Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for probability trees 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 Probability trees topic page.
When two (or more) things happen one after another and you want the probability of a combination.
When the question says 'at least one', 'at least once' or 'not both fail'.
For branches 1/2 and 1/3 the path probability is 1/6. Multiply.
Take two counters without replacement from 4 counters (3 red). The first is red with probability 3/4. Then 2 red are left in 3 counters, so the second is red with probability 2/3. Both red is 3/4 × 2/3 = 1/2.
With replacement, probabilities do not change on the second draw.
The probability of not taking a branch is 1 minus the branch probability.
With replacement means outcomes are independent; the denominators remain constant.
Without replacement, the total count drops by 1 on the second draw.
Read a route from left to right. Suppose the first branch is 1/2 and the next branch is 1/3. Both of those events have to happen for that end outcome, so the probability of the route is 1/2 × 1/3 = 1/6. Adding 1/2 + 1/3 would mix two choices that are not alternatives. They sit on the same journey, so they multiply.
The answer is smaller than either branch, which is what you should expect. Asking for two things to happen is stricter than asking for one of them. After you multiply, cancel if the top and bottom share a factor. Then look at the other routes only if the question accepts more than one ending. One route, one product.
Method: 1. Trace the route that matches the question. 2. Write the probability on each branch of that route. 3. Multiply those fractions. 4. Simplify the product.
One red and one blue can happen in two orders: red then blue, or blue then red. Those routes cannot happen together, so you multiply along each one and then add. From 7 red and 3 blue out of 10, red then blue is 7/10 × 3/9 and blue then red is 3/10 × 7/9. Each product is 21/90. Adding gives 42/90, which is 7/15. Missing one order cuts the answer in half.
At least one is often easier as a complement. Suppose you want at least one toffee. None means both draws miss the toffee, which is one route. Work out that route, then subtract it from 1. You are not adding a mystery extra chance. Every outcome is either no toffee or at least one toffee, so those two probabilities add to 1. Use this when the successful side has several routes and the failure side has one.
Method: 1. List the routes that match, including both orders. 2. Multiply along each route. 3. Add those products if the routes are separate. 4. For at least one, subtract the none route from 1.
In a tree diagram question you are often asked to complete the tree by filling in missing probabilities on the branches. The rule is that the branches leaving any one point always add up to 1. If one branch from a point is 0.35, the other is 1 - 0.35 = 0.65. With two bags, or with replacement, the second-stage branches are the same from every point. Without replacement the second-stage probabilities change, because both the total and the number of that colour are smaller.
Label each branch with its outcome and its probability, and write the outcome at the end of each path. A neat tree has the first event on the left and the second on the right. When counters are involved, write the probabilities as fractions with the right denominators. With 8 counters the first stage has denominator 8, and without replacement the second stage has denominator 7. Check that each set of branches from a point sums to 1.
Method: 1. Write the outcomes at the first stage and their probabilities, which add up to 1. 2. For each first-stage outcome, write the second-stage branches from that point. 3. Without replacement, take one counter off the total and one off the colour just taken. 4. Check that every set of branches from a point adds to 1.
Tip: Branches from the same point always add to 1. Paths from the start to the end, all added together, also give 1.
Read the wording carefully. "Both" means one path. "Exactly one" or "different colours" means two or more paths that are added. For two socks taken without replacement from 5 black and 3 white, different colours has two paths, black then white and white then black: 5/8 × 3/7 + 3/8 × 5/7 = 15/28. Multiply along each path, then add the paths that match the question.
"At least one" is easier with the complement, because at least one is everything except none. So P(at least one) = 1 - P(none). If the probability of rain on a day is 0.3, then the probability of no rain on two days is 0.7 × 0.7 = 0.49, so at least one rainy day has probability 1 - 0.49 = 0.51. Check by adding the paths directly: 0.09 + 0.21 + 0.21 = 0.51.
Method: 1. Decide which paths match the wording: both, exactly one, or at least one. 2. Multiply along each matching path. 3. Add the paths, or use 1 - the probability of none for "at least one". 4. Simplify the fraction or round the decimal as asked.
Some trees stop when something happens. A player has up to three attempts at a target and stops as soon as she hits it. If each attempt hits with probability 0.7, the paths are: hit on the first attempt, 0.7; miss then hit, 0.3 × 0.7 = 0.21; and miss, miss, then hit, 0.3 × 0.3 × 0.7 = 0.063. The tree has no branches after a hit, because she stops there.
To find the probability of hitting within three attempts, add the paths, 0.7 + 0.21 + 0.063 = 0.973. Or use the complement: all three miss with probability 0.3^3 = 0.027, so the answer is 1 - 0.027 = 0.973. The complement method has far less working. The probability of needing all three attempts is the probability of missing the first two, 0.3 × 0.3 = 0.09, whatever happens on the third.
Method: 1. Draw the branches only until the stopping event happens. 2. Multiply along each path to its end. 3. Add the paths for "within" questions, or subtract "none" from 1. 4. Check that all the end probabilities of the whole tree add to 1.
Higher questions sometimes give a probability and ask for a number of counters. Let r be the number of red counters in a bag of 8. Without replacement, the probability that both counters are red is r/8 × (r - 1)/7. If that equals 3/14, multiply out to get (r(r - 1))/56 = 3/14, so r(r - 1) = 12 and r^2 - r - 12 = 0.
Factorise to (r - 4)(r + 3) = 0, which gives r = 4 or r = -3. A number of counters cannot be negative, so r = 4. Say this rejection in words, because it earns the last mark. Check by putting r = 4 back in: 4/8 × 3/7 = 12/56 = 3/14. Always check that your answer is a whole number and does not exceed the total.
Method: 1. Let a letter stand for the unknown number of counters. 2. Write the second-stage probability using one fewer counter and one fewer of that colour. 3. Set the product equal to the given probability and rearrange into a quadratic. 4. Solve, reject any impossible value and check the answer.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.