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This page sets out the method for combined events in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Combined events topic page.
When two things happen (coin and spinner, two choices) and the number of outcomes is small.
When both events have several outcomes, especially two dice, use a grid.
The probability of A and then B, when they are independent, is P(A) × P(B).
A coin and a die do not affect each other.
Heads-then-a-six, or tails-then-a-six, are two paths.
Add those path probabilities. Do not add if the paths overlap.
Write the outcomes at the ends of the branches and the probabilities on the branches.
Each pair of branches from one point adds to 1.
With replacement, the second probability matches the first.
Without replacement, the second denominator drops by 1.
Two coins have four equally likely outcomes.
You can list them instead of drawing a tree, then count the successful ones.
All the end probabilities should add to 1.
If they do not, a branch probability is wrong.
Start by saying what one outcome looks like. For a coin and a dice, one outcome is a face of the coin together with a face of the dice. The coin has 2 faces and the dice has 6 faces, so there are 2 × 6 = 12 outcomes, all equally likely if both are fair. Adding 2 + 6 = 8 would mix two different experiments and miss most of the pairs. The product is the size of the list.
Count the successful pairs inside that list. Heads together with an even dice score is three outcomes if the even scores are 2, 4 and 6. The probability is 3/12, which simplifies to 1/4. Listing is slow for bigger experiments, but it shows why the fraction has that denominator. Use it whenever the numbers stay small enough to write down.
Method: 1. Name the outcome of each part. 2. Multiply the number of outcomes in each part. 3. Count the pairs that match the question. 4. Simplify the successful count over the total.
You multiply when both events have to happen and one does not change the other. A coin does not care what a spinner does. The probability of heads is 1/2. If the spinner has 4 equal sections and you want one named section, that probability is 1/4. Both together is 1/2 × 1/4 = 1/8. You are taking a fraction of a fraction. That is why the answer is smaller than either probability on its own.
The same idea is the chance of two sixes on fair dice: 1/6 × 1/6 = 1/36. There is only one cell, double six, in a 6 by 6 grid. Adding 1/6 + 1/6 would be the probability that a chosen dice shows a six, or that one named dice shows a six, not that both do. Read the word and. And means multiply for independent events.
Method: 1. Check that the events are independent. 2. Write each probability as a fraction. 3. Multiply the numerators and multiply the denominators. 4. Cancel any common factor in the product.
You add when the question says or, and the events cannot happen on the same outcome. One counter taken from a bag is either red or green or some other colour. It cannot be red and green at once, so those events are mutually exclusive. If 3 counters out of 10 are red and 4 out of 10 are green, the probability of red or green is 3/10 + 4/10 = 7/10. The denominators already match because it is the same bag.
Do not multiply those two fractions. Multiplying would mean red and green on the same draw, which cannot happen, so the probability would be 0, not the product. Also do not add events that can overlap. If a card can be both red and even, adding the two probabilities counts that card twice. Overlap belongs to a later method. Here, add only when the outcomes are separate.
Method: 1. Check the events cannot happen together. 2. Write both probabilities over the same total. 3. Add the numerators and keep the denominator. 4. Simplify the sum if a common factor appears.
From one point on a tree, the branches are the choices for that stage. One of them must happen, and they do not happen together, so their probabilities add to 1. If the probability of red is 2/5, the probability of not red on that same draw is 1 - 2/5 = 3/5. Then 2/5 + 3/5 = 1. If your pair of branches adds to something else, a probability has been copied down wrongly.
The ends of a full tree also add to 1, because they are all the routes and exactly one route happens. That is a useful check after you multiply along each route and add the routes you need. It does not mean you add along a single route. Along a route you still multiply, because every branch on that route has to happen. Add only when you are joining separate successful routes.
Method: 1. Write both probabilities that leave the same point. 2. Add them and check the total is 1. 3. Multiply the probabilities that lie on one route. 4. Add the routes only if more than one route succeeds.
A sample space shows every possible outcome of a combined event. For a coin and a spinner with three sections, list the outcomes in a fixed order: H red, H blue, H green, T red, T blue, T green. Listing in order avoids missing or repeating an outcome, which is how marks are lost in a "list all the possible outcomes" question. The number of outcomes equals the number of outcomes of the first event times the number of the second, so here it is 2 × 3 = 6.
For two dice or two spinners with numbers, use a table with one event along the top and the other down the side, and fill each cell with the score or total. Count the cells that match the question and divide by the total number of cells. For two 6-sided dice there are 36 cells, so the probability of a total of 7 is 6/36 = 1/6. Write "out of 36" in your working so that the denominator is clear.
Method: 1. Decide the two events and write the number of outcomes of each. 2. List or tabulate every combination in a fixed order. 3. Check the count equals the product of the two numbers of outcomes. 4. Count the favourable outcomes and write them over the total.
Tip: If your list does not match the product of the numbers of outcomes, an outcome is missing or repeated.
Two events are independent when one does not change the probability of the other, such as a coin and a dice. For independent events A and B, P(A and B) = P(A) × P(B). So the probability of tails and a number greater than 4 is 1/2 × 2/6 = 1/6. The word "and" joins two different events, and the answer is smaller than either probability on its own.
Events are mutually exclusive when they cannot both happen on the same turn, such as red and blue when one counter is taken. For mutually exclusive events A and B, P(A or B) = P(A) + P(B). A spinner with P(red) = 0.35 and P(blue) = 0.4 has P(red or blue) = 0.75. Check that your answer is not bigger than 1. Multiply for "and" with independent events, and add for "or" with mutually exclusive events.
Method: 1. Decide whether the events are separate (independent) or alternatives on one turn (exclusive). 2. For "and" with independent events, multiply the probabilities. 3. For "or" with exclusive events, add the probabilities. 4. Check the answer is between 0 and 1.
Worded questions give separate probabilities for two people or two days, such as "the probability that Ben scores a penalty is 0.7 and the probability that Chloe scores is 0.9". Treat the events as independent when the question says so. Both score: 0.7 × 0.9 = 0.63. Neither scores: 0.3 × 0.1 = 0.03, using 1 - 0.7 = 0.3 and 1 - 0.9 = 0.1. Exactly one scores means two separate cases, Ben only and Chloe only, which are added: 0.7 × 0.1 + 0.3 × 0.9 = 0.34.
Check by adding all the cases: both, neither and exactly one must total 1, and 0.63 + 0.03 + 0.34 = 1. Some Higher questions give an answer and ask for an unknown probability, for example the probability that Jo wins game 2 when she wins game 1 with probability 0.6 and both games with probability 0.42. Set up the product 0.6 × p = 0.42 and divide to get p = 0.7. Write each case on its own line so that method marks are visible.
Method: 1. Write the probability of each event and of its opposite, using 1 - p. 2. Identify each case the question describes, such as both, neither or exactly one. 3. Multiply along each case and add the cases that belong together. 4. Check that the cases of the whole event add up to 1.
Show-that questions give you the target, so your working has to reach it exactly. In a question that says show that the probability of yellow or green is 2/3, count the yellow and the green counters, add them, divide by the total and simplify to 2/3 on the last line. Mark schemes give a mark for each correct step and one more for the conclusion, so do not skip steps even when they seem obvious.
Always check that a probability lies between 0 and 1 and that all the cases of an event add to 1. If you get a probability bigger than 1, you have probably added where you should have multiplied, or counted an outcome twice. Write fractions in their simplest form unless told otherwise, and keep decimals exact, so 0.72 and not 0.7. For list questions, count your outcomes: if you listed 7 outcomes but should have 2 × 4 = 8, one is missing.
Method: 1. Write the favourable outcomes and the total number of outcomes separately. 2. For a show-that, finish on exactly the number given in the question. 3. Check the probability is between 0 and 1 and simplified. 4. Add up all the cases to see that they total 1.
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