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Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?

So far I have this:
Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132
Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132
--> $$\frac{55}{132} = \frac{5}{12}$$

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closed as off-topic by asdf, gt6989b, Parcly Taxel, Xander Henderson, Asaf Karagila Jul 26 '18 at 9:09

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    $\begingroup$ So the first ball is green or it is not green... $\endgroup$ – Robert Z Jul 25 '18 at 18:46
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    $\begingroup$ Yes, you are correct! $\endgroup$ – Robert Z Jul 25 '18 at 18:47
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    $\begingroup$ correction Wolf: : You have $$\frac{20 + 35}{132} = \frac{55}{132}= \frac 5{12}$$ $\endgroup$ – Namaste Jul 25 '18 at 18:48
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    $\begingroup$ $$P=P_1+P_2=\frac {5}{11}\frac {7}{12}+\frac {4}{11}\frac {5}{12}=\frac{5}{12 \times 11}(4+7)=\frac{5}{12 \times 11}(11)=\frac {5}{12}$$ $\endgroup$ – Isham Jul 25 '18 at 18:50
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    $\begingroup$ Possible duplicate of Second marble is of same color $\endgroup$ – Rahul Goswami Jul 25 '18 at 19:11
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Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.

Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina. The tree of possibilities is tedious and messy, but the answer is still $5/12$.

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Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $\dfrac{5}{12}$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:

$$\require{cancel} \dfrac{k}{n}\dfrac{k-1}{n-1}+\dfrac{n-k}{n}\dfrac{k}{n-1} = \dfrac{k(k-1)+k(n-k)}{n(n-1)} = \dfrac{k \cancel{(n-1)}}{n \cancel{(n-1)}} = \dfrac{k}{n}$$

Which is the same probability that Peter has for drawing a green ball.

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    $\begingroup$ "Angela's draw is actually independent from Peter's draw" - it's not. You're probably misusing the word "independent". The probability distribution for Angelina's draw is the same as it would be if Peter did not draw, but that's not what "independent" means. If they were independent, we would have equalities like P(Angelina draws red AND Peter draws red) = P(Angelina draws red)*P(Peter draws red). $\endgroup$ – user2357112 Jul 25 '18 at 22:04
  • $\begingroup$ @user2357112 I corrected my wording. Thank you for pointing it out. $\endgroup$ – InterstellarProbe Jul 25 '18 at 22:51
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HINT

Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.

Then group them together in one final expression.

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