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Probability Spaces with Infinitely Many Outcomes

To this point, we have focused entirely on probability spaces (S,P) with S a finite set. More generally, probability spaces are defined where S is an infinite set.

Probability Spaces with Infinitely Many Outcomes

To this point, we have focused entirely on probability spaces \((S,P)\) with \(S\) a finite set. More generally, probability spaces are defined where \(S\) is an infinite set. When \(S\) is countably infinite, we can still define \(P\) on the members of \(S\), and now \(\sum_{x\in S} P(x)\) is an infinite sum which converges absolutely (since all terms are non-negative) to \(1\). When \(S\) is uncountable, \(P\) is not defined on \(S\). Instead, the probability function is defined on a family of subsets of \(S\). Given our emphasis on finite sets and combinatorics, we will discuss the first case briefly and refer students to texts that focus on general concepts from probability and statistics for the second.

Example

Consider the following game. Nancy rolls a single die. She wins if she rolls a six. If she rolls any other number, she then rolls again and again until the first time that one of the following two situations occurs: (1) she rolls a six, which now this results in a loss or (2) she rolls the same number as she got on her first roll, which results in a win. As an example, here are some sequences of rolls that this game might take:

  1. \((4,2,3,5,1,1,1,4)\). Nancy wins!

  2. \((6)\). Nancy wins!

  3. \((5,2, 3,2,1,6)\). Nancy loses. Ouch.

So what is the probability that Nancy will win this game?

Nancy can win with a six on the first roll. That has probability \(1/6\). Then she might win on round\(n\) where \(n\ge2\). To accomplish this, she has a \(5/6\) chance of rolling a number other than six on the first roll; a \(4/6\) chance of rolling something that avoids a win/loss decision on each of the rolls, \(2\) through \(n-1\) and then a \(1/6\) chance of rolling the matching number on round\(n\). So the probability of a win is given by: \[\begin{aligned}\end{aligned}\]

Example

You might think that something slightly more general is lurking in the background of the preceding exampleand it is. Suppose we have two disjoint events \(A\) and \(B\) in a probability space \((S,P)\) and that \(P(A)+P(B)\lt 1\). Then suppose we make repeated samples from this space with each sample independent of all previous ones. Call it a win if event \(A\) holds and a loss if event \(B\) holds. Otherwise, it's a tie and we sample again. Now the probability of a win is: \[\begin{aligned}\end{aligned}\]

Nou jy Geen sakrekenaar los hierdie een nie, maar die stukke daarvan is kon nie wees kondifiseerbaar. Probeer een hieronder, of tik jou eie.

Hou jou eie werk

'n Vry rekening voeg notas by oor elke les, 'n verslag van wat jy klaar gemaak het, jou opgelos probleme op een plek, en' n tugmeester wat jy kan vra oor hierdie bladsy. Die wiskunde self is oop vir almal, geteken in of nie.

Meld op Meld aan

Simbools wat hier gebruik word

Maak enige simbool van die volledige definisie,'n prent en wat elke letter daarin beteken.

Vrae wat mense vra

What is the difference between probability and statistics?

Probability goes from a known model to what the data should look like; statistics goes from data back to the model. Probability theory is the deductive half.

What does the law of large numbers promise?

That the average of many independent samples converges to the expected value. It says nothing about any single trial.

Dele van hierdie bladsy word aangepas Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Verklaar en weer opklaar hier, foute is ons s'n.

Meer in Probability Theory

Sample spaces and the axiomsRandom variables and expectationThe common distributionsThe law of large numbers and the central limit theorem