Single-event probability

We are told the probabilities of events $A$ and $B$, and we are also told they are mutually exclusive.

$\text{P}(A)=0.35,\text{P}(B)=0.45$

We need $\text{P}(A\cup B)$, meaning the chance that $A$ happens or $B$ happens.

If two events are mutually exclusive, they cannot occur together. In set notation, that means there is no overlap between them, so:

$\text{P}(A\cap B)=0$

This makes finding the probability of “$A$ or $B$” straightforward.

From the IB Data Booklet, the addition rule for any two events is:

$\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)$

Because the events are mutually exclusive, $\text{P}(A\cap B)=0$, so we can simplify to:

$\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)$

This is a key idea in Combined-event probability.

Now replace $\text{P}(A)$ and $\text{P}(B)$ with the given values:

$\begin{array}{rl}\text{P}(A\cup B)& =0.35+0.45\\ & =0.80\end{array}$

Here’s why the other choices do not represent $\text{P}(A\cup B)$:

• Option A (0.10) comes from taking a difference like $0.45-0.35$, which is not how you find a union.
• Option B (1.00) would only make sense if $A$ and $B$ together covered every possible outcome (exhaustive), which is not stated.
• Option C (0.1575) is from $0.35\times 0.45$, which relates to an intersection for independent events, not a union of mutually exclusive events.

Our result is $0.80$, so the correct choice is Option D.

The wording mutually exclusive tells us $A$ and $B$ cannot happen at the same time (a key idea in Topic: Sets, Logic and Probability).

So there is no overlap between the events, meaning: $P(A\cap B)=0$

Think of the sample space as a rectangle, with one circle for $A$ and another for $B$.

Because the events are mutually exclusive, the circles are separated (disjoint), so there is no shared region.

The union $P(A\cup B)$ is the total area of both circles together. With no overlap to remove, we just combine the two probabilities by addition.

Option C ($0.1575$) comes from multiplying $0.35\times 0.45$.

Multiplication is used to find $P(A\cap B)$ for independent events, but here we want the union ($A\cup B$), and we already know the intersection is $0$ because the events are mutually exclusive.

Option A ($0.10$) is just the difference $0.45-0.35$, which is not the probability of “$A$ or $B$”.

Option B ($1.00$) would only be true if $A$ and $B$ were exhaustive and covered the whole sample space, but that information is not given.

Use the IB Data Booklet rule:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Since $P(A\cap B)=0$ for mutually exclusive events, this becomes:

$P(A\cup B)=0.35+0.45$

Adding gives the probability that either event occurs.

The total is $0.80$, which corresponds to Option D.

We have two events $A$ and $B$ with their probabilities given, and the important condition is that they are mutually exclusive.

For any events $A$ and $B$, the probability that at least one occurs is:

$\text{P}(A\cup B)=\text{P}(A)+\text{P}(B)-\text{P}(A\cap B)$

This is the general addition rule in the IB Mathematics Data Booklet.

Mutually exclusive events cannot happen together, so the intersection probability is:

$\text{P}(A\cap B)=0$

Substitute into the formula:

$\begin{array}{rl}\text{P}(A\cup B)& =0.35+0.45-0\\ \text{P}(A\cup B)& =0.80\end{array}$

The other options come from common wrong operations:

• Option A: subtraction ($0.45-0.35$) is not the union rule.
• Option B: assumes the probability must total 1.00, but $A$ and $B$ may not cover all outcomes.
• Option C: multiplication ($0.35\times 0.45$) is linked to intersections for independent events, not unions of mutually exclusive events.

Using the general addition rule with mutual exclusivity, $\text{P}(A\cup B)=0.80$.

The information gives a direct comparison between the chance of $A$ and the chance of its complement $A^{\prime }$.

So we can express it as: $P(A)=3P(A^{\prime })$

An event and its complement cover all outcomes, so their probabilities add to $1$ (this is part of Sets, Logic and Probability).

$\text{Fundamental Identity: }P(A)+P(A^{\prime })=1$

Replace $P(A)$ with $3P(A^{\prime })$ in the identity:

$\begin{array}{rl}3P(A^{\prime })+P(A^{\prime })& =1\\ 4P(A^{\prime })& =1\end{array}$

Now divide by $4$: $P(A^{\prime })=\frac{1}{4}=0.25$

Substitute $P(A^{\prime })=0.25$ into $P(A)=3P(A^{\prime })$:

$\begin{array}{rl}P(A)& =3\times 0.25\\ P(A)& =0.75\end{array}$

Check each option:

• Option A is $0.20$, which does not match our computed value.
• Option B is $0.25$, which is the value of $P(A^{\prime })$, not $P(A)$.
• Option C is $0.60$, which is not what we found.
• Option D equals $0.75$, which matches exactly.

Saying $P(A)$ is $3$ times $P(A^{\prime })$ means the probabilities are in the ratio:

$P(A):P(A^{\prime })=3:1$

So for every $3$ equal “shares” of probability for $A$, there is $1$ share for $A^{\prime }$.

Because $A$ and $A^{\prime }$ together make certainty, their probabilities must sum to $1$ (from Topic: Sets, Logic and Probability):

$\begin{array}{rl}P(A)+P(A^{\prime })& =1\end{array}$

In the ratio $3:1$, the total number of parts is:

$3+1=4\text{ parts}$

All $4$ parts together represent probability $1$, so one part is:

$\text{Value of 1 part}=\frac{1}{4}=0.25$

Use the part idea to judge the options:

• If $P(A)$ were $0.20$, it would not represent $3$ equal parts out of $4$. Eliminate Option A.
• If $P(A)$ were $0.25$, that would be just $1$ part, matching $P(A^{\prime })$, not $3$ times it. Eliminate Option B.
• If $P(A)$ were $0.60$, the remaining probability would be $0.40$, and $0.60$ is not $3$ times $0.40$. Eliminate Option C.

Since $A$ takes $3$ of the $4$ equal parts:

$\begin{array}{rl}P(A)& =3\times 0.25\\ P(A)& =0.75\end{array}$

That corresponds to the final option.

For any event $A$, the probabilities must satisfy:

$P(A)+P(A^{\prime })=1$

Also, the problem says the probability of $A$ is exactly 3 times the probability of its complement.

Assume $P(A)=0.20$. Then:

$\begin{array}{rl}P(A^{\prime })& =1-0.20=0.80\\ 3\times P(A^{\prime })& =3\times 0.80=2.40\end{array}$

Because $0.20\ne 2.40$, Option A cannot be correct.

Check the next two choices the same way:

• For Option B ($0.25$): $P(A^{\prime })=0.75$. Then $3\times 0.75=2.25$, not $0.25$.
• For Option C ($0.60$): $P(A^{\prime })=0.40$. Then $3\times 0.40=1.20$, not $0.60$.

Now test $P(A)=0.75$:

$\begin{array}{rl}P(A^{\prime })& =1-0.75=0.25\\ 3\times P(A^{\prime })& =3\times 0.25=0.75\end{array}$

This satisfies the condition exactly, so Option D is correct.

We are told two important details:

• The sample space $S$ has 12 equally likely outcomes.
• The probability of the complement of $A$ is $P(A^{\prime })=\frac{1}{3}$.

Our goal is to determine how many outcomes are in event $A$, written as $n(A)$. This connects to ideas from Sets, Logic and Probability.

Event $A$ and its complement $A^{\prime }$ together make up the whole sample space, so their probabilities add to 1.

$\begin{array}{rl}P(A)+P(A^{\prime })& =1\\ P(A)& =1-P(A^{\prime })\end{array}$

Substitute the given probability:

$P(A)=1-\frac{1}{3}=\frac{2}{3}$

We can quickly dismiss answers that clearly represent the wrong part of the sample space or the wrong fraction.

• Option B: 4. Since $n(S)=12$, $12\times \frac{1}{3}=4$. That matches $n(A^{\prime })$ (outcomes not in $A$), not $n(A)$.
• Option C: 6. This would give $P(A)=\frac{6}{12}=\frac{1}{2}$, which disagrees with $P(A)=\frac{2}{3}$.
• Option A: 10. This would give $P(A)=\frac{10}{12}=\frac{5}{6}$, also not equal to $\frac{2}{3}$.

With equally likely outcomes, probability is the fraction of favorable outcomes:

$P(A)=\frac{n(A)}{n(S)}$

Solve for $n(A)$:

$\begin{array}{rl}n(A)& =P(A)\times n(S)\\ n(A)& =\frac{2}{3}\times 12\end{array}$

Now compute the value.

Carrying out the multiplication:

$n(A)=\frac{24}{3}=8$

So there are 8 outcomes in event $A$, which corresponds to Option D.

The sample space contains 12 equally likely outcomes, so:

$n(S)=12$

Start by finding $n(A^{\prime })$ using $P(A^{\prime })=\frac{1}{3}$. Multiply this probability by the total outcomes:

$\begin{array}{rl}n(A^{\prime })& =P(A^{\prime })\times n(S)\\ n(A^{\prime })& =\frac{1}{3}\times 12\\ n(A^{\prime })& =4\end{array}$

So, 4 outcomes are not in event $A$.

Event $A$ is everything in the sample space except what’s in $A^{\prime }$, so:

$\begin{array}{rl}n(A)& =n(S)-n(A^{\prime })\\ n(A)& =12-4\\ n(A)& =8\end{array}$

Looking at the options:

• Option B is 4, but that is $n(A^{\prime })$, not $n(A)$.
• Options A and C are not equal to 8.

Therefore, the correct number of outcomes in event $A$ is 8.