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Consider the probabilities of the following letters: $$\begin{align}& A=0.1 & B=0.2 && C=0.3 && D=0.4 & \end{align}$$ What is the probability of picking 2 letters (without replacement) such that it does not contain $A$? Note that when a letter is picked, it cannot picked again.

The brute way of solving this would be: $$A[B/C/D] = 0.1 \left( \frac{0.2}{0.9}+\frac{0.3}{0.9}+\frac{0.4}{0.9}\right)\\ [B/C/D] A = 0.2 \times \frac{0.1}{0.8} + 0.3 \times \frac{0.1}{0.7} + 0.4 \times \frac{0.1}{0.6} $$ $$\therefore P(\text{No A}) = 1 - 0.1 - \frac{113}{840}=\frac{643}{840} \approx 77 \%$$

Is there way/formula that extends this intuitively, such as picking 3 letters (or a closed form of the example) and does this describe a type of a distribution?

I thought of $3 \choose 2$ the probabilities of $B,C,$ and $D$ but since they are unequal, I have no idea how to put it in practice.

Renaissance
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    The probabilities are discrete. I think you mean they're not identical (or equal). – saulspatz Apr 07 '21 at 14:53
  • @saulspatz is correct that they are discrete. If he's correct about what you did mean, you can fix the title to say "Non-uniform probability ....". Or just remove that part. – JonathanZ Apr 07 '21 at 15:01
  • Oh right, that is what I should've wrote. Also, I've noticed if it were uniform I imagine it would be tad easier to derive a formula however since it is not, am I better off brute forcing it? – Renaissance Apr 07 '21 at 15:05
  • Your calculations are correct. If it were uniform, it would be easy, but for non-uniform, I don't think you'll find a convenient formula. If the number of letters, and the number choices is large, I would use simulation. – saulspatz Apr 07 '21 at 15:10
  • What happens with the probabilities after the first draw? – user Apr 07 '21 at 15:16
  • For my understanding, one difficulty comes because the underlying process is not clear. What does "probabilities" of A, B and C mean? How does it change after draws (or why does it change in the manner you have given)? Are you modelling some other process by the process of picking letters? Sorry if these are obvious. – Rahul Madhavan Apr 07 '21 at 15:21
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    Its a similar case of drawing balls. That is after drawing one ball, all balls of that color are removed before drawing again. The question is after $n$ draws what is the probability of it has/it does not have a selected color. – Renaissance Apr 07 '21 at 15:24

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Direct calculation: $.2\times 7/8 +.3\times 6/7 +.4\times 5/6=0.765$ or $76.5$%.

herb steinberg
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