# What is the probability of getting a sum of either 7, 11, or 12 on a roll of two dice?

Feb 4, 2016

#### Answer:

The probability is 25%.

#### Explanation:

Let's first take a look at the probability for one of those sums.

There are $6 \times 6 = 36$ different results of a roll of two dice:

$\left(1 , 1\right) , \left(1 , 2\right) , \ldots , \left(1 , 6\right)$
$\left(2 , 1\right) , \left(2 , 2\right) , \ldots , \left(2 , 6\right)$
$\ldots$
$\left(6 , 1\right) , \left(6 , 2\right) , \ldots , \left(6 , 6\right)$

The probability of each one of those is $\frac{1}{36}$.

• How many possible combinations of two dice will give you a sum of $7$? There are $6$ combinations: $\left(1 , 6\right)$, $\left(6 , 1\right)$, $\left(2 , 5\right)$, $\left(5 , 2\right)$,  (3,4) and $\left(4 , 3\right)$.

$\implies P \left(\text{sum} = 7\right) = 6 \cdot \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$

• For a sum of $11$, there are $2$ combinations: $\left(5 , 6\right)$ and $\left(6 , 5\right)$.

$\implies P \left(\text{sum} = 11\right) = 2 \cdot \frac{1}{36} = \frac{2}{36} = \frac{1}{18}$

• For a sum of $12$, there is just $1$ combinations: $\left(6 , 6\right)$.

$\implies P \left(\text{sum} = 12\right) = \frac{1}{36}$

Now, how do you combine those three probabilities?

The events "$\text{sum} = 7$", "$\text{sum} = 11$" and "$\text{sum} = 12$" are independent events since neither of them can ever occur at the same time.

For independent events $A$ and $B$ it holds

$P \left(A \text{ or } B\right) = P \left(A\right) + P \left(B\right)$

Thus, our probability is

$P = P \left(\text{sum"=7) + P("sum"=11) + P("sum} = 12\right)$

$= \frac{6}{36} + \frac{2}{36} + \frac{1}{36} = \frac{9}{36}$

$= \frac{1}{4}$

= 25%

May 13, 2018

#### Answer:

$P \left(7 , 11 , 12\right) = \frac{9}{36} = \frac{1}{4}$

#### Explanation:

You can draw up a possibility space and count how many of the outcomes meet the requirements.

The following array shows the sum of two dice thrown.
(Created by Parzival)

$\left.\begin{matrix}\textcolor{w h i t e}{0} & \text{ "ul1" " & " "ul2" " & " "ul3" " & " "ul4" " & " "ul5" " & " "ul6" " \\ 1| & " "2" " & " "3" " & " "4" " & " "5" " & " "6" " & " "color(red)(7)" " \\ 2| & " "3" " & " "4" " & " "5" " & " "6" " & " "color(red)(7)" " & " "8" " \\ 3| & " "4" " & " "5" " & " "6" " & " "color(red)(7)" " & " "8" " & " "9" " \\ 4| & " "5" " & " "6" " & " "color(red)(7)" " & " "8" " & " "9" " & " "10" " \\ 5| & " "6" " & " "color(red)(7)" " & " "8" " & " "9" " & " "10" " & " "color(red)(11)" " \\ 6| & " "color(red)(7)" " & " "8" " & " "9" " & " "10" " & " "color(red)(11)" " & " "color(red)(12)" }\end{matrix}\right.$

There are $6 \times 6 = 36$ outcomes

Of these there are $9$ which add to $7 , 11 \mathmr{and} 12$

$P \left(7 , 11 , 12\right) = \frac{9}{36} = \frac{1}{4}$