GCSE Exhaustive and mutually exclusive events Practice Questions

Use the fact that the probabilities add to $1$

One of the outcomes must happen and no two can happen together, so the probabilities add to exactly $1$. The question gives $P(\text{red}) = \frac{1}{2}$ and $P(\text{blue}) = \frac{1}{5}$.

Subtract the given probabilities from $1$

Adding the probabilities given comes to $\frac{7}{10}$, and the rest of the certainty has to belong to green.

State the answer

So the probability is $\frac{3}{10}$.

Compare the events with the sample space

The sample space is $\{1, 2, 3, 4, 5, 6\}$. Two separate questions decide everything: does any outcome sit in two events (then they are not mutually exclusive), and does any outcome sit in no event (then they are not exhaustive)?

Read off the two properties

An outcome is in two events, and no outcome is left out. So the events are exhaustive but not mutually exclusive.

State the answer

So events $A$ and $B$ are exhaustive but not mutually exclusive.

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 $4$ of them and $B$ 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 $7$, $8$, $9$ and $10$ 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.

State the answer

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

Write down the sample space

There are $8$ equally likely outcomes, so every probability here is a count divided by $8$.

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

$A$ contains $3$ outcomes and $B$ contains $2$.

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 $5$ outcomes.

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

$5$ of the $8$ equally likely outcomes make it happen, so the probability is $\frac{5}{8}$.

Say whether adding would have worked

Because the events are mutually exclusive, adding the probabilities gives the same $\frac{5}{8}$ — 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 $\frac{5}{8}$

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

Rule out the statement giving $\frac{3}{8}$

That statement is false because the probability is really $\frac{5}{8}$, not $\frac{3}{8}$.

Rule out the statement giving $\frac{1}{4}$

That statement is false because the probability is really $\frac{5}{8}$, not $\frac{1}{4}$.

State the answer

So $P(A\text{ or }B) = \frac{5}{8}$, and the reason matters: the events are mutually exclusive, so the probabilities may be added.

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.