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Probability & Statistics/How would you find the probability with an undefined number of possible outcomes?


QUESTION: How would you find the probability with an undefined number of possible outcomes?

The text says probability = favorable outcomes / possible outcomes.

How would you find probability when you can't assign a number to possible outcomes? For instance, like if you have x marbles in the bag, and y are one color, what is the probability that they are all that color? How could you do that when there are a seemingly infinite number of possible colors? The text uses examples with coins, which is easy because you can only get a H or a T. What do you do when you could get a or b or c ad infinitum?

It seems possible to find probability without a definite number of possible outcomes. You always hear people say planes have a 1 in X chance of incidence. How is that possible when there is a seemingly infinite combination of possibilities that cause incidence? It could have broken ailerons, broken flaps, a broken rudder, a loss of hydraulic power, a loss of fuel, a and b, or all of the above.

ANSWER: There are two aspects of probability that your textbook does not take into account, and it is like to have explained this somewhere in the introduction of the text.

Basic probability theory starts with models of probability that make two assumptions:

1. There are a finite number of outcomes.
2. Each outcome is equally likely.

You are concerned about what happens when the first restriction is lifted. In this case, we consider the probability of an "event." For example, there are an infinite number of possible weights of a person -- even if we assume a maximum weight of 2000lbs for a person, there are infinitely many real numbers between 0 and 2000, even if we can't measure them all accurately.

Instead, we divide the interval 0 to 2000 into some intervals (for example, if a scale only reads out whole numbers, then the reading "214" actually means "any number x where 213<x≤214" or something like that). The "event" is that your weight is between 213 and 214, which is itself an infinite set of favorable outcomes among the infinite total set of outcomes.

The probability of that event will be some number, say 0.1%. However, the probability that your weight is exactly 214lbs is zero, because that is only one outcome out of infinitely many.

So, instead of single outcomes being considered, we consider bunches of outcomes (often, a contiguous interval of values) . This idea is called a probability "distribution."

You also have a different concern that sits behind some of your examples. If you are discussing some news or media source that states "there is a 1 in X chance of airplane failure," they are not using a theoretically computed probability where they have counted all possible outcomes (finite or infinite, that doesn't matter). It would be impossible to do that. They are using the statistics of what we know about plane failures to estimate this probability.

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QUESTION: I just saw a YouTube video where a guy used N! / n!(N-n)! Could that be used?

Like if you have 5 marbles and say you wanted the probability that all are clear. You have 5 objects and 2 possibilities per object, clear or not. Couldn't you say:

5! / 2! (5-2)!
5! / 2! (3)!
5x4x3x2x1 / 2x1x3x2x1
5x4 / 2x1
20 / 2

If that's right, couldn't they use the same thing on planes? Like count the objects in a plane relevant to its crashing, spoilers, ailerons, rudder, and do not count objects not relevant to its crashing, passengers, the beverage cart, etc. then assign the objects relevant to crashing to N and keep 2 as n, because it does cause an incidence of it doesn't?

I guess another way to word this question would be is 2 sufficient when you don't know how many object possibilities there are? Instead of saying the marble could be X or y or z, and the plane crash can happen because X or y or z, could you use 2 because the marble is clear or not and the plane crashes or it doesn't?

That is not correct because the probability of two objects colliding is not the same for every possible pair of objects, and because each object is not just in two possible states (causing a crash or not causing a crash). A highly complex system like an airplane cannot be analyzed by simply counting the outcomes -- either of the entire plane, or of a small list of individual components -- and assuming those outcomes are equally likely.

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Clyde Oliver


I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.


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