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Mathematics · Ch 14 — Probability

Probabilities of Equally Likely Outcomes

14.2.2

Probabilities of Equally Likely Outcomes

The Idea of Equally Likely Outcomes

When you toss a fair coin, you know that heads and tails are equally likely. But what does "equally likely" mean in a precise, mathematical sense? It means that every simple event in the sample space has the same probability of occurring. This is the simplest and most common situation in probability problems, and it leads directly to a clean formula for the probability of any event.

Suppose an experiment has a sample space SS containing nn distinct outcomes:

S={ω1,ω2,…,ωn}S = \{\omega_1, \omega_2, \dots, \omega_n\}

If all these outcomes are equally likely, then each simple event {ωi}\{\omega_i\} must have the same probability. Let that common probability be pp, where 0≤p≤10 \leq p \leq 1. So we have:

P(ω1)=P(ω2)=⋯=P(ωn)=pP(\omega_1) = P(\omega_2) = \cdots = P(\omega_n) = p

Now, the sum of the probabilities of all simple events in the sample space must equal 1. This gives:

P(ω1)+P(ω2)+⋯+P(ωn)=1P(\omega_1) + P(\omega_2) + \cdots + P(\omega_n) = 1

Since each term is pp, this sum is pp added to itself nn times, which is npnp. Therefore:

np=1np = 1

And so:

p=1np = \frac{1}{n}

This is the fundamental result: when outcomes are equally likely, each individual outcome has probability 1/n1/n.

Important

The condition 0≤p≤10 \leq p \leq 1 is automatically satisfied here because nn is a positive integer, so 1/n1/n is always between 0 and 1.

From Simple Events to Any Event

Now let EE be any event (a subset of the sample space SS). Suppose the total number of outcomes in SS is n(S)=nn(S) = n, and the number of outcomes favourable to EE is n(E)=mn(E) = m. Since each outcome is equally likely, the probability of EE is simply the sum of the probabilities of the outcomes that make up EE.

Each favourable outcome has probability 1/n1/n, and there are mm of them. So:

P(E)=1n+1n+⋯+1n⏟m times=mnP(E) = \underbrace{\frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n}}_{m \text{ times}} = \frac{m}{n}

This gives the classic formula:

P(E)=n(E)n(S)=Number of outcomes favourable to ETotal number of possible outcomesP(E) = \frac{n(E)}{n(S)} = \frac{\text{Number of outcomes favourable to } E}{\text{Total number of possible outcomes}}

This formula only works when all outcomes are equally likely. If they are not, you cannot use it — you would need the more general approach of assigning probabilities individually.

Watch out

A common mistake is to apply this formula without checking whether outcomes are equally likely. For example, when rolling two dice, the sum "2" and the sum "7" are not equally likely outcomes — but the 36 ordered pairs (1,1),(1,2),…,(6,6)(1,1), (1,2), \dots, (6,6) are equally likely. Always define your sample space carefully.

A Quick Example to Fix the Idea …