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 Santos Feb 13 at 8:42

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  • $\begingroup$ "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.;-) $\endgroup$ – tatan Feb 12 at 14:55
  • $\begingroup$ @tatan The not standard die has 6 faces: {1,1,2,2,3,3} $\endgroup$ – Daniel Mathias Feb 12 at 14:56
  • $\begingroup$ @DanielMathias She wrote in the question "two faces"... isn't it? $\endgroup$ – tatan Feb 12 at 14:57
  • $\begingroup$ @tatan Two faces marked 1, two faces marked 2, two faces marked 3 $\endgroup$ – Daniel Mathias Feb 12 at 14:57
  • $\begingroup$ @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 $\endgroup$ – tatan Feb 12 at 14:58

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

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

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

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