Probability basics: equally likely outcomes and counting

Favourable over total, with one condition

When every outcome of an experiment is equally likely, the probability of an event $A$ is

$$P(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in total}}.$$

The set of all outcomes is the sample space. The condition "equally likely" is not a formality: almost every wrong answer in this topic comes from counting outcomes that are not equally likely.

Take the sum of two dice. The possible sums are $2, 3, \dots, 12$, eleven values, but they are not equally likely: a sum of 2 happens only as $(1,1)$, while a sum of 7 happens six ways. The equally likely outcomes are the 36 ordered pairs (first die, second die). Treat the dice as different colours, even when the question does not, and you will always get the right sample space.

Three facts to keep in view

• Every probability lies between 0 and 1. An answer of $\frac{7}{6}$ or $-0.2$ is a signal to start again.
• The probabilities of all the outcomes add up to 1.
• Complement: $P(\text{not } A) = 1 - P(A)$. When "not $A$" is easier to count, as with "at least one", count it and subtract.

Probability by counting

With equally likely outcomes, a probability is a ratio of two counts, so the tools of counting problems carry straight over. The one rule: count the numerator and the denominator the same way. If the denominator counts ordered draws, so must the numerator; if it counts unordered groups, so must the numerator.

Drawing 2 balls at once from a bag of 5 red and 3 blue, the unordered sample space has $\binom{8}{2} = 28$ equally likely pairs, of which $\binom{5}{2} = 10$ are both red: $P = \frac{10}{28} = \frac{5}{14}$.

Probability step by step

The same answer comes from imagining the balls drawn one at a time and multiplying:

$$P(\text{red, then red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}.$$

The second fraction is $\frac{4}{7}$ because one red ball has gone: 4 red are left among 7 balls. This is drawing without replacement. With replacement, the ball goes back, the bag is the same for every draw, and the probability would be $\frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$.

Having two routes is useful in the exam room: do the problem one way and check it the other.

Independent repetitions

Tosses of a fair coin, rolls of a die, or draws with replacement do not affect each other. For such independent trials, the probability that several specified things all happen is the product of their probabilities: three heads in three tosses has probability $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.

When the question asks for a number of successes, not a fixed order, count the orders. Exactly two heads in three tosses can happen as HHT, HTH or THH, each with probability $\frac{1}{8}$, so $P = \frac{3}{8}$. In general the count of orders is $\binom{n}{k}$, and with probability $p$ of success at each trial,

$$P(\text{exactly } k \text{ successes in } n) = \binom{n}{k}\,p^k(1-p)^{n-k}.$$

When the outcomes are not symmetric

If the experiment is a list of options with given probabilities (a spinner with unequal sectors, a loaded die), there is no counting: you add the probabilities of the outcomes in the event. "Favourable over total" only works once the outcomes are equally likely.

A routine for every question

1. Name the experiment and write, or picture, a sample space of equally likely outcomes.
2. Decide: ordered or unordered, with or without replacement.
3. Count the event, or its complement if that is shorter.
4. Check that the result is between 0 and 1 and roughly the size you expected; then check by the other route.