Probability of getting at least 1 red or green ball I am working on the following problem:  

We have 2 bags that each contain 3 yellow, 4 blue, 5 red, 6 green and
  2 black balls.   In a simultaneous draw what is the chance of getting
  at least 1 red or 1 green ball?  

My approach:
Find probability of getting at least 1 green ball:
$(\frac{14}{20})^2 = (\frac{7}{10})^2= \frac{49}{100}$ is the probability of not getting a green ball in the draw hence $1 - \frac{49}{100} = \frac{51}{100}$ is the probability of getting at least 1 green ball.  
Find probability of getting at least 1 red ball:
$(\frac{15}{20})^2 = (\frac{3}{4})^2 = \frac{9}{16}$ is the probability of not getting a red ball in the draw hence $1 - \frac{9}{16} = \frac{7}{16}$ is the probability of getting at least 1 red ball.  
Therefore what we are looking for is the sum of these probabilities i.e.
$\frac{51}{100} + \frac{7}{16} = \frac{816 + 700}{1600} =\frac{1516}{1600} = \frac{379}{400}$  
But my notes say $\frac{319}{400}$
Is this a typo or is my solution wrong?

 A: You can use counting by complement to greatly simplify this problem. The original statement asks for at least 1 red ($\geq 1$ red) OR at least 1 green ($\geq 1$ green). 
Condition:
$\geq 1$ green OR $\geq 1$ red
Negating this, we get cases where this condition is violated (by DeMorgan's Law):
$0$ green AND $0$ red
There are $20$ choices in total for each bag. If you eliminate red and green selections, you are left with $9$ balls to choose from each bag for a probability of $\frac{9}{20}$ * $\frac{9}{20}$ = $\frac{81}{400}$.
The cases where this is NOT violated, then, is $\frac{400-81}{400}$ = $\frac{319}{400}$
A: You must subtract the probability of getting exactly one green ball and one red ball, since this was counted twice.
The probability of Ball 1 being green and Ball 2 being red is $\frac6{20}\cdot\frac5{20}=\frac{30}{400}$ 
The probability of Ball 2 being green and Ball 1 being red is the same, so in total you are subtracting $\frac{60}{400}$.
A: Your solution is wrong because you forgot to subtract the joint probability of drawing a red ball and a green one. However, here's the best way to do this problem.
The probability of at least $1$ red or green ball is $1$ minus the probability of no red or green ball on either draw. Thus:
$1-(\frac{9}{20}\cdot\frac{9}{20}) = 1-\frac{81}{400}=\frac{319}{400}$
Does that make sense?
