You roll a number cube twice. What is the probability of rolling an even number and a 5?

May 4, 2018

$\frac{1}{6}$

Explanation:

P(even)=$\frac{1}{2}$

P(5)=$\frac{1}{6}$

So the probability of $\left(e v e n \cap 5\right) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$
But you could also get $\left(5 \cap e v e n\right) \frac{1}{12}$

$\frac{1}{12} + \frac{1}{12} = \frac{1}{6}$

May 4, 2018

$\frac{6}{36} = \frac{1}{6}$

Explanation:

Before we calculate an answer, drawing a possibility space allows us to see all the $36$ outcomes of throwing a cube twice.
The grid shows the sum of the two throws.

The red values are those which have a $5$ and and an even number

$6 | \text{ "7" "8" "9" "10" "color(red)(11)" } 12$
$5 | \text{ "6" "color(red)(7)" "8" "color(red)(9)" "10" } \textcolor{red}{11}$
$4 | \text{ "5" "6" "7" "8" "color(red)(9)" } 10$
$3 | \text{ "4" "5" "6" "7" "8" } 9$
$2 | \text{ "3" "4" "5" "6" "color(red)(7)" } 8$
$1 | \underline{\text{ "2" "3" "4" "5" "6" } 7}$
$\textcolor{w h i t e}{\times \times} 1 \text{ "2" "3" "4" "5" } 6$

$P \left(5 \mathmr{and} \text{even}\right) = \frac{6}{36} = \frac{1}{6}$

Using a calculation:

$P \left(5 \mathmr{and} \text{even") = P(5, "even") " OR " P("even} , 5\right)$

$= \left(\frac{1}{6} \times \frac{3}{6}\right) + \left(\frac{3}{6} \times \frac{1}{6}\right)$

$= \frac{3}{36} + \frac{3}{36}$

$= \frac{6}{36}$

$= \frac{1}{6}$