Combinations

[cluelessstudent]

My teacher never explained how to do this with combinations. I understand simple combinations. And i know how to solve these drawing a tree. But how would you do this using combinations:

5 women and 3 men are trying out for a 4 member bowling team. What is the probability that the team consists of 2 women and 2 men?


and would the strategy to do that question be used for this one:

A bag contains 5 marbles, 2 white and 3 red. You are going to chose three marbles. What is the probablity of chosing a white marble first, and two red following?

 

[pka]

For #1, \(\displaystyle \L \frac {{5 \choose 2} {3 \choose 2}} {{8 \choose 4}}\).

For #2 \(\displaystyle \L
\frac{2}{5} \times \frac{3}{4} \times \frac{2}{3}\)

 

[cluelessstudent]

For #2 \(\displaystyle \L
\frac{2}{5} \times \frac{3}{4} \times \frac{2}{3}\)

That's what i thought, but i remember her saying something abotu putting it in nCr form. I was unsure about that. Did i just hear her wrong? Or is there a way to put it in that form?

 

[stapel]

i remember her saying something abotu putting it in nCr form....Did i just hear her wrong?

There is no way for us to know what your instructor might have said.

If you need to solve this exercise in some other manner, please reply showing what you have tried. Thank you.

Eliz.

 

[cluelessstudent]

There is no way for us to know what your instructor might have said.

If you need to solve this exercise in some other manner, please reply showing what you have tried. Thank you.

Eliz.

I know that you have no way to know what she said, i guess i was "thinking outloud". I wish i could reply with what i have tried, i just have no clue on where to begin. Like with #2, i'm unsure if i need to put:

5C2 x 4C3 x 3C2 or 5C2 x 2C1 x 5C3 x 3C2

Can it be done in this form?
I know how to do it like the poster above, but i'm pretty sure it has to be in nCr form. And i have no clue on where to start with this.

 

[JakeD]

Using the hypergeometric distribution shows how to put #2 in nCr form similar to #1.

After the draw of the first marble, the next two draws are from a hypergeometric distribution with 1 object of the first type and 3 objects of the second type. (The hypergeometric is used with sampling without replacement, which this is.) Using the formula for selecting 2 objects of the second type from that distribution, the probability is

\(\displaystyle \Large \frac{{1 \choose 0}{3 \choose 2}}{{4 \choose 2}}\).

And the total probability including the first draw is

\(\displaystyle \Large \frac{2}{5} \times \frac{{1 \choose 0}{3 \choose 2}}{{4 \choose 2}}\).

Evaluating that yields the same probability as pka.

If you were not taught the hypergeometric distribution, it's possible to derive that formula using a direct argument (but no one actually would).

 

[Denis]

5 women and 3 men are trying out for a 4 member bowling team. What is the probability that the team consists of 2 women and 2 men?

Probability is 0; all 3 men will be chosen, since they're better bowlers :wink:

"5C2 x 4C3 x 3C2 or 5C2 x 2C1 x 5C3 x 3C2"

5
2 : that's same as 5C2 ; just a different way of "showing"