This note covers the probability of a single event: how to measure chance on a 0 to 1 scale, how to calculate probabilities when outcomes are equally likely, how to estimate them from experiments using relative frequency, and how to predict how many times something should happen. Probability is used in insurance, quality control, medicine and games.
Key ideas
An event is something that might happen, such as rolling an even number. An outcome is one possible result, such as rolling a 4. Probability measures how likely an event is, on a scale from 0 to 1.
- Probability 0 means the event is impossible.
- Probability 1 means the event is certain.
- Probability \(\tfrac12\) (or 0.5, or 50%) means even chance.
Probabilities can be written as fractions, decimals or percentages. They are never negative and never more than 1.
Equally likely outcomes. When every outcome has the same chance (a fair die, a fair coin, a well-mixed bag), use:
\[ P(\text{event}) = \frac{\text{number of outcomes that give the event}}{\text{total number of possible outcomes}} \]
For example, a fair six-sided die has 6 outcomes. The probability of rolling a prime number (2, 3 or 5) is \(\tfrac36 = \tfrac12\).
Complementary events. The event "not A" happens exactly when A does not. Because one or the other must occur:
\[ P(\text{not } A) = 1 - P(A) \]
Mutually exclusive events cannot both happen at the same time (for example, rolling a 2 and rolling a 5 in one throw). For mutually exclusive events, add the probabilities to find the chance that either happens. The probabilities of all the possible outcomes of an experiment add up to 1. This helps you find a missing probability.
Relative frequency. Sometimes outcomes are not equally likely, or you cannot work the probability out by counting (a biased coin, a faulty machine, tomorrow's weather). You can estimate the probability by doing an experiment or using past data:
\[ \text{relative frequency} = \frac{\text{number of times the event happened}}{\text{total number of trials}} \]
The more trials you have, the more reliable the estimate. A fair coin tossed 10 times might give 7 heads, but after 1000 tosses the relative frequency will be much closer to 0.5.
Expected number of outcomes. If you repeat a situation \(n\) times and the probability of the event each time is \(p\), then you expect the event about \(n \times p\) times. This is an average prediction, not a promise. Actual results will usually differ a little.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P(A)\) | The probability that event \(A\) happens, with \(0 \le P(A) \le 1\). |
| \(P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}\) | Use when all outcomes are equally likely. |
| \(P(\text{not } A) = 1 - P(A)\) | Find the probability that an event does not happen. |
| Sum of all probabilities \(= 1\) | Find a missing probability when you know the others. |
| Relative frequency | Estimate of probability from an experiment or data. |
| Expected number \(= n \times p\) | Predict how many times an event happens in \(n\) trials. |
Worked examples
Example 1 A bag contains 5 red counters, 3 green counters and 2 yellow counters. One counter is taken out at random. Find the probability that it is (a) red, (b) not red, (c) green or yellow.
There are \(5 + 3 + 2 = 10\) counters, all equally likely to be chosen.
(a) \(P(\text{red}) = \dfrac{5}{10} = \dfrac12\).
(b) \(P(\text{not red}) = 1 - \dfrac12 = \dfrac12\). Check by counting: \(3 + 2 = 5\) counters are not red, and \(\dfrac{5}{10} = \dfrac12\).
(c) Green and yellow cannot both happen, so \(P(\text{green or yellow}) = \dfrac{3}{10} + \dfrac{2}{10} = \dfrac{5}{10} = \dfrac12\).
Answer: (a) \(\tfrac12\). (b) \(\tfrac12\). (c) \(\tfrac12\).
Example 2 A factory tests 400 phones and finds that 18 are defective. (a) Estimate the probability that a phone chosen at random is defective. (b) A batch has 5000 phones. How many would you expect to be defective?
(a) Relative frequency \(= \dfrac{18}{400} = 0.045\).
(b) Expected number \(= 5000 \times 0.045 = 225\).
Answer: (a) 0.045. (b) 225 phones.
Example 3 A biased spinner has four sections numbered 1 to 4. The probability of 1 is 0.2 and the probability of 2 is 0.35. The probability of 4 is twice the probability of 3. (a) Find the probability of 3 and of 4. (b) The spinner is spun 200 times. How many times is a 4 expected?
(a) Let \(P(3) = x\). Then \(P(4) = 2x\). All the probabilities add to 1:
\(0.2 + 0.35 + x + 2x = 1\), so \(0.55 + 3x = 1\), \(3x = 0.45\) and \(x = 0.15\).
So \(P(3) = 0.15\) and \(P(4) = 2 \times 0.15 = 0.3\).
(b) Expected number of 4s \(= 200 \times 0.3 = 60\).
Answer: (a) \(P(3) = 0.15\), \(P(4) = 0.3\). (b) 60 times.
Common errors
- Common error: using "favourable out of total" when the outcomes are not equally likely, such as treating a biased spinner as fair.
- Common error: writing a probability greater than 1 or a negative probability. If you get one, there is a mistake.
- Common error: writing probabilities as ratios such as "3 : 7", or in words such as "3 in 10 chance" with no number. Use a fraction, decimal or percentage.
- Common error: thinking the expected number must happen exactly. It is a prediction, and may not be a whole number of events in one experiment.
- Common error: forgetting to include the item you are looking for in the total, for example using 9 instead of 10 counters.
Exam tip: Use \(P(\text{not } A) = 1 - P(A)\) whenever the question says "not" or "at least one", because counting the opposite is often quicker. Give fractions in simplest form unless asked otherwise.
Check yourself
- A fair die is rolled. What is the probability of a number greater than 4?
Answer
The numbers are 5 and 6, so 2 outcomes out of 6: \(\tfrac26 = \tfrac13\). - The probability that a train is late is 0.12. What is the probability that it is not late?
Answer
\(1 - 0.12 = 0.88\). - In 80 shots, a basketball player scores 56 times. Estimate the probability that she scores with her next shot, and the number of scores expected in 150 shots.
Answer
Relative frequency \(= \dfrac{56}{80} = 0.7\). Expected scores \(= 150 \times 0.7 = 105\). - A bag has only blue and white beads. \(P(\text{blue}) = \tfrac37\). There are 28 beads in total. How many are white?
Answer
\(P(\text{white}) = 1 - \tfrac37 = \tfrac47\). Number of white beads \(= \tfrac47 \times 28 = 16\).