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

N

Navyguy

Junior Member
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Jul 24, 2006
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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
 
Calculate \(\displaystyle \frac{{\left( {\begin{array}{c}
4 \\
2 \\
\end{array}} \right)\left( {\begin{array}{c}
5 \\
1 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
9 \\
3 \\
\end{array}} \right)}}.\)
 
Re: ok

Navyguy said:
(6)(6)/12=3, is this right.1/3
Click to expand...
WRONG!
You must learn how to calculate binomial coefficients.
\(\displaystyle \L
\left( {\begin{array}{c}
N \\
k \\
\end{array}} \right) = \frac{{N!}}{{\left( {N - k} \right)!\left( {k!} \right)}}\)
 
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.
 
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
 
thanks

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