GCSE Conditional probability Practice Questions

Restrict the universe to the group you are choosing from

You are told the student is one of the girls. Every other student in the survey is now irrelevant: the choice is made from the $12$ girls only. So $12$ is the denominator — NOT $24$.

Count how many of that group are in the event, then write the fraction

Of those $12$ girls, $7$ walk to school — they are the $7$ girls who walk to school. So the probability is $\frac{7}{12} = \frac{7}{12}$.

State the answer

So given that the student is one of the girls, the probability that the student is also one of the students who walk to school is $\frac{7}{12}$.

Write the two stages as a product

$P(\text{1st} \cap \text{2nd}) = P(\text{1st}) \times P(\text{2nd} \mid \text{1st})$. The second fraction is CONDITIONAL on the first: the bag has lost a green token, so it holds $11$ tokens of which $8$ are purple.

Multiply and simplify

Multiply the tops and multiply the bottoms, then cancel: $\frac{8}{33}$.

State the answer

So the probability is $\frac{8}{33}$.

Read what the statements are all claiming

Every option is a conditional probability. Each one names the group the adult is known to be in, and then asks how likely it is that the adult is in a second group as well. So each is a count from inside the first group, divided by the size of that group.

Work out the denominator of the true statement

The condition is that the adult is in $A$. There are $16$ of them, so $16$ is the size of the restricted universe.

Form the fraction and simplify

Of the $16$ in the conditioning group, $6$ are in $B$ as well, so the probability is $\frac{3}{8}$.

Refute the statement that quotes the transposed conditional

$\frac{2}{5}$ is the same $6$ divided by the $15$ in the SECOND group. That is the probability the other way round. Conditioning on the first group means dividing by $16$, not $15$, so that statement is false.

Refute the remaining statements

$\frac{1}{5}$ divides by the whole survey of $30$ and so ignores the condition; $\frac{5}{8}$ is the complement, the chance of the opposite outcome inside the same restricted universe. Neither is the probability asked for.

State the answer

So the correct statement is the one giving the probability as $\frac{3}{8}$.

Read what the statements are all claiming

Every option is a conditional probability. Each one names the group the teenager is known to be in, and then asks how likely it is that the teenager is in a second group as well. So each is a count from inside the first group, divided by the size of that group.

Work out the denominator of the true statement

The condition is that the teenager is in $A$. There are $20$ of them, so $20$ is the size of the restricted universe.

Work out the numerator of the true statement

Of those $20$, exactly $7$ are in $B$ as well.

Form the fraction and simplify

So the true statement is the one that gives this probability as $\frac{7}{20}$.

Refute the statement that quotes the transposed conditional

$\frac{7}{16}$ is the same $7$ divided by the $16$ in the SECOND group. That is the probability the other way round. Conditioning on the first group means dividing by $20$, not $16$, so that statement is false.

Refute the statement that quotes the joint probability

$\frac{7}{40}$ is $7$ out of the whole survey of $40$. That is the probability of landing in both groups with nothing assumed — it ignores the condition completely.

Refute the statement with the two groups swapped over

This option turns the condition round and then still quotes $\frac{7}{20}$. Conditioning the other way divides by $16$ and gives $\frac{7}{16}$, so the statement is false. $P(A \mid B)$ and $P(B \mid A)$ are different numbers.

Refute the statement that quotes the complement

$\frac{13}{20}$ is the probability of the OPPOSITE outcome inside the same restricted universe: the $13$ of the $20$ that are not in $B$.

Do NOT transpose the conditional

Turning the condition round asks a different question. The numerator is the same $7$ teenagers — the overlap does not care which way round you read it — but the denominator becomes the $16$ teenagers who are in $B$, giving $\frac{7}{16}$. $P(A \mid B) \ne P(B \mid A)$.

State the answer

So the correct statement is the one giving the probability as $\frac{7}{20}$.

Read what the statements are all claiming

Every option is a conditional probability. Each one names the group the worker is known to be in, and then asks how likely it is that the worker is in a second group as well. So each is a count from inside the first group, divided by the size of that group.

Work out the denominator of the true statement

The condition is that the worker is in $A$. There are $25$ of them, so $25$ is the size of the restricted universe.

Work out the numerator of the true statement

Of those $25$, exactly $10$ are in $B$ as well.

Form the fraction and simplify

So the true statement is the one that gives this probability as $\frac{2}{5}$.

Refute the statement that quotes the transposed conditional

$\frac{5}{11}$ is the same $10$ divided by the $22$ in the SECOND group. That is the probability the other way round. Conditioning on the first group means dividing by $25$, not $22$, so that statement is false.

Refute the statement that quotes the joint probability

$\frac{1}{5}$ is $10$ out of the whole survey of $50$. That is the probability of landing in both groups with nothing assumed — it ignores the condition completely.

Refute the statement with the two groups swapped over

This option turns the condition round and then still quotes $\frac{2}{5}$. Conditioning the other way divides by $22$ and gives $\frac{5}{11}$, so the statement is false. $P(A \mid B)$ and $P(B \mid A)$ are different numbers.

Refute the statement that quotes the complement

$\frac{3}{5}$ is the probability of the OPPOSITE outcome inside the same restricted universe: the $15$ of the $25$ that are not in $B$.

Do NOT transpose the conditional

Turning the condition round asks a different question. The numerator is the same $10$ workers — the overlap does not care which way round you read it — but the denominator becomes the $22$ workers who are in $B$, giving $\frac{5}{11}$. $P(A \mid B) \ne P(B \mid A)$.

Compare with the unconditional probability

Out of everybody, $\frac{11}{25}$ of the workers are in $B$. Restricting to the workers who are in $A$ makes it $\frac{2}{5}$ — different from the unconditional probability, so the two events are not independent.

Test independence by exact arithmetic

Independence means $n(A \cap B) \times N = n(A) \times n(B)$. Compare $500$ with $550$: the events are not independent. Never judge this by eye — the two products settle it exactly.

The opposite event, in the same restricted universe

Of the $25$ workers the condition leaves you with, $15$ are not in $B$. The universe has not moved, so the two conditional probabilities still add to $1$.

Check with the multiplication rule

$P(A) \times P(B \mid A) = P(A \cap B)$. The $25$ cancels, leaving $10$ out of $50$ — the overlap count, straight off the diagram. The conditional probability is consistent.

Note the mistake to avoid

Dividing by $50$ here would ignore the condition entirely. The word "given" has already removed $25$ workers from the sample space; only $25$ remain.

State the answer

So the correct statement is the one giving the probability as $\frac{2}{5}$.