1. ## Probability

In a box there are 22 cubes, 12 red and 10 blue.
7 cubes are chosen at random.
Find the probability for the following events:

a. Among the seven cubes there are 4 blue cubes and 3 red cubes

There are others, but I've tried to do them, and for a I get something like..
0.00780 or something like that.
Which I think might be wrong..
So if anyone could just show me what to do, I think I can do the rest myself, thanks

2. Hi Fnus,

The probability of getting one blue cube is $\displaystyle \frac{10}{22}$ . Due to the fact that the cubes aren't being replaced the probability of getting another blue cube is $\displaystyle \frac{9}{21}$ and so on and so fourth.

This gives an answer of $\displaystyle \frac{10}{1539} \approx 0.0065$

3. Hmm, I think that was what I did..

And I'd multiply them, right?

Maybe I just pressed something wrong.
I'll try again, thanks.

4. Here is the correct answer:
$\displaystyle \frac {{10 \choose 4}{12 \choose 3}}{{22 \choose 7}} =0.270897832817337$

5. Originally Posted by Moo
I thought it was that...

Oohh...I'm gonna have to deal with that guy and all these other weird distributions this fall in my hardcore stats class...once I survive probability

--Chris

6. Originally Posted by Plato
$\displaystyle \frac {{10 \choose 4}{12 \choose 3}}{{22 \choose 7}} =0.270897832817337$