Skip to content

Mathematics · Ch 14 — Probability

Probability of an Event

14.2.1

Probability of an Event

Probability of an Event

When we talk about the probability of an event, we are really asking: "What is the chance that the outcome of an experiment falls into a particular collection of outcomes?" That collection is the event. The definition is straightforward once we have a sample space and a probability assigned to each individual outcome.

Consider the experiment of examining three consecutive pens produced by a machine, classifying each as Good (G, non-defective) or Bad (B, defective). The sample space SS has eight equally likely outcomes:

S={BBB,  BBG,  BGB,  GBB,  BGG,  GBG,  GGB,  GGG}S = \{BBB,\; BBG,\; BGB,\; GBB,\; BGG,\; GBG,\; GGB,\; GGG\}

Suppose the probabilities assigned to these outcomes are all equal — each outcome gets probability 18\frac{1}{8}. This is a natural assignment when the machine produces pens independently and each pen is equally likely to be good or bad.

Now define two events:

  • Event A: exactly one defective pen.
  • Event B: at least two defective pens.

From the sample space:

A={BGG,  GBG,  GGB}A = \{BGG,\; GBG,\; GGB\}

B={BBG,  BGB,  GBB,  BBB}B = \{BBG,\; BGB,\; GBB,\; BBB\}

The probability of an event is simply the sum of the probabilities of all the outcomes that belong to that event. So:

P(A)=P(BGG)+P(GBG)+P(GGB)=18+18+18=38P(A) = P(BGG) + P(GBG) + P(GGB) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}

P(B)=P(BBG)+P(BGB)+P(GBB)+P(BBB)=18+18+18+18=48=12P(B) = P(BBG) + P(BGB) + P(GBB) + P(BBB) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}

This is the core idea: the probability of an event is the sum of the probabilities of its constituent sample points.

Note

This works because the outcomes in a sample space are mutually exclusive — no two can happen at the same time. So the probability that any one of them occurs is just the sum of their individual probabilities.


A Second Example: Unequal Probabilities

Now consider tossing a coin twice. The sample space is:

S={HH,  HT,  TH,  TT}S = \{HH,\; HT,\; TH,\; TT\}

But this time the probabilities are not equal. Suppose they are assigned as:

P(HH)=14,P(HT)=17,P(TH)=27,P(TT)=928P(HH) = \frac{1}{4},\quad P(HT) = \frac{1}{7},\quad P(TH) = \frac{2}{7},\quad P(TT) = \frac{9}{28}

Check that these sum to 1:

14+17+27+928=728+428+828+928=2828=1\frac{1}{4} + \frac{1}{7} + \frac{2}{7} + \frac{9}{28} = \frac{7}{28} + \frac{4}{28} + \frac{8}{28} + \frac{9}{28} = \frac{28}{28} = 1

So this assignment satisfies the axiomatic definition of probability.

Define event EE: "Both tosses yield the same result." Then:

E={HH,  TT}E = \{HH,\; TT\}

P(E)=P(HH)+P(TT)=14+928=728+928=1628=47P(E) = P(HH) + P(TT) = \frac{1}{4} + \frac{9}{28} = \frac{7}{28} + \frac{9}{28} = \frac{16}{28} = \frac{4}{7}

Define event FF: "Exactly two heads." Then:

F={HH}F = \{HH\}

P(F)=P(HH)=14P(F) = P(HH) = \frac{1}{4}

Watch out

A common mistake is to assume all outcomes are equally likely. In the coin-toss example above, the outcomes are not equally likely — the probabilities are different. Always check the probability assignment before summing.


The General Definition …