How is probability changed when an experiment is repeated? This is a simple question, but I'm in a beginning level math class and my teacher is not very clear. We're studying probability, namely the equation (# of ways the event can occur)/(total # of possible outcomes). This is assuming each outcome is equally likely. My teacher said that when an experiment is repeated twice, the total number of outcomes is equal to the number of outcomes in experiment 1 multiplied by the number of outcomes in experiment 2. This describes the denominator of the experiment--what is done with the numerator?
For example, if something has a 1/1000 chance of happening in a day, what are the chances that it will happen either today or tomorrow? The experiment is repeated and the probability should go up, but I'm unsure how to do this. 
 A: The set of possible outcomes is now the set of pairs (what happens today, what happens tomorrow). What you want to count is the number of such pairs in which the event happens. I think that an example will make it more clear.
Experiment: roll a 6-sided die. Number of outcomes: 6. Probability of rolling a $1$: 1/6.
Double experiment: roll today, roll tomorrow. Number of outcomes is now $6\cdot 6 = 36$, corresponding to the 36 pairs of outcomes (1,1), (1,2), ..., (1,6), (2,1), (2,2), ..., (2,6), ..., (6,1), (6,2), ..., (6,6). Probability of rolling a $1$ today only is still 1/6; but now I could see this as 6/36, because the total number of outcomes, taking both days into account, in which I roll a $1$ today, is 6, since I could roll any of 6 different things tomorrow: (1,1), (1,2), ..., (1,6). Likewise, there are 6 outcomes (taking both days into account) in which I roll a $1$ tomorrow, since I could roll anything today: (1,1), (2,1), ..., (6,1).
At first glance, this makes it seem that the number of outcomes in which I roll a $1$ on either day is 6+6=12, but actually it is 11, because I have to be careful not to double count the outcome (1,1), in which I roll a $1$ today and tomorrow. There are really only 11 outcomes in which I roll a $1$ on one or both days: the five in which I roll it today and not tomorrow, the five in which I roll it tomorrow and not today, and the one in which I roll it on both days.
Thus, the total probability of rolling a 1 on either day is 11/36.
Similarly, in your example, if there is a 1/1000 chance of something happening in the experiment, and I try today and tomorrow, then out of the $1000\times 1000$ possible pairs of outcomes, in 999 the event happens today and not tomorrow; in 999 it happens tomorrow and not today; and in 1 case it happens on both days. The total probability of the event taking place is thus
$$\frac{999+1+999}{1000\cdot 1000} = \frac{1,999}{1,000,000}$$
Does that clarify?
