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Suppose that you have a bag of marked marbles, initially 12 red and 12 black.

A total of 21 marbles will be withdrawn, one at a time, without replacement. Suppose the probability of 5 particular red marbles being drawn out is $\frac{1}{61440}$ and that this event occurs as part of the first 20 draws.

What is the probability of the 21st marble being any one of the black marbles?

I've gone at this a few different directions and I think I've managed to overthink it entirely.

There are $_{(24-5)}C_{(20-5)} = 3876$ total ways to select the first 20 marbles if we treat the required event separately.

Of those 3876 ways there are $_{12}C_{11}+_{12}C_{10}+_{12}C_{9}+_{12}C_{8}=12+66+220+495=793$ ways to have at least 1 black marble left in the bag. This means there is a $\frac{793}{3876}\approx 0.20459$ probability that there is at least 1 black marble left in the bag on the 21st draw.

I know that the probability of picking 1 black marble out of "at least 1" is not 25% so I can't just do $\frac{1}{61440} \times \frac{793}{3876} \times 0.25$ to get my final probability. I somewhat feel like I could say $0.25 * 0.50 * 0.75 * 1.0 = 0.09375$ might be more correct corresponding to the cases of picking 1 black marble from 1, 2, 3 or 4 remaining.

I would appreciate confirmation on the steps as well as a suggestion on how to complete this problem.

This is not homework.

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    I'm confused about your 'certainly markings' statement. Are we to treat the balls as all distinguishable, yet we do not care *which* black marble we get on the 21st draw, so long as it is black? – 2017-01-30
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    @TheCount Yes, the marbles are distinguishable but we don't care which of the black marbles we obtain on the 21st draw. There is a requirement that 5 particular red marbles are drawn as part of the 1st 20 with the stated probability. – 2017-01-30
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    I thought this would be a quick answer, but it is giving me some difficulty! – 2017-01-30
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    That's what I said :> – 2017-01-30

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Don't over complicate things.

We seek the probability that the twenty-first marble drawn is one from the $12$ black marbles under the condition that $5$ particular red marbles (from $12$) have been drawn among the first twenty positions of consecutive draws without replacement.

Observe that wheresoever those particular marbles are among the first twenty positions, the remaining $19$ marbles each have equal probability of being in the favoured position, and $12$ of these are black.

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Remark: Although this assumes that there is no bias in selecting among the remaining marbles, which is contraindicated by the stated probability of $1/61440$ for selecting the particular marbles into the first twenty positions, rather than the unbiased probability $\binom{19}{15}/\binom{24}{20}$ or $646/1771$.

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    Selecting the 5 particularly marked red ones is 1/61440, we're simply indicating that this event has happened and so we're interested in the probability of the black draw occuring as indicated subsequently. I had wondered if this could be expressed in the idea $\frac{1}{61440}\times (1-\frac{12/19})^{15}\times\frac{1/19}$ but I know this is lacking. – 2017-01-31