Anna thinks of a number between one and $\scriptsize 10$. Draw a Venn diagram to answer the following questions.

1. What is the probability that the number is a multiple of two?
2. What is the probability that the number is a multiple of three?
3. What is the probability that the number is a multiple of two or three?
4. What is the probability that the number is a multiple of two and three?
5. What is the probability that the number is NOT a multiple of two or three?

Solutions

Step 1: Draw the Venn diagram
The Venn diagram should show the sample space containing all numbers from one to $\scriptsize 10$. Let $\scriptsize \text{A}$ be the event that contains all the multiples of two, $\scriptsize \text{A}=\{2;\text{ }4;\text{ }6;\text{ }8;\text{ }10\}$. Let $\scriptsize \text{B}$ be the event that contains all the multiples of three, $\scriptsize \text{B}=\{3;\text{ 6};\text{ 9}\}$ .

We see that there are ten outcomes. The intersection of the two events is $\scriptsize 6$ and the outcomes $\scriptsize 1;\text{ }7;\text{ }5$ are not part of either event.

Step 2: Calculate the probabilities

Remember that the probability of an event is the number of outcomes in the event set divided by the number of outcomes in the sample space.

$\scriptsize \displaystyle \begin{align*}\text{P}(\text{E})&=\text{ }\displaystyle \frac{{\text{number of favourable outcomes}}}{{\text{number of possible outcomes}}}\\&=\displaystyle \frac{{n(\text{E})}}{{n(\text{S})}}\end{align*}$

1. Since there are five outcomes in event $\scriptsize \text{A}$, the probability of a multiple of two is:
   $\scriptsize \displaystyle \begin{align*}\text{P}(\text{A})&=\displaystyle \frac{5}{{10}}\\&=\displaystyle \frac{1}{2}\end{align*}$
2. Since there are three outcomes in event $\scriptsize \text{B}$, the probability of a multiple of three is:
   $\scriptsize \displaystyle \text{P}(\text{B})=\displaystyle \frac{3}{{10}}$
3. The event that the number is a multiple of two or three is the union of the above two event sets. There are seven elements in the union of the event sets, so the probability is $\scriptsize \displaystyle \frac{7}{{10}}$. This can be calculated as:
   $\scriptsize \displaystyle \begin{align*}P(\text{A or B})&=P(\text{A})+P(\text{B)}-P(\text{A}\cap \text{B})\\&=\displaystyle \frac{5}{{10}}+\displaystyle \frac{3}{{10}}-\displaystyle \frac{1}{{10}}\\&=\displaystyle \frac{7}{{10}}\end{align*}$
4. The event that the number is a multiple of two and three is the intersection of the two event sets. There is one element in the intersection of the event sets, so $\scriptsize \displaystyle P(\text{A}\cap \text{B})=\displaystyle \frac{1}{{10}}$.
5. The event ‘NOT a multiple of two or three’ is the set of all numbers not in events $\scriptsize \text{A}$ or $\scriptsize \text{B}$. This is the complement of events $\scriptsize \text{A}$ or $\scriptsize \text{B}$.
   $\scriptsize \begin{align*}P(\text{not A or B)}&=1-\displaystyle \frac{7}{{10}}\\&=\displaystyle \frac{3}{{10}}\end{align*}$

In a group of $\scriptsize \displaystyle 50$ learners, $\scriptsize 35$ take mathematics and $\scriptsize 30$ take science, while $\scriptsize 12$ take neither of the two subjects. Draw a Venn diagram representing this information. If a learner is chosen at random from this group, what is the probability that they take both mathematics and science?

Solution

Step 1: Draw an outline of Venn diagram

Let $\scriptsize \text{M}$ be the event ‘takes mathematics’.

Let $\scriptsize \text{SC}$ be the event ‘takes science’.

We need to do some calculations before drawing the full Venn diagram, but with the information given we can already draw the outline.

Step 2: Write down sizes of the event sets, their union and intersection

We are told that $\scriptsize 12$ learners take neither of the two subjects. Graphically we can represent this outside the two events in the Venn diagram.

Since there are $\scriptsize \displaystyle 50$ learners in the sample space, we can see that there are $\scriptsize 50-12=38$ elements in $\scriptsize \text{M or SC}$. So far we know:

$\scriptsize n(\text{M)}=35$

$\scriptsize n(\text{SC)}=30$

$\scriptsize n(\text{M or SC)}=38$

We need to find the intersection of the two events to complete the Venn diagram.

From the addition rule:

$\scriptsize \begin{align*}n(\text{M }\cup \text{ SC)}&=n(\text{M)}+n(\text{SC)}-n(\text{M}\cap \text{SC})\\\therefore n(\text{M}\cap \text{SC})&=35+30-38\\&=27\end{align*}$

Step 3: Draw the final Venn diagram

Step 4: Calculate the probability

$\scriptsize P(\text{M}\cap \text{SC)}=\displaystyle \frac{{27}}{{50}}$

Use the following information to draw a Venn diagram. Then using the Venn diagram, find$\scriptsize P\left( {\text{B and (not A)}} \right)$.

$\scriptsize \begin{align*}P(\text{A})&=0.3\\P(\text{A and B)}&=0.2\\P(\text{B})&=0.7\end{align*}$

Solution

Step 1: Calculate the probabilities of each event only

Start with the intersection and work out the probability of each event only.

Since $\scriptsize P(\text{A and B)}=0.2$

$\scriptsize \begin{align*}P(\text{A only})&=0.3-0.2\\&=0.1\\P(\text{B only})&=0.7-0.2\\&=0.5\end{align*}$

Therefore the probability of not being in event $\scriptsize \text{A}$or $\scriptsize \text{B}$ is $\scriptsize 1-(0.1+02+0.5)=0.2$.

Step 2: Draw the Venn diagram showing the probabilities of the events

Step 3: Write down the solution

$\scriptsize P\left( {\text{B and (not A)}} \right)$i s the same as probability of $\scriptsize \text{B}$ only.

$\scriptsize P\left( {\text{B and (not A)}} \right)=0.5$