# Finding probability when two dice are rolled [closed]

You have two dice. Die one is a standard die with the six faces marked from 1 to 6. The second die has two faces marked with 1, two faces marked with 2 and two faces marked with 3. Both dice are rolled. The probability that the sum of values on the top face of the two dice is greater than 6 is:

A. 6/36

B. 8/36

C. 10/36

D. 12/36

## closed as off-topic by tatan, GNUSupporter 8964民主女神 地下教會, MathOverview, darij grinberg, José Carlos SantosFeb 13 at 8:42

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• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – tatan, GNUSupporter 8964民主女神 地下教會, MathOverview, darij grinberg, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• "Die $2$ has $2$ faces each marked with $1,2,3$"... Can you explain? A die cannot have two faces. A coin has 2 faces. And what do you mean by each face is marked by $1,2,3$. Also, welcome to MSE!!! To get a good response please include your our thoughts ;otherwise no one will bother to asnwer and you question will be closed. This is not a homework help site.;-) – tatan Feb 12 at 14:55
• @tatan The not standard die has 6 faces: {1,1,2,2,3,3} – Daniel Mathias Feb 12 at 14:56
• @DanielMathias She wrote in the question "two faces"... isn't it? – tatan Feb 12 at 14:57
• @tatan Two faces marked 1, two faces marked 2, two faces marked 3 – Daniel Mathias Feb 12 at 14:57
• @DanielMathias Okay... I see... lack of punctuations is making it hard to follow the English... Please go on and edit the question and add proper punctuations – tatan Feb 12 at 14:58

## 2 Answers

Daniel Mathias already added a good asnwer but I am posting my asnwer since I already began writing it- $$\text{First Die(Left)-SecondDie(Right)}$$ $$6-\{1,1,2,2,3,3\}$$ $$5-\{2,2,3,3\}$$ $$4-\{3,3\}$$

So $$P(Sum>6)=(\frac16\times1)+(\frac16\times\frac46)+(\frac16\times\frac26)=\frac{12}{36}$$

• That should be $\frac{12}{36}$ – Daniel Mathias Feb 12 at 15:53
• it should be 12/36 – Minaxi Joshi Feb 12 at 15:55
• I think I should revise class 1 addition once more ;-P @MinaxiJoshi – tatan Feb 12 at 16:01
• @DanielMathias Edited ;-) – tatan Feb 12 at 16:01

Make a table of possible outcomes: $$\begin{array}{c|cc} &1&2&3&4&5&6\\ \hline 1&2&3&4&5&6&\boxed7\\ 1&2&3&4&5&6&\boxed7\\ 2&3&4&5&6&\boxed7&\boxed8\\ 2&3&4&5&6&\boxed7&\boxed8\\ 3&4&5&6&\boxed7&\boxed8&\boxed9\\ 3&4&5&6&\boxed7&\boxed8&\boxed9\\ \end{array}$$ Count the number of outcomes with sum greater than $$6$$. (There are $$12$$ of them, as marked)

The probability is therefore $$\frac{12}{36}$$

• (+1) for creating the wonderful visual ;-) – tatan Feb 12 at 16:01