Question

Medal
Two six-sided dice are rolled. Find the probability of the given outcome (express your answers as fractions in lowest terms).
(a) The sum of the dots on the upward faces is 11.


(b) The sum of the dots on the upward faces is greater than 5.

Answer 1

(A) Think of the possibilities that could lead to 11:
\(5+6=11\)
\(6+5=11\)

There are 2 possibilities.
There are two dices with 6 sides. To find the total possibilities, simply multiply the numbers together.

\(6*6=36\)
So there are 2 posibilities we want, out of 36. This gives:
\[\frac{2}{36}\]\[=\frac{1}{16}\]

We could also find the percentage
\[Percentage=\frac{1}{16}*100\]
\[Percentage=6.25%\]

Answer 2

so part a is 1/16 in lowest terms

Answer 3

Yes

Answer 4

2/36 doesn't reduce to 1/16

Answer 5

it's close though

Answer 6

Sorry

Answer 7

Its \[\frac{1}{18}\]

Answer 8

correct

Answer 9

Do the same thing for (B)
There are a total of 26 combinations that equal to greater than 5
\[\frac{26}{36}\]
\[=\frac{13}{18}\]

Answer 10

13/18 is in lowest terms?

Answer 11

Combinations for (B)
1+5,
2+4,
3+3,
4+2,
5+1,
1+6,
2+5,
3+4,
4+3,
5+2,
6+1,
2+6,
3+5,
4+4,
5+3,
6+2,
3+6,
4+5,
5+4,
6+3,
4+6,
5+5,
6+4,
6+5.
5+6,
6+6

Answer 12

@harz360 Yes

Answer 13

something like this may come in handy

http://bestcase.files.wordpress.com/2011/01/dicediagram.jpg

Answer 14

Here's a visual for the sample space we're working with (where the first row and column are the values of the sides of the two dice and each entry in the table is then the sum of the dice).

\[\begin{array}{c|c|c|c|c|c|c} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\hline 5 & 6 & 7 & 8 & 9 & 10 & 11\\\hline 6 & 7 & 8 & 9 & 10 & 11 & 12\end{array}\]
In part (a), we notice that the following entries add to 11: \[\begin{array}{c|c|c|c|c|c|c} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\hline 5 & 6 & 7 & 8 & 9 & 10 & \color{red}{11}\\\hline 6 & 7 & 8 & 9 & 10 & \color{red}{11} & 12\end{array}\]
There are two such values, and since there are 36 total outcomes here, we now see that the probability that the sum of the two dice is 11 is \(\dfrac{2}{36} = \dfrac{1}{18}\).

In part (b), the entries that sum to values larger than 5 are highlighed:

\[\begin{array}{c||c|c|c|c|c|c} & 1 & 2 & 3 & 4 & 5 & 6\\\hline\hline 1 & 2 & 3 & 4 & 5 & \color{red}{6} & \color{red}{7}\\\hline 2 & 3 & 4 & 5 & \color{red}{6} & \color{red}{7} & \color{red}{8}\\\hline 3 & 4 & 5 & \color{red}{6} & \color{red}{7} & \color{red}{8} & \color{red}{9}\\ \hline 4 & 5 &\color{red}{6} & \color{red}{7} & \color{red}{8} & \color{red}{9} & \color{red}{10} \\\hline 5 & \color{red}{6} & \color{red}{7} & \color{red}{8} & \color{red}{9} & \color{red}{10} & \color{red}{11}\\\hline 6 & \color{red}{7} & \color{red}{8} & \color{red}{9} & \color{red}{10} & \color{red}{11} & \color{red}{12}\end{array}\] Can you take things from here and find the probability that the sum is larger than 5?

Answer 15

This could also be useful:

http://www.freemathhelp.com/rolling-dice.html

Answer 16

oh ok thank you so part a is 1/18 and part b is 13/18

Answer 17

@harz360, Yes :)

Answer 18

@ChristopherToni, nice diagram :)

Answer 19

The power of \(LaTeX\) never seems to end

Answer 20

It was annoying coloring each entry individually. \(\LaTeX\) certainly has it's pros and cons. >_>

Answer 21

thanks everyone for their help :)

Answer 22

what would be the median of this 73, 80, 90, 72, 83, 79, 91, 75, 74, 88, 76, and 79 , would it be 73+79/2 = 118.5?

Answer 23

79+79/2 = 118.5 *

Answer 24

Yes @ChristopherToni, yes it does. I am pretty happy I can do matrices right now ;)
\[\begin{bmatrix}
Y& E& A& H \\
\end{bmatrix}\]

Answer 25

the median is not 118.5

Answer 26

then 79

Answer 27

yes because (79+79)/2 = 79

Answer 28

ok thank you

Answer 29

First of all, organize the numbers:
\[73, 80, 90, 72, 83, 79, 91, 75, 74, 88, 76,79\]\(\text{ORGANIZED}\)
\[72, 73, 74, 75, 76, 79, 79, 80, 83, 88, 90, 91\]
Use this equation to find the median:
\[Median=\frac{n+1}{2}\]\[Median=\frac{12+1}{2}\]\[Median=\frac{13}{2}\]\[Median=6.5th\]
Its the 6.5th number. This will be your median:
72, 73, 74, 75, 76, \(79\), \(79\), 80, 83, 88, 90, 91\[Median=\frac{79+79}{2}\]\[Median=79\]
Therefore, @harz360, the median is 79

Answer 30

thank soo much :)

Answer 31

No problem @harz360 :D