Probability measures how likely an event is, on a scale from 0 to 1. This note covers probability from equally likely outcomes and from relative frequency, expected numbers, mutually exclusive and independent events, sample space diagrams, Venn diagrams and tree diagrams for independent events. These ideas are used in insurance, medicine, games and quality control.
Key ideas
An event is a set of outcomes we are interested in. The probability of an event \(A\), written \(P(A)\), is a number from 0 to 1. A probability of 0 means the event is impossible and 1 means it is certain. Probabilities can be written as fractions, decimals or percentages, but never as ratios or with a value above 1.
When all outcomes are equally likely:
\[P(A)=\frac{\text{number of outcomes in }A}{\text{total number of possible outcomes}}\]
- Complement: the probability that \(A\) does not happen is \(P(\text{not }A)=1-P(A)\). The probabilities of all the possible outcomes add up to 1.
- Relative frequency: when outcomes are not equally likely (for example a biased coin), estimate the probability from an experiment: relative frequency \(=\dfrac{\text{number of times the event happens}}{\text{total number of trials}}\). More trials give a better estimate.
- Expected number: if an experiment is repeated \(n\) times, the expected number of times \(A\) happens is \(n\times P(A)\). It is an average over many trials, not a promise of what will happen.
Combining events.
- Events are mutually exclusive if they cannot happen at the same time. Then \(P(A\text{ or }B)=P(A)+P(B)\). This is the addition rule.
- Events are independent if one happening does not change the probability of the other. Then \(P(A\text{ and }B)=P(A)\times P(B)\). This is the multiplication rule. Examples: separate coin tosses, or picking with replacement.
- If events can overlap, then \(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\). The overlap is subtracted so that it is not counted twice.
Sample space diagrams. To list the outcomes of two events (for example two dice), draw a grid with one event along the top and the other down the side. Each cell is an outcome. Counting cells gives the probability for equally likely outcomes. Two fair dice give 36 outcomes.
Venn diagrams. A Venn diagram shows how a group splits into overlapping sets. The region where two circles overlap is "both". Fill in the overlap first, then work outwards by subtracting. The total of all regions, including those outside the circles, is the size of the whole group.
Tree diagrams. A tree diagram shows the outcomes of events in order. Each pair of branches shows the possible results of one event, with the probability written on every branch. The probabilities on the branches from one point add up to 1. Multiply along a path to find the probability of that combination ("and"), and add the paths that give the result you want ("or"). This note uses trees where the events are independent, so the same probabilities appear at every stage.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(P(A)=\dfrac{\text{favourable}}{\text{total}}\) | Equally likely outcomes |
| \(P(\text{not }A)=1-P(A)\) | Complement. Useful for "at least one" questions |
| Relative frequency | Estimate of probability from an experiment |
| Expected number \(=n\times P(A)\) | How many times to expect \(A\) in \(n\) trials |
| \(P(A\text{ or }B)=P(A)+P(B)\) | Only if \(A\) and \(B\) are mutually exclusive |
| \(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\) | General case, with overlap |
| \(P(A\text{ and }B)=P(A)\times P(B)\) | Only if \(A\) and \(B\) are independent |
Worked examples
Example 1 A bag contains 5 red, 7 blue, 3 green and 5 yellow counters. One counter is taken at random. Find the probability that it is (a) blue or green, (b) not red. The counter is replaced and the process is repeated 80 times in total. (c) How many red counters would you expect?
The total number of counters is \(5+7+3+5=20\).
(a) Blue and green cannot both happen, so the events are mutually exclusive. \(P=\tfrac{7}{20}+\tfrac{3}{20}=\tfrac{10}{20}=\tfrac12\).
(b) \(P(\text{red})=\tfrac{5}{20}=\tfrac14\), so \(P(\text{not red})=1-\tfrac14=\tfrac34\).
(c) Expected number \(=80\times\tfrac14=20\).
Answer: (a) \(\tfrac12\), (b) \(\tfrac34\), (c) 20 red counters.
Example 2 Each morning, the probability that Maya's bus is on time is 0.85 and the probability that her train is on time is 0.9. The two are independent. Find the probability that (a) both are on time, (b) exactly one is on time, (c) at least one is on time.
Draw a tree. First branches: bus on time 0.85, late 0.15. From each, train on time 0.9, late 0.1.
(a) Multiply along the path: \(0.85\times0.9=0.765\).
(b) Two paths give exactly one on time. Bus on time and train late: \(0.85\times0.1=0.085\). Bus late and train on time: \(0.15\times0.9=0.135\). Add them: \(0.085+0.135=0.22\).
(c) The opposite of "at least one on time" is "both late". \(P(\text{both late})=0.15\times0.1=0.015\), so the answer is \(1-0.015=0.985\).
Check: \(0.765+0.22+0.015=1\).
Answer: (a) 0.765, (b) 0.22, (c) 0.985.
Example 3 Two fair dice are rolled and the scores are added. Find the probability that the total is (a) 8, (b) at least 10, (c) a prime number.
The sample space has \(6\times6=36\) equally likely outcomes.
(a) A total of 8 comes from (2,6), (3,5), (4,4), (5,3), (6,2). That is 5 outcomes, so \(P=\tfrac{5}{36}\).
(b) Totals of 10, 11 or 12: (4,6), (5,5), (6,4) for 10; (5,6), (6,5) for 11; (6,6) for 12. That is 6 outcomes, so \(P=\tfrac{6}{36}=\tfrac16\).
(c) The prime totals are 2, 3, 5, 7 and 11. The numbers of ways are 1, 2, 4, 6 and 2, which add up to 15. \(P=\tfrac{15}{36}=\tfrac{5}{12}\).
Answer: (a) \(\tfrac{5}{36}\), (b) \(\tfrac16\), (c) \(\tfrac{5}{12}\).
Common errors
- Adding probabilities when the question says "and", or multiplying when it says "or". Along a tree path you multiply; between paths you add.
- Using the addition rule \(P(A)+P(B)\) when events overlap, without subtracting \(P(A\text{ and }B)\).
- Writing probabilities as "1 in 4" or "1:4" instead of \(\tfrac14\), 0.25 or 25%, or giving an answer greater than 1.
- Treating events as independent when they are not. Check whether the first result changes the second probability.
- Writing a sample space with outcomes missing, for example listing (2,6) but not (6,2). The order matters with two dice.
Exam tip: For "at least one", work out the opposite event ("none") and subtract from 1. This is much quicker than adding many paths of a tree.
Check yourself
- A biased coin is flipped 200 times and lands on heads 124 times. Estimate \(P(\text{heads})\), and the expected number of heads in 500 flips.
Answer
Relative frequency \(=124\div200=0.62\). Expected heads in 500 flips \(=500\times0.62=310\). - The probability of rain tomorrow is 0.35. What is the probability that it does not rain?
Answer
\(1-0.35=0.65\). - Events \(A\) and \(B\) are independent with \(P(A)=0.3\) and \(P(B)=0.5\). Find \(P(A\text{ and }B)\) and \(P(A\text{ or }B)\).
Answer
\(P(A\text{ and }B)=0.3\times0.5=0.15\). \(P(A\text{ or }B)=0.3+0.5-0.15=0.65\). - In a class of 40 students, 25 study French, 18 study Spanish and 8 study both. A student is chosen at random. Find the probability that the student studies French or Spanish.
Answer
Number studying French or Spanish \(=25+18-8=35\). The probability is \(\tfrac{35}{40}=\tfrac78\).