Probability Is a Model, Not a Prophecy

Suppose we toss a coin and it lands heads. What was the probability of that outcome?

You might be tempted to say 50 percent. But that answer rests on an assumption: the coin is fair, it was tossed in a particular way, and our model treats the toss as random. A heavily weighted coin would not have the same probabilities.

Probability is a mathematical description of uncertainty under specified assumptions. It is not a guarantee that reality will behave exactly as expected over a handful of observations.

What counts as an outcome?

A sample space, often written $\Omega$, lists the possible outcomes of an experiment. For a single coin toss, $\Omega=\{H,T\}$. An event is a collection of outcomes. In a toss of two coins, the event “at least one head” consists of $HH$, $HT$, and $TH$.

For an event $A$, we assign a probability $P(A)$ satisfying

$$0\leq P(A)\leq 1,\qquad P(\Omega)=1.$$

For mutually exclusive events, the probability of their union is the sum of their probabilities.

Under our fair-coin model, all four two-toss outcomes are equally likely, so the probability of at least one head is $3/4$.

But notice what made the calculation possible: the model assigning equal probability to the four outcomes.

Complements are often easier

The complement of event $A$, written $A^c$, is everything in the sample space that is not in $A$.

$$P(A^c)=1-P(A).$$

For example, if each of five independent tosses has a one-half chance of tails, the probability of at least one head is easiest to obtain by subtracting the probability of all tails:

$$P(\text{at least one head})=1-\left(\frac12\right)^5=\frac{31}{32}.$$

This does not mean that five tosses are guaranteed to include a head. It means the all-tails outcome has probability $1/32$ under the model.

Addition and overlap

For arbitrary events $A$ and $B$,

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

That subtraction prevents us from counting the intersection twice.

If 40 percent of a group likes tea, 35 percent likes coffee, and 15 percent likes both, then 60 percent likes tea or coffee (or both):

$$.40+.35-.15=.60.$$

Independent is not mutually exclusive

Two events are independent if knowing that one occurred does not change the probability of the other:

$$P(A\cap B)=P(A)P(B).$$

Two events are mutually exclusive if they cannot both happen:

$$P(A\cap B)=0.$$

These are very different ideas. Drawing a heart and drawing a spade on a single card draw are mutually exclusive, not independent. If one happened, the other definitely did not.

Repeated independent fair-coin tosses are independent by the model’s assumptions, but outcomes on different tosses can both be heads.

Exercise

1. In three independent fair-coin tosses, what is the probability of at least one tail?
2. Events $A$ and $B$ have probabilities .6 and .3, with overlap .2. Find $P(A\cup B)$.
3. Can two events with positive probability be both independent and mutually exclusive?

Worked answers

1. Subtract the all-heads probability: $1-(1/2)^3=7/8$.

2. $0.6+0.3-0.2=0.7$.

3. No. Mutual exclusivity requires $P(A\cap B)=0$, but independence with positive probabilities requires $P(A\cap B)=P(A)P(B)>0$.

Read further

The free OpenStax probability chapter provides additional examples and exercises.

State the probability model before using its rules

A sample space lists possible outcomes; events are subsets of that space; a probability assignment is nonnegative, totals one, and adds over disjoint events. For a finite equally likely sample space, probability is favorable-count divided by total-count. Equal likelihood is an assumption, not something established by naming outcomes.

For synthetic events A and B with probabilities 0.4 and 0.5 and intersection probability 0.1, inclusion–exclusion gives $P(A\cup B)=0.8$. The chance neither occurs is $0.2$. The conditional probability $P(A\mid B)=0.1/0.5=0.2$, so these events are not independent. Simply multiplying 0.4 by 0.5 would impose a different intersection model.

For independent repetitions of an event with probability $p$, the chance of at least one occurrence in $n$ trials is $1-(1-p)^n$. With $p=0.1$ and $n=3$, it is $0.271$, not $0.3$. Adding the single-event probabilities double-counts outcomes containing multiple occurrences. Without independence, the complement formula still holds but the product for the no-occurrence probability may fail.

Additional practice

1. If $P(A)=0.6$, $P(B)=0.5$, and $P(A\cap B)=0.3$, calculate union probability and test independence.
2. Can two positive-probability mutually exclusive events be independent?
3. Give a model assumption needed to interpret a coin probability as a long-run experimental frequency.

Worked additional solutions

1. Union $0.8$; the intersection equals $0.6(0.5)$, so the pair is independent.
2. No. Their intersection is zero, whereas the product is positive.
3. Specify a stable repeated-trial mechanism, with suitable independence or dependence assumptions for the long-run theorem being used. A single toss cannot validate that mechanism.