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Probability · Basic probability

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Probability of a single event

Why a probability is a share of the whole — so it lives between 0 and 1, why P(A) = favourable ÷ total needs equally likely outcomes, and how the "it all adds to 1" rule drives complements, missing table entries and expected numbers.

Probability · Basic probability

Probability of a single event

Why a probability is a share of the whole — so it lives between 0 and 1, why P(A) = favourable ÷ total needs equally likely outcomes, and how the "it all adds to 1" rule drives complements, missing table entries and expected numbers.

Why it works

A probability is not a score out of ten and it is not a feeling. It is a share of the whole. Picture every possible thing that could happen laid out as slices of one complete cake. The probability of an event is simply how much of that cake the event covers.

Two facts fall straight out of that picture.

First, the scale is 0 to 1. Nothing can cover less than none of the cake or more than all of it, so every probability satisfies 0≤P≤10 \leq P \leq 1. A probability of 00 means impossible, 11 means certain, and 0.50.5 means the event covers exactly half of everything that could happen. This makes the scale a free error-check: an answer of 1.41.4, or −0.2-0.2, or 32\frac{3}{2} is not a slightly wrong probability — it is not a probability at all, and it is telling you to go back and look at the working.

Second, the slices add to the whole cake. If you list outcomes that cannot happen together (mutually exclusive) and which between them cover every possibility (exhaustive), then their probabilities add to exactly 11.

P(outcome1)+P(outcome2)+⋯+P(outcomen)=1.P(\text{outcome}_1) + P(\text{outcome}_2) + \dots + P(\text{outcome}_n) = 1.

Where P(A)=favourabletotalP(A) = \frac{\text{favourable}}{\text{total}} comes from. This formula is not a rule handed down from nowhere — it is the cake cut fairly. Suppose a spinner has 88 identical sectors. By symmetry no sector is favoured over any other, so the 11 has to be shared equally: each sector gets 18\frac{1}{8}. If 33 of those sectors are red, red's share is three lots of 18\frac{1}{8}, which is 38\frac{3}{8}. So "favourable over total" really means count how many of the equal slices belong to your event.

That word equally likely is load-bearing. A biased spinner with 33 red sectors out of 88 does not give P(red)=38P(\text{red}) = \frac{3}{8}, because its slices are not equal. And "either it rains tomorrow or it doesn't, so P(rain)=12P(\text{rain}) = \frac{1}{2}" fails for exactly the same reason: two outcomes, but nobody said they were equally likely.

The trap, made concrete. A bag holds 33 red counters and 55 blue counters, and you are asked for P(red)P(\text{red}). An enormous number of students write 35\frac{3}{5}. That number answers a different question — it is the ratio of red to blue, 3:53 : 5. A probability compares red with everything, and everything here is 3+5=83 + 5 = 8 counters, so

P(red)=38=0.375.P(\text{red}) = \frac{3}{8} = 0.375.

The sum test exposes the error instantly. If P(red)=35P(\text{red}) = \frac{3}{5} then P(blue)P(\text{blue}) would be 53\frac{5}{3}, which is bigger than 11 — impossible — and the two would total 3415\frac{34}{15}, not 11. Whereas 38+58=1\frac{3}{8} + \frac{5}{8} = 1. ✓ So every time you write a probability, glance at the denominator and ask: *is that the total number of things that could happen?*

One number, three costumes. 38=0.375=37.5%\frac{3}{8} = 0.375 = 37.5\%. Fractions, decimals and percentages are just different ways of writing the size of the slice, and a question may hand you one and ask for another. Trouble only starts when you mix them mid-calculation: 1−35%1 - 35\% is meaningless as written. Convert first, then work: either 1−0.35=0.651 - 0.35 = 0.65, or 100%−35%=65%100\% - 35\% = 65\%. The whole is 11 if you are in fractions and decimals, and 100100 if you are in percentages — pick one and stay in it.

The complement. Every trial produces exactly one outcome, so "AA happens" and "AA does not happen" split the cake into two pieces with no overlap and nothing left over. Therefore P(A)+P(not A)=1P(A) + P(\text{not } A) = 1, which rearranges to

P(not A)=1−P(A).P(\text{not } A) = 1 - P(A).

This is worth far more than it looks, because "not" is very often the easier half to count. Finding the probability that a card drawn from a pack is not a picture card is one subtraction; counting the 4040 cards that aren't picture cards is a chore.

The missing entry in a table. A table of probabilities for a set of outcomes that cannot overlap and that cover every case must total 11. So if three of four probabilities are 0.20.2, 0.350.35 and 0.150.15, the fourth is

1−(0.2+0.35+0.15)=1−0.7=0.3.1 - (0.2 + 0.35 + 0.15) = 1 - 0.7 = 0.3.

Notice the two-step shape: add up what you have, then subtract from 1. Stopping at 0.70.7 answers nothing — that is the probability of not getting the missing outcome.

The algebraic version is the identical rule wearing algebra. If the probabilities of three outcomes are 3x3x, 2x2x and xx, then because they must total 11,

3x+2x+x=1⇒6x=1⇒x=16.3x + 2x + x = 1 \quad \Rightarrow \quad 6x = 1 \quad \Rightarrow \quad x = \tfrac{1}{6}.

And here is the sting: xx is almost never what the question asked for. If it wants the probability of the outcome written as 2x2x, that is 2×16=132 \times \frac{1}{6} = \frac{1}{3}. Solve for xx, then go back and re-read the question.

Expected number of occurrences. If P(head)=12P(\text{head}) = \frac{1}{2} and you flip a coin 200200 times, you expect about 100100 heads:

expected number=P×number of trials.\text{expected number} = P \times \text{number of trials}.

Why multiply? Because a probability is a share, and a share of 200200 trials is that fraction of 200200 — the same "of means multiply" you use to find 12\frac{1}{2} of 200200. It is a multiplication, never a division.

Two health warnings. "Expected" is a forecast, not a promise: 100100 is the long-run average, and a real run of 200200 flips will usually land near it rather than exactly on it. And an expected number needn't be a whole number — a fair dice rolled 100100 times gives 100×16≈16.7100 \times \frac{1}{6} \approx 16.7 sixes, which is a perfectly good estimate even though you cannot roll 0.70.7 of a six.