric probability distributions

Well, I can't just type the one question since I need help with unit in general. But thanks for nothing.