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An urn contains 5 red and 5 black balls. A ball is drawn at random and it's color is noted and is then returned to the urn. Moreover,2 additional balls of the same color are put in the urn and then the ball is drawn at random.What is the probability that the second ball is red?

Here is the approach I am using but I am getting confused.The above problem is based on the concept of Law of Total Probability. Given that events are mutually exclusive and exhaustive.

E1 = Both balls are black

E2 = Both balls are red

A = The second Ball is red

Sample Space S = Two balls drawn at random (after 2 balls of same colour are added) = $\dbinom{12}{2}$

Is my initial approach correct? If not, how do I approach this?

Any help will be greatly appreciated. Thanks!

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  • $\begingroup$ Just $\frac{1}{2}$ be symmetry $\endgroup$ Sep 11, 2021 at 9:15

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This problem is an equivalent of Polya urn if you find it useful to find more information.

I think that your approach in searching the sample space might be a little wrong because there is only four cases: both same color (2 cases), one red then one black and one black then one red.

The probability of taking a red ball in the first draw is $\frac{1}{2}$ because there are 5 red balls and 5 black, so there are 10 in total, given that it is a random choice the avery event is equiprobable and the probability we are searching for is favorable cases over all cases that is $\frac{5}{10}=\frac{1}{2}$, note that this is true also for the probability of taking a black one in the first draw

Then for finding the probability of E1, E2 are equivalent and i will do it like this:

Because we are in the case of both being the same color we know that we add two more balls of the first color draw, then the favorable cases will be $7=(5+2)$ and total will be $12= (10+2)$, using that is a random choice the probability of the second one being the same color as the first one is $\frac{7}{12}$ ,and using the probability of the first one being red we have that $P(E1)=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$ and in the analogous case on E2 we have $P(E2)=\frac{1}{2}\cdot\frac{7}{12}=\frac{7}{24}$.

With and similar approach we can find A, we all ready have the probability of the second ball being red if the first one is red $\frac{7}{24}$ that is because it is condicional to the first one being red, then we are only missing what happened if the first one is black, we will have that the favorable cases in the second draw will still be 5 and the total 12 (as we added two black balls) and it is conditional to the first draw being black so the probability will be $\frac{1}{2}\cdot\frac{5}{12}=\frac{5}{24}$

Finally we have that this two cases are disjoint and add up to the total so the probability of A will be $\frac{5}{24}+ \frac{7}{24}=\frac{12}{24}$.

And useful way to see if it's okay you can add up the probability of the second one being black (that is easy to see that is equivalent to this one) and it should add 1, and it does it (it doesn't prove anything its just to check).

Hope this is helpful

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It helps to draw a tree diagram. For the first ball, there is a $50\%$ chance it is red and $50\%$ chance it is black.

If the first ball is red, the chance the second ball is red is $7/12$, and the chance it's black is $5/12$.

If the first ball is black, the chance the second ball is red is $5/12$, and the chance it's black is $7/12$.

All together there are four outcomes (red-red, red-black, black-red, and black-black). You need to find the probabilities of "red-red" and "black-red" and add them together.

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