GCSE Tree diagrams Practice Questions

Use the rule that the branches from a node add to $1$

At the first node the spinner either lands on red or it does not — there is no third possibility. Every set of branches leaving a node must add to exactly $1$, so these two probabilities add to $1$.

Subtract from $1$

$1 - \frac{1}{4} = \frac{3}{4}$.

State the answer

So the probability that the spinner does not land on red is $\frac{3}{4}$, and $\frac{1}{4} + \frac{3}{4} = 1$ as a check.

Write down the probabilities on every branch

Beth wins round 1 with probability $\frac{1}{4}$, so loses it with probability $1 - \frac{1}{4} = \frac{3}{4}$; the same for round 2. The rounds are INDEPENDENT, so the round-2 branches carry the same numbers whichever way round 1 went — that is what independence means.

Multiply along the branches of the path

MULTIPLY along the two branches of this path: $\frac{1}{4} \times \frac{3}{5} = \frac{3}{20}$.

State the answer

So the probability is $\frac{3}{20}$.

Write down the two probabilities along the path

The first branch is $\frac{7}{12}$: $7$ red cubes out of $12$. The first cube was red and it was NOT put back, so only $11$ cubes are left and there is one fewer red cube: $6$ red and $5$ green. The two draws are DEPENDENT — the denominator drops from $12$ to $11$. So the second branch of this path is $\frac{5}{11}$.

Multiply along the branches of the path

To follow one path you MULTIPLY along its branches: $\frac{7}{12} \times \frac{5}{11} = \frac{35}{132}$. That is the probability that the first is red and the second is green.

Check that the branches from every node add to $1$

Something must happen at each stage, so every set of branches leaving a node has to add to exactly $1$. All three sets do here, so the tree is drawn correctly. If a set does not add to $1$, stop — the rest of the working cannot be right.

Say what is being assumed

Because the cube is NOT put back, the second draw is made from a different bag. The draws are DEPENDENT: the total falls from $12$ to $11$, and whichever colour went first has one fewer left. This is why the two second-stage nodes carry different numbers.

Work out the probability of every path

Multiplying along each of the four paths in turn gives the probability of each of the four possible outcomes of the experiment: red then red: $\frac{7}{22}$, red then green: $\frac{35}{132}$, green then red: $\frac{35}{132}$ and green then green: $\frac{5}{33}$.

State the answer

So the probability is $\frac{35}{132}$.

Decide whether the two draws are independent

The disc is put back, so the bag is exactly the same for the second draw and the first result cannot affect the second. The two draws ARE independent.

Write down the first pair of branches

Before anything is taken there are $7$ discs: $5$ blue and $2$ green. These two branches add to $1$.

Work out the second-stage branch for a blue disc

The disc is put back, so the bag is exactly as it was: $7$ discs, $5$ of them blue and $2$ of them green. The two draws are INDEPENDENT, so the second set of branches carries the same probabilities as the first. So the probability that the second disc is blue is $\frac{5}{7} = \frac{5}{7}$.

Rule out the statements that get the independence the wrong way round

Two of the statements claim the draws are not independent. The question says the disc is put back, so that claim is false whatever number follows it. A statement is only correct if BOTH halves of it are.

Rule out the number the wrong assumption gives

$\frac{2}{3}$ is precisely the probability you would get by assuming the disc were taken out and kept. Under the assumption this question actually states, the answer is $\frac{5}{7}$. This is the whole point of the question: the two assumptions give different numbers, so you must use the one you are told.

Rule out the statement about a green second disc

$\frac{1}{3}$ is the probability that the second disc is green, not blue. It is a true statement about a different branch, and the wrong answer to this question.

Check the second-stage branches add to $1$

Whatever was taken first, the second disc must be blue or green, so the two branches out of that node add to exactly $1$. They do.

Compare the two nodes of the second stage

The two second-stage nodes carry the SAME numbers. That is the signature of independence — and the reason it does not matter what came first.

Write the probability as a decimal

$\frac{5}{7}$ is about $0.714$.

State the answer

The draws are independent, and the probability that the second disc is blue is $\frac{5}{7}$. Only one statement says both of those things.

Turn "at least one green" into its opposite

"At least one green" is true on THREE of the four paths. Its opposite — no green at all — is the single yellow-then-yellow path. One multiplication and one subtraction beats three multiplications and two additions, and there is far less to get wrong.

Write down the first pair of branches

There are $10$ marbles in the bag, $4$ green and $6$ yellow. The two branches add to $1$.

Work out the second-stage branch after a yellow marble

The marble is put back, so the bag is exactly as it was: $10$ marbles, $4$ of them green and $6$ of them yellow. The two draws are INDEPENDENT, so the second set of branches carries the same probabilities as the first.

Multiply along the yellow-then-yellow path to get P(no green)

To follow one path you MULTIPLY along its branches: $\frac{6}{10} \times \frac{6}{10} = \frac{9}{25}$. That is the probability that the first is yellow and the second is yellow.

Subtract from $1$

No green marble at all has probability $\frac{9}{25}$, so at least one green marble has probability $1 - \frac{9}{25} = \frac{16}{25}$.

Rule out the answer the WRONG assumption gives

$\frac{2}{3}$ is what "at least one green" comes to if you assume the marble is NOT put back — the opposite of what the question says. The two assumptions give different second-stage branches and therefore different answers, so the replacement sentence has to be read, not skimmed.

Rule out $\frac{4}{25}$

$\frac{4}{25}$ is the probability that BOTH marbles are green. "At least one" also allows exactly one, so it must be the bigger number.

Rule out $\frac{12}{25}$

$\frac{12}{25}$ is the probability of EXACTLY one green marble. "At least one" includes the case of two, so $\frac{12}{25}$ leaves out the green-then-green path — and indeed $\frac{12}{25} + \frac{4}{25} = \frac{16}{25}$, which is a neat check that the answer is right.

Rule out $\frac{2}{5}$

$\frac{2}{5}$ only looks at the first marble and forgets that the second one could be green instead. It is the first branch, not a whole event.

Check the answer the long way, by adding the three paths

Adding ACROSS the three paths that contain at least one green gives $\frac{16}{25}$ — the same answer, confirming that $1 - P(\text{none})$ was right. It also shows why the short way is worth knowing.

Check that the branches from every node add to $1$

Every node adds to exactly $1$, so the tree is correctly labelled.

Check the four paths add to $1$

The four paths cover everything that can happen, so they must add to $1$.

Say what is being assumed

The item is put back, so the draws are INDEPENDENT and the second-stage branches repeat the first-stage ones.

Write the answer as a decimal

$\frac{16}{25}$ is about $0.64$ — a high probability, which is what you would expect when only $6$ of the $10$ marbles are not green.

State the answer

So the probability that at least one of the marbles is green is $\frac{16}{25}$.