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

]]>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:

]]>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

]]>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:

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.

]]>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.

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

]]>I am not getting .38

]]>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

]]>Bob

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