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My question is based the scenario in Non-Uniform Probability Without Replacement.

Suppose we have probabilities of four letters $$P(A) = 0.1 \quad P(B) = 0.2 \quad P(C) = 0.3 \quad P(D) = 0.4$$

If I were to draw two letters without replacement, and the first letter is A, what is the probability that I will get B on the second draw? That is, what is $P(B | A)$?

I guess that $P(B | A) = P(B)/(1 - P(A)) = 0.2/0.9$. But why is that so? Where does the formula come from? Can it be derived from simple conditional probability formula?

note: I also tried to simulate this problem, and the relative frequency does approach $0.2/0.9.$ But I still confused where the formula comes from.

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  • $\begingroup$ I expanded my answer below. $\endgroup$
    – ryang
    Commented Feb 28, 2023 at 7:35

2 Answers 2

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If one is just given the four probabilities stated in your question, it is a little unclear what happens once the first letter is chosen and then becomes unavailable for subsequent draws. In the solution to the other question you referenced, the assumption seems to be that once a letter is removed from the pool, the probabilities of the other letters will remain unchanged relative to each other.

One way to simulate that would be to have a bag with with the following Scrabble tiles: one A, two B's, three C's and four D's. Then, after the first letter is drawn, all tiles with the same letter are removed before the next draw.

Now suppose A is the first letter drawn. Then for the next draw, there are nine tiles left, two of which are B's, so the probability of drawing a B given that the first letter drawn was an A would be $\frac{2}{9} = \frac{0.2}{0.9}$.

This doesn't really require the application of the conditional probability formula, however one would still get the same answer as follows (here I am using the notations $\text{A}_1$ and $\text{B}_2$ to denote A being drawn first and B being drawn second):

$$ \begin{align} \text{P}[\text{B}_2 | \text{A}_1] &= \frac{\text{P}[\text{A}_1 \cap \text{B}_2]}{\text{P}[\text{A}_1 ]} \\[2mm] &= \frac{0.1 \times \frac{0.2}{0.9}}{0.1} \\[2mm] &= \frac{0.2}{0.9} \end{align} $$

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Suppose we have probabilities of four letters $$P(A) = 0.1 \quad P(B) = 0.2 \quad P(C) = 0.3 \quad P(D) = 0.4$$

If I were to draw two letters without replacement, and the first letter is A, what is the probability that I will get B on the second draw? That is, what is $P(B | A)$?

I guess that $P(B | A) = P(B)/(1 - P(A)) = 0.2/0.9$. Can it be derived from simple conditional probability formula?

What you require: the probability of 'b' on the second draw given that 'a' has been drawn without replacement.

Your problem arises from the above events not having been well-defined:

  • if $A$ is the event of obtaining an 'a' in the first draw, i.e., $$P(A)=P(ab,ac,ad),$$ then your required probability is not $P(B|A),$ which has value $0,$ but $$\frac{P(ab)}{P(A)}=\frac{\frac1{10}\times\frac29}{0.1}=\frac29,$$ as you have obtained above;
  • if $A$ is the event of obtaining an 'a' at all over two draws, i.e., $$P(A)=P(ab,ac,ad,ba,ca,da),$$ then your required probability is still not $P(B|A),$ which equals $\dfrac{P(ab,ba)}{P(ab,ac,ad,ba,ca,da)},$ but $$\frac{P(ab)}{P(ab,ac,ad)}=\frac{?}{?}.$$

P.S. It's helpful to be clear about the terminology. Our probability experiment has two dependent trials; its outcome $ac$ denotes drawing 'a' and 'c' in trials 1 and 2, respectively; the outcome 'a' of trial 1 corresponds to the event $\{ab,ac,ad\},$ which contains three experiment outcomes.

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