# In a single roll of a six-sided dice, what is the probability of rolling a three or an even number?

Oct 15, 2017

$\frac{1}{12}$

#### Explanation:

You have 2 events: A & B.
$\textcolor{m a \ge n t a}{A}$: rolling a three
$\textcolor{b l u e}{B}$: rolling an even number

$\Rightarrow P \left(A \mathmr{and} B\right) = P \left(A\right) \cdot P \left(B\right)$

Probability of rolling a three $\textcolor{m a \ge n t a}{A}$$\rightarrow \frac{1}{6}$
Probability of rolling an even number $\textcolor{b l u e}{B}$: you either get a $2 , 4$ or $6$. $\rightarrow \frac{3}{6}$

$P \left(A\right) \cdot P \left(B\right) = \frac{1}{6} \cdot \frac{3}{6}$
$= \frac{1}{12}$

Oct 17, 2017

$\textcolor{b l u e}{\frac{2}{3}}$

#### Explanation:

We have two events A and B:

A being an even number 2 4 6.
B being a 3.

$P \left(A\right) = \frac{3}{6} = \frac{1}{2}$

$P \left(B\right) = \frac{1}{6}$

Since this is an A or B event occurring, we have a union of events. i.e. $A \cup B$. This strictly means A or B or both. In this particular case A and B cannot occur simultaneously since they are mutually exclusive events. You can't throw a 3 and an even number, since 3 is an odd number.

So we have:

$P \left(A \mathmr{and} B\right) = P \left(A\right) + P \left(B\right) \implies \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \textcolor{b l u e}{\frac{2}{3}}$

We could have obtained this result directly by considering the 2 events as one event:

An even number and 3 is four favourable outcomes, so:

$\frac{4}{6} = \textcolor{b l u e}{\frac{2}{3}}$