Hard GCSE Exhaustive and mutually exclusive events Questions

Write down the sample space

There are $10$ equally likely outcomes, so every probability here is a count divided by $10$.

Work out $P(A)$ and $P(B)$

$A$ contains $4$ outcomes and $B$ contains $3$.

Look for outcomes in both events

No outcome is in both events, so the events ARE mutually exclusive.

List the outcomes that make "$A$ or $B$" happen

That is $7$ outcomes.

Work out $P(A\text{ or }B)$

$7$ of the $10$ equally likely outcomes make it happen, so the probability is $\frac{7}{10}$.

Say whether adding would have worked

Because the events are mutually exclusive, adding the probabilities gives the same $\frac{7}{10}$ — the OR rule $P(A$ or $B) = P(A) + P(B)$ holds exactly when no outcome lies in both events.

Rule out the statement giving $\frac{7}{10}$

That statement is false because the events are mutually exclusive, so that reason is the wrong way round.

Rule out the statement giving $\frac{2}{5}$

That statement is false because the probability is really $\frac{7}{10}$, not $\frac{2}{5}$.

Rule out the statement giving $\frac{3}{10}$

That statement is false because the probability is really $\frac{7}{10}$, not $\frac{3}{10}$.

Rule out the statement giving $\frac{3}{25}$

That statement is false because the probability is really $\frac{7}{10}$, not $\frac{3}{25}$.

Work out $P(A)$

Every number is equally likely, so $P(A)$ is the number of outcomes in $A$ divided by the $10$ outcomes altogether: $\frac{4}{10} = \frac{2}{5}$.

Work out $P(B)$

Every number is equally likely, so $P(B)$ is the number of outcomes in $B$ divided by the $10$ outcomes altogether: $\frac{3}{10} = \frac{3}{10}$.

Add the probabilities of the events

The probabilities of the events add to $\frac{7}{10}$. Be careful with what this does and does not tell you: for MUTUALLY EXCLUSIVE events a total of $1$ is exactly what exhaustive means, but if the events overlap the total can reach $1$ — or pass it — while outcomes are still left out.

Look for outcomes in more than one event

No outcome belongs to two of the events, so the events are mutually exclusive: they cannot happen together.

State the answer

So $P(A\text{ or }B) = \frac{7}{10}$, and the reason matters: the events are mutually exclusive, so the probabilities may be added.

List the outcomes in $A$ or $B$

"$A$ or $B$" happens when the number is in $A$, or in $B$, or in both. Writing the two events out and putting them together gives $1$, $2$, $3$, $4$, $5$ and $6$. The numbers $3$ are in BOTH events, so they must be listed only once.

Count them

There are $6$ different numbers that make "$A$ or $B$" happen.

Divide by the size of the sample space

All $10$ numbers are equally likely, so the probability is $\frac{6}{10} = \frac{3}{5}$.

See why you cannot simply add the probabilities

Adding gives $\frac{7}{10}$, but the true answer is $\frac{3}{5}$. The difference is the $1$ outcome(s) in both events, which the sum counts twice. $P(A) + P(B)$ is only equal to $P(A$ or $B)$ when the events are MUTUALLY EXCLUSIVE.

Work out $P(A)$

Every number is equally likely, so $P(A)$ is the number of outcomes in $A$ divided by the $10$ outcomes altogether: $\frac{3}{10} = \frac{3}{10}$.

Work out $P(B)$

Every number is equally likely, so $P(B)$ is the number of outcomes in $B$ divided by the $10$ outcomes altogether: $\frac{4}{10} = \frac{2}{5}$.

Add the probabilities of the events

The probabilities of the events add to $\frac{7}{10}$. Be careful with what this does and does not tell you: for MUTUALLY EXCLUSIVE events a total of $1$ is exactly what exhaustive means, but if the events overlap the total can reach $1$ — or pass it — while outcomes are still left out.

Look for outcomes in more than one event

The outcomes $3$ belong to more than one event, so the events are NOT mutually exclusive — they can happen together.

Look for outcomes in no event at all

The outcomes $7$, $8$, $9$ and $10$ belong to no event at all, so the events are NOT exhaustive: it is possible for none of them to happen.

Say what "mutually exclusive" means

Mutually exclusive events have no outcome in common: if one happens the other cannot. It says nothing at all about whether they cover everything.

Say what "exhaustive" means

Exhaustive events between them cover every outcome in the sample space, so at least one of them must happen. It says nothing at all about whether they overlap.

Say why the two ideas are different

These are two separate tests and a pair of events can pass either one without the other. Only when events are mutually exclusive AND exhaustive do their probabilities have to add to exactly $1$.

Check by listing the sample space once more

The sample space is the $10$ equally likely numbers $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ and $10$. Both tests are decided by comparing the events with this list — no probabilities are needed to decide either one.

Summarise the method

List the outcomes in each event. Ask whether any outcome appears twice (if not, mutually exclusive). Ask whether any outcome appears not at all (if not, exhaustive). The two answers are independent.

State the answer

So $P(A\text{ or }B) = \frac{3}{5}$.

List the outcomes in $A$ or $B$

"$A$ or $B$" happens when the number is in $A$, or in $B$, or in both. Writing the two events out and putting them together gives $1$, $2$, $3$, $4$, $5$, $6$ and $7$. The numbers $4$ and $5$ are in BOTH events, so they must be listed only once.

Count them

There are $7$ different numbers that make "$A$ or $B$" happen.

Divide by the size of the sample space

All $12$ numbers are equally likely, so the probability is $\frac{7}{12} = \frac{7}{12}$.

See why you cannot simply add the probabilities

Adding gives $\frac{3}{4}$, but the true answer is $\frac{7}{12}$. The difference is the $2$ outcome(s) in both events, which the sum counts twice. $P(A) + P(B)$ is only equal to $P(A$ or $B)$ when the events are MUTUALLY EXCLUSIVE.

Work out $P(A)$

Every number is equally likely, so $P(A)$ is the number of outcomes in $A$ divided by the $12$ outcomes altogether: $\frac{5}{12} = \frac{5}{12}$.

Work out $P(B)$

Every number is equally likely, so $P(B)$ is the number of outcomes in $B$ divided by the $12$ outcomes altogether: $\frac{4}{12} = \frac{1}{3}$.

Add the probabilities of the events

The probabilities of the events add to $\frac{3}{4}$. Be careful with what this does and does not tell you: for MUTUALLY EXCLUSIVE events a total of $1$ is exactly what exhaustive means, but if the events overlap the total can reach $1$ — or pass it — while outcomes are still left out.

Look for outcomes in more than one event

The outcomes $4$ and $5$ belong to more than one event, so the events are NOT mutually exclusive — they can happen together.

Look for outcomes in no event at all

The outcomes $8$, $9$, $10$, $11$ and $12$ belong to no event at all, so the events are NOT exhaustive: it is possible for none of them to happen.

Say what "mutually exclusive" means

Mutually exclusive events have no outcome in common: if one happens the other cannot. It says nothing at all about whether they cover everything.

Say what "exhaustive" means

Exhaustive events between them cover every outcome in the sample space, so at least one of them must happen. It says nothing at all about whether they overlap.

Say why the two ideas are different

These are two separate tests and a pair of events can pass either one without the other. Only when events are mutually exclusive AND exhaustive do their probabilities have to add to exactly $1$.

Check by listing the sample space once more

The sample space is the $12$ equally likely numbers $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, $11$ and $12$. Both tests are decided by comparing the events with this list — no probabilities are needed to decide either one.

Summarise the method

List the outcomes in each event. Ask whether any outcome appears twice (if not, mutually exclusive). Ask whether any outcome appears not at all (if not, exhaustive). The two answers are independent.

State the answer

So $P(A\text{ or }B) = \frac{7}{12}$.

Write down the sample space

There are $12$ equally likely outcomes. Both of the tests in this question are made by comparing the events with this list.

Write down the outcomes in each event

Listing the events as sets of outcomes is what makes both tests easy: $A$ contains $4$ of them and $B$ contains $3$ of them and $C$ contains $3$ of them.

Test for mutually exclusive: is any outcome in two events?

The outcomes $4$ appear in more than one event, so the events CAN happen together: they are NOT mutually exclusive.

Test for exhaustive: is any outcome in no event?

The outcomes $10$, $11$ and $12$ are in no event at all, so it is possible for none of the events to happen: they are NOT exhaustive.

Put the two tests together

The two tests are independent of each other, and here they give "no" and "no". So the events are neither mutually exclusive nor exhaustive.

Rule out "mutually exclusive but not exhaustive"

The events are really neither mutually exclusive nor exhaustive, so "mutually exclusive but not exhaustive" is false: it gets at least one of the two tests the wrong way round.

Rule out "exhaustive but not mutually exclusive"

The events are really neither mutually exclusive nor exhaustive, so "exhaustive but not mutually exclusive" is false: it gets at least one of the two tests the wrong way round.

Rule out "both mutually exclusive and exhaustive"

The events are really neither mutually exclusive nor exhaustive, so "both mutually exclusive and exhaustive" is false: it gets at least one of the two tests the wrong way round.

Rule out the statement about the total probability

The probabilities of the events really add to $\frac{5}{6}$, not $\frac{1}{2}$, so that statement is false as it stands. It is also the wrong test to be applying: a total of $1$ only proves the events are exhaustive if you already know they are mutually exclusive.

Work out $P(A)$

Every number is equally likely, so $P(A)$ is the number of outcomes in $A$ divided by the $12$ outcomes altogether: $\frac{4}{12} = \frac{1}{3}$.

Work out $P(B)$

Every number is equally likely, so $P(B)$ is the number of outcomes in $B$ divided by the $12$ outcomes altogether: $\frac{3}{12} = \frac{1}{4}$.

Work out $P(C)$

Every number is equally likely, so $P(C)$ is the number of outcomes in $C$ divided by the $12$ outcomes altogether: $\frac{3}{12} = \frac{1}{4}$.

Add the probabilities of the events

The probabilities of the events add to $\frac{5}{6}$. Be careful with what this does and does not tell you: for MUTUALLY EXCLUSIVE events a total of $1$ is exactly what exhaustive means, but if the events overlap the total can reach $1$ — or pass it — while outcomes are still left out.

Look for outcomes in more than one event

The outcomes $4$ belong to more than one event, so the events are NOT mutually exclusive — they can happen together.

State the answer

So events $A$, $B$ and $C$ are neither mutually exclusive nor exhaustive.

Write down the sample space

There are $10$ equally likely outcomes. Both of the tests in this question are made by comparing the events with this list.

Write down the outcomes in each event

Listing the events as sets of outcomes is what makes both tests easy: $A$ contains $3$ of them and $B$ contains $3$ of them and $C$ contains $4$ of them.

Test for mutually exclusive: is any outcome in two events?

No outcome appears in more than one event, so the events cannot happen together: they ARE mutually exclusive.

Test for exhaustive: is any outcome in no event?

Every outcome is in at least one event, so one of the events is bound to happen: they ARE exhaustive.

Put the two tests together

The two tests are independent of each other, and here they give "yes" and "yes". So the events are both mutually exclusive and exhaustive.

Rule out "mutually exclusive but not exhaustive"

The events are really both mutually exclusive and exhaustive, so "mutually exclusive but not exhaustive" is false: it gets at least one of the two tests the wrong way round.

Rule out "exhaustive but not mutually exclusive"

The events are really both mutually exclusive and exhaustive, so "exhaustive but not mutually exclusive" is false: it gets at least one of the two tests the wrong way round.

Rule out "neither mutually exclusive nor exhaustive"

The events are really both mutually exclusive and exhaustive, so "neither mutually exclusive nor exhaustive" is false: it gets at least one of the two tests the wrong way round.

Rule out the statement about the total probability

The probabilities of the events really add to $1$, not $\frac{1}{2}$, so that statement is false as it stands. It is also the wrong test to be applying: a total of $1$ only proves the events are exhaustive if you already know they are mutually exclusive.

Work out $P(A)$

Every number is equally likely, so $P(A)$ is the number of outcomes in $A$ divided by the $10$ outcomes altogether: $\frac{3}{10} = \frac{3}{10}$.

Work out $P(B)$

Every number is equally likely, so $P(B)$ is the number of outcomes in $B$ divided by the $10$ outcomes altogether: $\frac{3}{10} = \frac{3}{10}$.

Work out $P(C)$

Every number is equally likely, so $P(C)$ is the number of outcomes in $C$ divided by the $10$ outcomes altogether: $\frac{4}{10} = \frac{2}{5}$.

Add the probabilities of the events

The probabilities of the events add to $1$. Be careful with what this does and does not tell you: for MUTUALLY EXCLUSIVE events a total of $1$ is exactly what exhaustive means, but if the events overlap the total can reach $1$ — or pass it — while outcomes are still left out.

Look for outcomes in more than one event

No outcome belongs to two of the events, so the events are mutually exclusive: they cannot happen together.

State the answer

So events $A$, $B$ and $C$ are both mutually exclusive and exhaustive.