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amuthamku01 · Jun 21, 2010, 11:59 PM

a box contains six red balls and four green balls. Four balls are selected at random from the box.what is the probability that two of the selected balls balls are red and two are green?

galactus · Jun 22, 2010, 03:36 AM

Originally Posted by amuthamku01

a box contains six red balls and four green balls. four balls are selected at random from the box.what is the probability that two of the selected balls balls are red and two are green?

$\frac{C(6,2)C(4,2)}{C(10,4)}$

ebaines · Jun 22, 2010, 06:25 AM

Galactus: I think you need to multiply your answer by C(4,2), to accoyunt for the freds and greens being selected in any order. I approach the problem this way:

Start by thinking through the probability of picking two reds follwed by 2 greens: that probability is:

$<br /> \frac {6 \cdot 5} {10 \cdot 9} \cdot \frac {4 \cdot 3} {8 \cdot 7}<br />$

I think this is the probability that you calculated. But, realizing that these 4 balls could be selected in C(4,2) ways, we need to multiply this by C(4,2):

$<br /> \frac {6 \cdot 5} {10 \cdot 9} \cdot \frac {4 \cdot 3} {8 \cdot 7} \cdot \frac {4 \cdot 3} 2<br />$

galactus · Jun 22, 2010, 06:37 AM

Are you sure about that, ebaines?

Doing it another way I get the same thing. My previous way, it is already taken into account is it not?

My answer is the same as yours. Here are two different methods.

$\frac{C(6,2)C(4,2)}{C(10,4)}=\frac{3}{7}$

Another way:

$6(\frac{6}{10})(\frac{5}{9})(\frac{4}{8})(\frac{3} {7})=\frac{3}{7}$

Here we have to multiply by 6 to account for the arrangements of 4 items with two of each the same $\frac{4!}{2!2!}=6$