Hard GCSE Tree diagrams Questions

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 $\frac{16}{25}$.

Turn "at least one red" into its opposite

"At least one red" is true on THREE of the four paths. Its opposite — no red at all — is the single blue-then-blue 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 $8$ counters in the bag, $5$ red and $3$ blue. The two branches add to $1$.

Work out the second-stage branch after a blue counter

The first counter was blue and it was NOT put back, so only $7$ counters are left and there is one fewer blue counter: $5$ red and $2$ blue. The two draws are DEPENDENT — the denominator drops from $8$ to $7$.

Multiply along the blue-then-blue path to get P(no red)

To follow one path you MULTIPLY along its branches: $\frac{3}{8} \times \frac{2}{7} = \frac{3}{28}$. That is the probability that the first is blue and the second is blue.

Subtract from $1$

No red counter at all has probability $\frac{3}{28}$, so at least one red counter has probability $1 - \frac{3}{28} = \frac{25}{28}$.

Rule out the answer the WRONG assumption gives

$\frac{55}{64}$ is what "at least one red" comes to if you assume the counter is 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 $\frac{5}{14}$

$\frac{5}{14}$ is the probability that BOTH counters are red. "At least one" also allows exactly one, so it must be the bigger number.

Rule out $\frac{15}{28}$

$\frac{15}{28}$ is the probability of EXACTLY one red counter. "At least one" includes the case of two, so $\frac{15}{28}$ leaves out the red-then-red path — and indeed $\frac{15}{28} + \frac{5}{14} = \frac{25}{28}$, which is a neat check that the answer is right.

Rule out $\frac{5}{8}$

$\frac{5}{8}$ only looks at the first counter and forgets that the second one could be red 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 red gives $\frac{25}{28}$ — 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 not put back, so the draws are DEPENDENT and each second-stage node has its own branch probabilities.

Write the answer as a decimal

$\frac{25}{28}$ is about $0.893$ — a high probability, which is what you would expect when only $3$ of the $8$ counters are not red.

State the answer

So the probability that at least one of the counters is red is $\frac{25}{28}$.

Decide whether the two draws are independent

The cube 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 $11$ cubes: $7$ red and $4$ black. These two branches add to $1$.

Work out the second-stage branch for a red cube

The cube is put back, so the bag is exactly as it was: $11$ cubes, $7$ of them red and $4$ of them black. The two draws are INDEPENDENT, so the second set of branches carries the same probabilities as the first. So the probability that the second cube is red is $\frac{7}{11} = \frac{7}{11}$.

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 cube 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

$\frac{3}{5}$ is precisely the probability you would get by assuming the cube were taken out and kept. Under the assumption this question actually states, the answer is $\frac{7}{11}$. 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 black second cube

$\frac{2}{5}$ is the probability that the second cube is black, not red. 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 cube must be red or black, 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

$\frac{7}{11}$ is about $0.636$.

Say what independence means in one line

Two events are independent exactly when knowing that the first happened does not change the probability of the second. Replacing the item makes this true; not replacing it makes it false.

Note what would happen to the whole tree

Because the draws are independent, both second-stage nodes carry the same pair of probabilities, and the tree is symmetric.

Note the mistake to avoid

On the second set of branches the denominator is $11$. It does NOT drop, because the item went back.

Check the answer is a possible probability

$\frac{7}{11}$ lies strictly between $0$ and $1$.

Work out the probability of the opposite

The probability that the second cube is NOT red is $\frac{4}{11}$, and the two add back to $1$ — which is the branch rule again.

State the answer

The draws are independent, and the probability that the second cube is red is $\frac{7}{11}$. Only one statement says both of those things.

Decide whether the two draws are independent

The ball is NOT put back, so the bag has changed by the time the second ball is taken. The first result therefore does affect the second: the two draws are NOT independent.

Write down the first pair of branches

Before anything is taken there are $10$ balls: $6$ blue and $4$ yellow. These two branches add to $1$.

Work out the second-stage branch for a blue ball

The first ball was blue and it was NOT put back, so only $9$ balls are left and there is one fewer blue ball: $5$ blue and $4$ yellow. The two draws are DEPENDENT — the denominator drops from $10$ to $9$. So the probability that the second ball is blue is $\frac{5}{9} = \frac{5}{9}$.

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

Two of the statements claim the draws are independent. The question says the ball is not 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

$\frac{3}{5}$ is precisely the probability you would get by assuming the ball were put back. Under the assumption this question actually states, the answer is $\frac{5}{9}$. 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 yellow second ball

$\frac{2}{5}$ is the probability that the second ball is yellow, 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 ball must be blue or yellow, 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 DIFFERENT numbers. That is the signature of dependence: what came first genuinely changes the second probability, so the two nodes cannot share one set of branches.

Write the probability as a decimal

$\frac{5}{9}$ is about $0.556$.

Say what independence means in one line

Two events are independent exactly when knowing that the first happened does not change the probability of the second. Replacing the item makes this true; not replacing it makes it false.

Note what would happen to the whole tree

Because the draws are dependent, each second-stage node needs its own pair of probabilities, worked out from what is left in the bag after that particular first draw.

Note the mistake to avoid

On the second set of branches the denominator is $9$, not $10$, because one ball has gone and is not coming back.

Check the answer is a possible probability

$\frac{5}{9}$ lies strictly between $0$ and $1$.

Work out the probability of the opposite

The probability that the second ball is NOT blue is $\frac{4}{9}$, and the two add back to $1$ — which is the branch rule again.

State the answer

The draws are not independent, and the probability that the second ball is blue is $\frac{5}{9}$. Only one statement says both of those things.

Write down the first pair of branches

The first token comes from all $8$ tokens, so the first branch of the path is $\frac{5}{8}$. The two first-stage branches add to $1$.

Decide what the second-stage branches must be

The token is put back, so the bag is exactly as it was: $8$ tokens, $5$ of them red and $3$ of them blue. The two draws are INDEPENDENT, so the second set of branches carries the same probabilities as the first.

Multiply along the branches of the path

To follow one path you MULTIPLY along its branches: $\frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$. That is the probability that the first is red and the second is red.

Check the answer is a possible probability

$\frac{25}{64}$ lies strictly between $0$ and $1$.

Rule out the answer the WRONG assumption gives

$\frac{5}{14}$ is exactly what you get by assuming the opposite of what the question says — that the token is NOT put back. It is a perfectly good answer to a different question, and it is the single most tempting wrong option here. The question tells you which assumption applies; read it.

Rule out $\frac{25}{56}$

This drops the TOTAL from $8$ to $7$ but leaves the number of red tokens at $5$. If a red token has been removed, both numbers change; if none has been removed, neither does. Changing one and not the other is never right.

Rule out $\frac{5}{16}$

This is the mirror-image slip: the number of red tokens drops to $4$ but the total stays at $8$. The token cannot leave the bag without changing the total.

Rule out $\frac{9}{64}$

$\frac{9}{64}$ is the probability that both tokens are blue, not red. It answers the wrong question.

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

Every set of branches leaving a node adds to exactly $1$, so the completed tree is right and any path can now be read off it safely.

Work out the probability of every path

All four paths, for completeness: red then red: $\frac{25}{64}$, red then blue: $\frac{15}{64}$, blue then red: $\frac{15}{64}$ and blue then blue: $\frac{9}{64}$.

Check the four paths add to $1$

The four paths are everything that can happen, so they must add to $1$ — which checks every multiplication at once.

Say what is being assumed

Putting the counter back makes the two draws INDEPENDENT: the first result tells you nothing about the second.

Note the mistake to avoid: along and across

This event is a single path, so it is one multiplication. Adding the two branch probabilities instead would give a number bigger than $1$.

Write the answer as a decimal

$\frac{25}{64}$ is about $0.390625$.

State the answer

So the probability that both tokens are red is $\frac{25}{64}$.