# 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?

• Where do you get $14\over20$ from? Also those events aren't mutually exclusive so you can't add them together. If 1 bag had all greens and the other had all reds, then you'd end up with 2 as your answer. Commented Jul 10, 2017 at 18:23
• Remember that the probability of $A$ or $B$ is the sum of the individual probabilities minus the joint probability of $A$ and $B$. There's an easier way to do this, though. Commented Jul 10, 2017 at 18:24
• In probability questions which say something similar to 'find $P(A)$ where $A$ is the event of getting at least 1 XXX', then you should almost always consider if $P(A^C)=1-P(A)$ is easier to find. Commented Jul 10, 2017 at 18:27
• @Shuri2060 That's where the $\frac{14}{20}$ came from, it's the complement of getting a green ball. Commented Jul 10, 2017 at 18:31
• @browngreen Yeah - didn't realise at first (and there's a typo on that line), but as others have mentioned, getting no reds and no greens is the way to go Commented Jul 10, 2017 at 18:32

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?

• Somehow I got confused now. Why doesn't the 1 minus the probability of no red or green ball on either draw includes the probability of drawing a red ball and a green one right? So how come we don't need to subtract it?
– Jim
Commented Jul 11, 2017 at 18:44
• Because you're not double counting it this way. When you add the probability of getting a red to the probability of getting a green, you're counting the event of getting both red and green twice. That's why you have to subtract in that case, to make up for the double counting. Commented Jul 11, 2017 at 18:48
• A selection that has a red and a green is acceptable right? So the probability to get at least 1 green indeed has the case of red + green included. And the probability to get at least 1 red indeed has the option of red + green included. So if I remove from each these cases (like mentioned by browngreen) won't I be removing completely the case of red + green from counting it? But it is an acceptable condition no?
– Jim
Commented Jul 11, 2017 at 18:52
• That's why you don't remove it from both cases. You only subtract it once. Commented Jul 11, 2017 at 18:53
• I think I got you. I got confused by the fact that we are counting as distinct combinations red from bag1, green from bag2 and red from bag2 and green from bag1. So the "at least 1 green" includes both these 2 combinations right? That's why we care about the order?
– Jim
Commented Jul 11, 2017 at 20:04

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 $\frac{400-81}{400}$ = $\frac{319}{400}$

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 $\frac{60}{400}$.

• Why are we counting the probability of Ball1,Ball2 as distinct from Ball2,Ball1 and we add to get the $\frac{60}{400}$?
– Jim
Commented Jul 11, 2017 at 10:37
• I am confused on this a bit. Why are we counting the sequence ball1, ball2 plus ball2,ball1 for getting a red and green?
– Jim
Commented Jul 11, 2017 at 18:28
• @Jim There are two distinct ways to get one green and one red, each with a probability of $30\over 400$. (Ball1-green, Ball2-red) is a different scenario from (Ball1-red, Ball2-green) so you have to add the probabilities. Commented Jul 11, 2017 at 19:58