Hi. How would I calculate a throw with two dice if they both consist out of six sides that are numbered with 1, 2, 2, 4, 6, 6? I need to calculate the probability that the sum is greater than 8. I think the possibilities for the specific situation are (4, 6), (6, 4), (6, 6) but I'm not sure quite how many possibilities there actually are in total.
(Oh also could someone verify if this is correct? I had to calculate the possibility for a sum greater than 8 with a normal dice. Would it be 10/36 = 5/18 ?)
+118609
Use this table, or one the same to work out all your possibilities.
The top heading row is the number on dice 1
The left heading row it the number on dice 2
You fill out the sums and answer the question/s accordingly.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | ||||||
| 2 | ||||||
| 3 | ||||||
| 4 | ||||||
| 5 | ||||||
| 6 |
Melody gave you a table for 2 REGULAR 6-sided dice. But yours are NOT normal dice. If you made a table of 36 possible sums of your 2 dice, you would get the following totals:
(2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 12, 12, 12, 12) = 36 sums
Notice the that sums of 10 and 12 are: 4 + 4 = 8. Therefore, the probability of obtaining a sum greater than 8 is:
8 / 36 =2 / 9
+118609 Thanks answering guest.
Sorry asker guest, I did not read your question properly but you can still use the table only instead of using the headings
1,2,3,4,5,6
jus use
1,2,2,4,6,6
It will work well :)
You can count the number of boxes that make it true and put that over 36 because there are 36 squares (in the grid) in total.
Thank you so much for the responses! And it's alright, mistakes happen :) So it'd look something like this then? I'll definitely keep it in mind to draw assignments out in the future!
| + | 1 | 2 | 2 | 4 | 4 | 6 | 6 |
| 1 | 2 | 3 | 3 | 5 | 5 | 7 | 7 |
| 2 | 3 | 4 | 4 | 6 | 6 | 8 | 8 |
| 2 | 3 | 4 | 4 | 6 | 6 | 8 | 8 |
| 4 | 5 | 6 | 6 | 8 | 8 | 10 | 10 |
| 4 | 5 | 6 | 6 | 8 | 8 | 10 | 10 |
| 6 | 7 | 8 | 8 | 10 | 10 | 12 | 12 |
| 6 | 7 | 8 | 8 | 10 | 10 | 12 | 12 |
Yes, it will make it much easier for you to visualize it. However, in this attempt you made a small mistake of having a:
7 x 7 table instead of 6 x 6. You added one extra 4 that shouldn't be there. That is why you have 4 extra 10s.
+26367 How would I calculate a throw with two dice if they both consist out of six sides that are numbered with \(\mathbf{1, 2, 2, 4, 6, 6}\)?
I need to calculate the probability that the sum is greater than 8.
I think the possibilities for the specific situation are (4, 6), (6, 4), (6, 6)
but I'm not sure quite how many possibilities there actually are in total.
\(\mathbf{1, 2, 2, 4, 6, 6}\):
\(\begin{array}{|c|c|c|c|c|c|c|} \hline + &\mathbf{1} & \mathbf{2}& \mathbf{2}& \mathbf{4}& \mathbf{6}& \mathbf{6} \\ \hline \mathbf{1} &2 & 3& 3& 5& 7& 7 \\ \hline \mathbf{2} &3 & 4& 4& 6& 8& 8 \\ \hline \mathbf{2} &3 & 4& 4& 6& 8& 8 \\ \hline \mathbf{4} &5 & 6& 6& 8& \color{red}10& \color{red}10 \\ \hline \mathbf{6} &7 & 8& 8& \color{red}10& \color{red}12& \color{red}12 \\ \hline \mathbf{6} &7 & 8& 8& \color{red}10& \color{red}12& \color{red}12 \\ \hline \end{array} \qquad ~\dfrac{\color{red}8}{36} \)
Normal dice:
\(\begin{array}{|c|c|c|c|c|c|c|} \hline + &\mathbf{1} & \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}& \mathbf{6} \\ \hline \mathbf{1} &2 & 3& 4& 5& 6& 7 \\ \hline \mathbf{2} &3 & 4& 5& 6& 7& 8 \\ \hline \mathbf{3} &4 & 5& 6& 7& 8& \color{red}9 \\ \hline \mathbf{4} &5 & 6& 7& 8& \color{red}9& \color{red}10 \\ \hline \mathbf{5} &6 & 7& 8& \color{red}9& \color{red}10& \color{red}11 \\ \hline \mathbf{6} &7 & 8& \color{red}9& \color{red}10& \color{red}11& \color{red}12 \\ \hline \end{array} \qquad ~ \dfrac{\color{red}10}{36} \)
