Probability when sample space decreases

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 $29/34$.

• 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 – Phil H May 10 '18 at 20:10

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argue the same result by making use of conditional probability by definining $H_1$ to be the event that the first ball withdrawn is white and let $H_2$ be the event that the second ball withdrawn is white the required probability is then $$\mathbf{P}(H_1\cap H_2) = \mathbf{P}(H_1)\mathbf{P}(H_2|H_1) = \frac{30}{35}\cdot\frac{29}{34}$$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 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. – Atif Farooq May 10 '18 at 20:52