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Topic review (newest first)

bob bundy
2012-10-06 06:19:02

Thanks.  yes, that's my answer too.  My tree is still ok but the probabilities on the branches are not equal.

Skyblast72: Sorry.  Revised diagram below.

I had disregarded the number of males and females available.

There are six outcomes but the probability of each is not 1/16.

Let's say the first chosen is a male.  There are 20 to choose from out of 35 students altogether.  So P(M) = 20/35.

Now suppose the next chosen is also male.  There are 19 to choose from out of 34 students so P(M) = 19/34

Then if you were to choose a female next that would be P(F) = 15/33

And the last female would be P(F) = 15/32

Thus P(MMFF in that order) = (20x19x15x14)/(35x34x33x32)

If you choose in a different order you get the same calculation but the numbers come in a different order.

eg P(MFMF) = 20/35 x 15/34 x 19/33 x 14/32

All six outcomes have this same calculation so  P(two Ms and two Fs in any order) =



Hope I haven't confused you with my earlier post.

Bob

bobbym
2012-10-06 05:52:32

Hi Bob;

I am using a committee problem template.

http://www.analyzemath.com/statistics/s … ity_3.html
problem #2

http://www.algebra.com/algebra/homework … 52126.html

The ordinary tree does not contain the structure of a committee.

See post #4 for the answer. Simpler to use the formula. The answer is:

bob bundy
2012-10-06 05:43:56

Hi Skyblast72 and bobbym,

I think 6/16 is correct.  What are you getting please, bobbym (+ method) ?

Wait a mo.  Did you get 0.0635...

I may have made an error here.  Whoops.

My re-calc is 0.3810

If bobbym agrees I'll have to explain how to modify the tree diagram,  (it still works but the probabilities are more tricky to work out)

Bob

bobbym
2012-10-06 04:42:24

Hold on, there are many incorrect assumptions in those statements.

I'll just be happy to pass this class, it's so frustrating.

Math is hard. Get that sunk in first. It takes a lot of effort and practice. I have been doing it for 100 years and still am lost.

Wish my mind worked like yours!!

What good would that do you? Then there would be two dummies on this forum. Stick with your own mind you are better off. Check this out:

http://media.hamptonroads.com/cache/files/images/blogs/133901.jpg

Who says I am right? In mathematics we prove things for ourselves. We do not ever take anyone's word no matter how great.

There are 4 ways of doing things. The right way, the wrong way, my way and everyone else's way.

If you want I will give you my solution. You will have to decide which one you want. See post #4 for my way.

Skyblast72
2012-10-06 04:37:03

Ok I used the diagram.  I never know when to use which formula!!! I'll just be happy to pass this class, it's so frustrating.  Wish my mind worked like yours!!

bobbym
2012-10-06 04:33:15

4)
A school club consist of 20 male students and 15 female students.  If 4 students are selected to represent the club in the student government, what is the probability 2 will be female and 2 will be male?

That is a commitee problem and there is a formula for those.

Skyblast72
2012-10-06 04:31:41

4th one with tree diagram

bobbym
2012-10-06 04:22:07

What question are you working on?

Skyblast72
2012-10-06 04:18:59

6/16=.38

6 possible outcomes 16 total outcomes is how I figured it.  .38 is one of the answer choices.

bobbym
2012-10-06 03:37:05

Hi;

I am not getting .38

Skyblast72
2012-10-06 01:59:29

.38 is what I got  Thanks

Skyblast72
2012-10-06 01:50:05

I get 16 using the tree diagram

bob bundy
2012-10-04 04:10:41

That's what I got too.  smile

Now for Q4.

4)
A school club consist of 20 male students and 15 female students.  If 4 students are selected to represent the club in the student government, what is the probability 2 will be female and 2 will be male?

There are formulas for this but I think you will get a better understanding if you see a diagram.

You've got to choose 4 students and at each choice there are 2 possibiliities for the outcome, M or F.

The diagram is called a Tree Diagram.  For choice number one you draw two branches, label one M and the other F

From each end you draw another set of two branches, again marking them M and F, giving four outcomes so far {MM, MF, FM, FF}

Continue like this for two more choices.  The final diagram now has 2 x 2 x 2 x 2 = 16 outcomes.

Below I've given you a start with the diagram but I've left some labels and outcomes for you to complete.

You should be able to do



Later edit.  This isn't right!  See my post 34 for the corrected version.



Bob

Skyblast72
2012-10-04 01:00:08

I got m=10

bob bundy
2012-10-03 17:56:05

Ok.  I'll still be here.

Bob

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