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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?

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    $\begingroup$ 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. $\endgroup$
    – Shuri2060
    Jul 10, 2017 at 18:23
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    $\begingroup$ 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. $\endgroup$ Jul 10, 2017 at 18:24
  • $\begingroup$ 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. $\endgroup$
    – Shuri2060
    Jul 10, 2017 at 18:27
  • $\begingroup$ @Shuri2060 That's where the $\frac{14}{20}$ came from, it's the complement of getting a green ball. $\endgroup$
    – browngreen
    Jul 10, 2017 at 18:31
  • $\begingroup$ @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 $\endgroup$
    – Shuri2060
    Jul 10, 2017 at 18:32

3 Answers 3

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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?

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  • $\begingroup$ 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? $\endgroup$
    – Jim
    Jul 11, 2017 at 18:44
  • $\begingroup$ 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. $\endgroup$ Jul 11, 2017 at 18:48
  • $\begingroup$ 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? $\endgroup$
    – Jim
    Jul 11, 2017 at 18:52
  • $\begingroup$ That's why you don't remove it from both cases. You only subtract it once. $\endgroup$ Jul 11, 2017 at 18:53
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    $\begingroup$ 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? $\endgroup$
    – Jim
    Jul 11, 2017 at 20:04
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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 $\frac{400-81}{400}$ = $\frac{319}{400}$

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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 $\frac{60}{400}$.

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  • $\begingroup$ Why are we counting the probability of Ball1,Ball2 as distinct from Ball2,Ball1 and we add to get the $\frac{60}{400}$? $\endgroup$
    – Jim
    Jul 11, 2017 at 10:37
  • $\begingroup$ I am confused on this a bit. Why are we counting the sequence ball1, ball2 plus ball2,ball1 for getting a red and green? $\endgroup$
    – Jim
    Jul 11, 2017 at 18:28
  • $\begingroup$ @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. $\endgroup$
    – browngreen
    Jul 11, 2017 at 19:58

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