Hard GCSE Conditional probability Questions

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 $\frac{2}{5}$.

Read what the statements are all claiming

Every option is a conditional probability. Each one names the group the person is known to be in, and then asks how likely it is that the person 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 person is one of the men. 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 $9$ are one of the people who chose coffee as well.

Form the fraction and simplify

So the true statement is the one that gives this probability as $\frac{9}{25}$.

Refute the statement that quotes the transposed conditional

$\frac{9}{22}$ is the same $9$ 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

$\frac{3}{17}$ is $9$ out of the whole survey of $51$. 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 $\frac{9}{25}$. Conditioning the other way divides by $22$ and gives $\frac{9}{22}$, so the statement is false. $P(A \mid B)$ and $P(B \mid A)$ are different numbers.

Refute the statement that quotes the complement

$\frac{16}{25}$ is the probability of the OPPOSITE outcome inside the same restricted universe: the $16$ of the $25$ that are not one of the people who chose coffee.

Do NOT transpose the conditional — work out the other one and compare

$P(A \mid B)$ and $P(B \mid A)$ are DIFFERENT questions. Both count the same $9$ men who chose coffee on top, but this one divides by the $22$ people who chose coffee instead of the $25$ men, and gives $\frac{9}{22}$. Swapping the two round is the commonest mistake in this topic, and here it changes the answer.

Compare the conditional probability with the unconditional one

Without knowing anything, the probability of being one of the people who chose coffee is $\frac{22}{51}$. Being told the person is one of the men changes it to $\frac{9}{25}$. That comparison is exactly the test for independence.

Test whether the two events are independent

$A$ and $B$ are independent exactly when $P(A \cap B) = P(A) \times P(B)$, which for whole counts means $n(A \cap B) \times N = n(A) \times n(B)$. Here $459$ against $550$, so the events are NOT independent — which is why the conditional probability differs from the unconditional one.

Work out the opposite event inside the SAME restricted universe

Of the $25$ men, $16$ are not people who chose coffee. The two conditional probabilities still add to $1$, because the universe has not changed — it is still the $25$ men.

Check the answer with the multiplication rule

$P(A \cap B) = P(A) \times P(B \mid A)$. Multiplying back gives $\frac{3}{17}$, which is exactly $9$ out of $51$ — the joint count. The conditional probability is consistent with the table.

Check the numerator really is part of the denominator

Every one of the $9$ men who chose coffee is one of the $25$ men, so the numerator cannot exceed the denominator and the fraction cannot exceed $1$. If yours does, you have divided by the wrong group.

State the answer

So the correct statement is the one giving the probability as $\frac{9}{25}$.

Write down what independence means

Two events are independent when knowing that one of them happened tells you nothing about the other — that is, when conditioning on $B$ leaves the probability of $A$ exactly where it was.

Read the three counts you need off the table

There are $30$ men, $20$ members who use the pool, and $12$ who are both. The survey has $50$ members in it.

Work out the probability of A

$30$ of the $50$ members are men.

Work out the probability of B

$20$ of the $50$ members are members who use the pool.

Work out the probability of A and B together

$12$ of the $50$ members are in both groups at once.

Multiply the probability of A by the probability of B

If the events were independent this product would have to come out equal to $P(A \cap B)$.

Compare the two and decide

The two are equal, so $A$ and $B$ are independent. This is decided by exact fractions, never by how close the two numbers look.

Check it the other way: work out the probability of A given B

Restrict the universe to the $20$ members who use the pool and recount: $12$ of them are men. That gives $\frac{3}{5}$, which agrees with $P(A) = \frac{3}{5}$ — the same verdict as before, as it must be.

Do not confuse the probability of A given B with the probability of B given A

The two conditionals share the numerator $12$ but divide by different groups — $30$ men against $20$ members who use the pool. They come out as $\frac{2}{5}$ and $\frac{3}{5}$. Any option quoting $\frac{2}{5}$ as $P(A \mid B)$ is wrong before you even read its verdict.

Refute the option with the right numbers but the wrong verdict

One option quotes both of these correctly and then draws the opposite conclusion. The rule is the whole point: $P(A \mid B)$ equals $P(A)$, so the events are independent, whatever the option says.

Refute the options that quote the transposed conditional

Two of the options give $P(A \mid B)$ as $\frac{2}{5}$. That number is $P(B \mid A)$ — the same $12$ people divided by the $30$ men instead of the $20$ members who use the pool. It answers a different question, so those options are false.

Refute the option that quotes the probability of B in place of the probability of A

The last option quotes $P(A)$ as $\frac{2}{5}$, which is really $P(B)$. The independence test compares $P(A \mid B)$ with $P(A)$, not with $P(B)$.

Say what independence would have required of the table

For independence the overlap would have to be exactly $12$. The table says $12$. That single comparison settles the question.

Say what the verdict means in plain words

Knowing that the member is one of the members who use the pool tells you nothing new about whether they are one of the men.

State the answer

So $A$ and $B$ are independent.

Write down what independence means

Two events are independent when knowing that one of them happened tells you nothing about the other — that is, when conditioning on $B$ leaves the probability of $A$ exactly where it was.

Read the three counts you need off the table

There are $24$ girls, $23$ students who revised for the test, and $14$ who are both. The survey has $48$ students in it.

Work out the probability of A

$24$ of the $48$ students are girls.

Work out the probability of B

$23$ of the $48$ students are students who revised for the test.

Work out the probability of A and B together

$14$ of the $48$ students are in both groups at once.

Multiply the probability of A by the probability of B

If the events were independent this product would have to come out equal to $P(A \cap B)$.

Compare the two and decide

The two are not equal, so $A$ and $B$ are not independent. This is decided by exact fractions, never by how close the two numbers look.

Check it the other way: work out the probability of A given B

Restrict the universe to the $23$ students who revised for the test and recount: $14$ of them are girls. That gives $\frac{14}{23}$, which differs from $P(A) = \frac{1}{2}$ — the same verdict as before, as it must be.

Do not confuse the probability of A given B with the probability of B given A

The two conditionals share the numerator $14$ but divide by different groups — $24$ girls against $23$ students who revised for the test. They come out as $\frac{7}{12}$ and $\frac{14}{23}$. Any option quoting $\frac{7}{12}$ as $P(A \mid B)$ is wrong before you even read its verdict.

Refute the option with the right numbers but the wrong verdict

One option quotes both of these correctly and then draws the opposite conclusion. The rule is the whole point: $P(A \mid B)$ does not equal $P(A)$, so the events are not independent, whatever the option says.

Refute the options that quote the transposed conditional

Two of the options give $P(A \mid B)$ as $\frac{7}{12}$. That number is $P(B \mid A)$ — the same $14$ people divided by the $24$ girls instead of the $23$ students who revised for the test. It answers a different question, so those options are false.

Refute the option that quotes the probability of B in place of the probability of A

The last option quotes $P(A)$ as $\frac{23}{48}$, which is really $P(B)$. The independence test compares $P(A \mid B)$ with $P(A)$, not with $P(B)$.

Say what independence would have required of the table

For independence the overlap would have to be exactly $\frac{23}{2}$. The table says $14$. That single comparison settles the question.

Say what the verdict means in plain words

Knowing that the student is one of the students who revised for the test changes what you should believe about whether they are one of the girls.

State the answer

So $A$ and $B$ are not independent.

Write down what independence means

Two events are independent when knowing that one of them happened tells you nothing about the other — that is, when conditioning on $B$ leaves the probability of $A$ exactly where it was.

Read the three counts you need off the table

There are $25$ men, $18$ workers who cycle to work, and $10$ who are both. The survey has $45$ workers in it.

Work out the probability of A

$25$ of the $45$ workers are men.

Work out the probability of B

$18$ of the $45$ workers are workers who cycle to work.

Work out the probability of A and B together

$10$ of the $45$ workers are in both groups at once.

Multiply the probability of A by the probability of B

If the events were independent this product would have to come out equal to $P(A \cap B)$.

Compare the two and decide

The two are equal, so $A$ and $B$ are independent. This is decided by exact fractions, never by how close the two numbers look.

Check it the other way: work out the probability of A given B

Restrict the universe to the $18$ workers who cycle to work and recount: $10$ of them are men. That gives $\frac{5}{9}$, which agrees with $P(A) = \frac{5}{9}$ — the same verdict as before, as it must be.

Do not confuse the probability of A given B with the probability of B given A

The two conditionals share the numerator $10$ but divide by different groups — $25$ men against $18$ workers who cycle to work. They come out as $\frac{2}{5}$ and $\frac{5}{9}$. Any option quoting $\frac{2}{5}$ as $P(A \mid B)$ is wrong before you even read its verdict.

Refute the option with the right numbers but the wrong verdict

One option quotes both of these correctly and then draws the opposite conclusion. The rule is the whole point: $P(A \mid B)$ equals $P(A)$, so the events are independent, whatever the option says.

Refute the options that quote the transposed conditional

Two of the options give $P(A \mid B)$ as $\frac{2}{5}$. That number is $P(B \mid A)$ — the same $10$ people divided by the $25$ men instead of the $18$ workers who cycle to work. It answers a different question, so those options are false.

Refute the option that quotes the probability of B in place of the probability of A

The last option quotes $P(A)$ as $\frac{2}{5}$, which is really $P(B)$. The independence test compares $P(A \mid B)$ with $P(A)$, not with $P(B)$.

Say what independence would have required of the table

For independence the overlap would have to be exactly $10$. The table says $10$. That single comparison settles the question.

Say what the verdict means in plain words

Knowing that the worker is one of the workers who cycle to work tells you nothing new about whether they are one of the men.

State the answer

So $A$ and $B$ are independent.