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I have a bag containing $20$ red balls and $16$ blue balls. I uniformly randomly take balls out from the bag without replacement until all balls of both colors have been removed. If the probability that the last ball I took was red can be represented as $\frac{p}{q}$, where $p$ and $q$ are coprime positive integers. Find $p+q$

My solution: as the last ball is red, that means we have drawn $19$ red balls and $16$ blue balls so far. The total no of ways of doing it is $35 \choose19$. Now total no of ways of drawing all the balls is $36 \choose 20$. The probability is

$$\frac{35\choose19}{36\choose20}=\frac{5}{9}$$ But the answer is $\frac{4}{9}$. Can you please correct my solution?

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This is a called a hypergeometric function. If you have N objects and K desired objects in a bag and if you remove n objects, the probability of choosing k desired objects is $$P(X=k) = \frac{\binom{n}{k}\binom{N-K}{n-k}}{\binom{N}{n} }.$$ In your case, you have 20 red balls and 16 blue balls and you're looking for the last one to be red. This is equivalent 19 of the first 35 being red. Therefore, the answer can be written as $$P(X=19) = \frac{\binom{20}{19}\binom{16}{16}}{\binom{36}{35}} = \frac{20 \cdot 1}{36} = \frac{5}{9}.$$ You're sure the answer wasn't $5+9=14$ or it wasn't asking for the probability of the last ball being blue? The answer $\frac{4}{9}$ is absolutely nonsensical. If you have more red balls than blue balls and no prior information, there's no logical way for the final ball to have a higher likelihood of being blue.

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Lets look at another perpective . Assume that we are arranging all balls in line . Moreover , this line is constructed according to selection order of the balls.

Then, how many ways are there to align all of the balls ? . The answer is $\frac{36!}{20! \times 16!}$.

Now assume that the last ball is red and we placed one red ball in the last place , so we need to align $19$ red balls and $16$ blue balls in a straight line (their positons are determined by selection order).Then , we can arrange them by $\frac{35!}{19! \times 16!}$ ways.

As a result , $\frac{\frac{35!}{19! \times 16!}}{\frac{36!}{20! \times 16!}}=\frac{20}{36} = \frac{5}{9}$

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I'm going to add another answer since my other one is still correct, but the original question is slightly different.

The way you phrased it, $\frac{5}{9}$ is definitely correct. The original question says, "all balls of a color have been removed." From that perspective, it's asking what is the probability of there being at least one blue ball when all red balls have been drawn.

The "trick" is to argue that the last ball (it you kept going) will either be red or blue. If it's red, then all blue balls will have already been drawn. If it's blue, all red balls will have been drawn. Since those outcomes are disjoint, the probability of all red balls being drawn is equal to the probability that the final ball drawn is blue which is $1-\frac{5}{9}=\frac{4}{9}$.

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