# If you roll a pair of dice, what is the probability of rolling either a single 3 or a sum that is an odd number?

Nov 6, 2015

The probability is: $\frac{11}{18}$

#### Explanation:

In this task we have to calculate the probability of sum of 2 events (i.e. rolling a single 3 or rolling an odd sum).

To do this we must use the following formula:

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

$| \Omega | = 36$

Event $A$ is "rolling a single 3", so:

$A = \left\{\begin{matrix}3 & 1 \\ 3 & 2 \\ 3 & 4 \\ 3 & 5 \\ 3 & 6 \\ 1 & 3 \\ 2 & 3 \\ 4 & 3 \\ 5 & 3 \\ 6 & 3\end{matrix}\right\}$

$| A | = 10$, $P \left(A\right) = \frac{10}{36}$

Event $B$ is "rolling an odd sum", so:

$B = \left\{\begin{matrix}1 & 2 \\ 1 & 4 \\ 1 & 6 \\ 2 & 1 \\ 2 & 3 \\ 2 & 5 \\ 3 & 2 \\ 3 & 4 \\ 3 & 6 \\ 4 & 1 \\ 4 & 3 \\ 4 & 5 \\ 5 & 2 \\ 5 & 4 \\ 5 & 6 \\ 6 & 1 \\ 6 & 3 \\ 6 & 5\end{matrix}\right\}$

$| B | = 18$

$P \left(B\right) = \frac{18}{36}$

Event $A \cap B$ is "rolling a single 3 and an odd sum", so

$A \cap B = \left\{\begin{matrix}3 & 2 \\ 3 & 4 \\ 3 & 6 \\ 2 & 3 \\ 4 & 3 \\ 6 & 3\end{matrix}\right\}$

$| A \cap B | = 6$

$P \left(A \cap B\right) = \frac{6}{36}$

Now we can use the first formula:

$P \left(A \cup B\right) = \frac{10}{36} + \frac{18}{36} - \frac{6}{36} = \frac{22}{36} = \frac{11}{18}$