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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.

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    $\begingroup$ The probabilities are discrete. I think you mean they're not identical (or equal). $\endgroup$
    – saulspatz
    Apr 7, 2021 at 14:53
  • $\begingroup$ @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. $\endgroup$ Apr 7, 2021 at 15:01
  • $\begingroup$ 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? $\endgroup$ Apr 7, 2021 at 15:05
  • $\begingroup$ 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. $\endgroup$
    – saulspatz
    Apr 7, 2021 at 15:10
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    $\begingroup$ 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. $\endgroup$ Apr 7, 2021 at 15:24

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

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  • $\begingroup$ Nice shortcut ! $\endgroup$ Apr 7, 2021 at 16:45

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