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I have one very, very simple question. How can I calculate the probability of an event if the sample space decreases with every "iteration" of an event?

I'll give an example to make it clear. Suppose that there is a jar of marbles. 30 marbles are white and 5 marbles are red. If I randomly take out two marbles, what is the probability that both of them are white?

Notice that in the first iteration the probability is $30/35$, but in the second one it's $29/34$.

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  • $\begingroup$ These are called dependent events, whereby the outcome of the first event effects the probability of the second event and the third etc. Whenever there is no replacement of objects after selection, the probability of the next selection is affected. With replacement the probabilities remain unchanged. In your case the probability is 30/35 times 29/34 = .731 $\endgroup$ – Phil H May 10 '18 at 20:10
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Let $S = R\cup W = \{R_1,R_2,\dots,R_5\}\cup\{W_1,W_2,W_3,\dots, W_{30}\}$ denote the set of all marbels in addition let $E$ denote the event that of the two balls randomly selected both of them are white.

If we take a combinatorial approach then $$\mathbf{P}(E) = \mathbf{P(}\{X\subseteq W:|X| = 2\}) = \frac{\binom{30}{2}}{\binom{35}{2}}$$ alternatively we may argue the same result by making use of conditional probability by definining $H_1$ to be the event that the first ball withdrawn is white and let $H_2$ be the event that the second ball withdrawn is white the required probability is then $$\mathbf{P}(H_1\cap H_2) = \mathbf{P}(H_1)\mathbf{P}(H_2|H_1) = \frac{30}{35}\cdot\frac{29}{34}$$

the latter approach reflects more clearly your notion of a reduced sample space.

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  • $\begingroup$ I understand the combinatorial approach, but I don't stand the latter one. Why do we multiply? $\endgroup$ – Hanlon May 10 '18 at 20:48
  • $\begingroup$ The multiplication is a consequence of the fact that $\mathbf{P}(H_1|H_2) = \frac{\mathbf{P}(H_1\cap H_2)}{H_1}$ there is no special combinatorial interepretation of multiplication here. $\endgroup$ – Atif Farooq May 10 '18 at 20:52
  • $\begingroup$ Is it correct to think about it like this: iteration #1 will happen 30/35 times, and iteration #2 29/34 times for every 30/35 times. So we multiply (30/35)(29/34)? $\endgroup$ – Hanlon May 10 '18 at 21:00
  • $\begingroup$ What do you mean by" iteration #2 29/34 times out of 30/35 times" ? $\endgroup$ – Atif Farooq May 10 '18 at 21:05
  • $\begingroup$ Iteration is the picking of a marble randomly (in my example). In my example, event of picking two white marbles is composed out of 2 iterations: iteration #1 that picks the first marble and iteration #2 that picks the second marble. Statistically, out of 35 picking I should 30 times get the white marble (I return the marble after picking to the jar). If after getting the white marble I stop and pick again, I have 34 marbles in the jar and 29 of them are white. So, for every picked white marble, I have probability of 29/34 that the next one will be white too. $\endgroup$ – Hanlon May 10 '18 at 21:14

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