box of balls: probability two red and one green are drawn is

[Navyguy]

A box contains 4 red and 5 green balls. Three balls are drawn. The probability 2 red and 1 green are drawn is:

1. 1/3
2. 4/9
3. 5/14
4. 3/5

Ok well you have 9 balls all together so 4/9=.44 and 5/9=.55 so my answer would be 4out 0f 9 times 4/9. I this answer right. thanks for checking

 

[pka]

Calculate \(\displaystyle \frac{{\left( {\begin{array}{c}
4 \\
2 \\
\end{array}} \right)\left( {\begin{array}{c}
5 \\
1 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
9 \\
3 \\
\end{array}} \right)}}.\)

 

[Navyguy]

ok

(6)(6)/12=3, is this right.1/3

 

[pka]

Re: ok

(6)(6)/12=3, is this right.1/3

WRONG!
You must learn how to calculate binomial coefficients.
\(\displaystyle \L
\left( {\begin{array}{c}
N \\
k \\
\end{array}} \right) = \frac{{N!}}{{\left( {N - k} \right)!\left( {k!} \right)}}\)

 

[Navyguy]

ok, well thinks for the formula

I will try and use this formula with this problem. Funny that my text book has no such fomula. thanks for helping.

 

[galactus]

Another way to look at it is the chance of drawing the first red is 4/9

The chance of drawing the second red is 3/8

The chance of drawing the 1 green is 5/7

(4/9)(3/8)(5/7)=5/42

Multiply by 3 to account for the combinations.

3(5/42)=5/14

 

[Navyguy]

thanks

I like the way you put it, it looks like a better way of doing this problem and moe.