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I'm having trouble solving this problem:

There's a dice with 14 faces, of which 6 are squared, and 8 are triangular. If the probability of getting a squared face is twice the probability of getting a triangular one, then what is the probability of getting two squared sides when throwing the dice twice?

My answer is $\frac49$ (but it's not even in the alternatives), since the probability of getting a squared side is $P(S)=2P(T)$, and I can either get a squared or a triangular one, then the probability of getting a squared side is: $$\frac{2\cdot P(T)}{3\cdot P(T)}=\frac23$$

The solution that is proved to me is $\frac9{25}$, but it doesn't even make sense to me, because that implies that the probability of getting a squared size is $\frac35$. So I want to know if either I'm wrong, or if the solution is correct. Thanks for your help.

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    $\begingroup$ It should be a "die", not a "dice". $\endgroup$ – Ben Grossmann Jul 24 '17 at 0:58
  • $\begingroup$ You're correct in that this question is poorly phrased. I read it the same way you did. $\endgroup$ – Jared Goguen Jul 24 '17 at 1:32
  • $\begingroup$ perhaps the total surface area has to do with the probability of getting a triangle or square, as well. Physically speaking. $\endgroup$ – pwned Jul 24 '17 at 7:32
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The critical thing is the interpretation of the question. You read the question as the total probability of getting one of the square faces is twice the total probability of getting one of the triangular faces, which tells us that the probability of getting some square face is $\frac 23$ and your calculation is correct. In your reading the number of square and triangular faces is extraneous information. The intended reading is that the chance of getting any one particular square face is twice the chance of getting any one particular triangular face. Now the number of each type of face matters and the probability of some square face is $\frac 35$. I find the problem as written ambiguous on how it should be read.

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  • $\begingroup$ I'm translating it from Spanish, so I wrote my interpretation, which could skew the perception, a more thorough translation of the problematic part would be: "When throwing this dice, getting a squared face is "doubly" probable of getting a triangular face, the probability of getting two squared faces is:..." Which I think supports the second interpretation. $\endgroup$ – Nick Cassol Jul 24 '17 at 1:14
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Let $s$ denote the probability $P(S)$ of getting a square side, and let $t$ denote the probability of getting a triangular side. Assuming that all square sides are equally likely and that all triangular sides are equally likely, we have $$ 6s + 8t = 1 \implies\\ 6(2t) + 8t = 1 \implies\\ t = \frac 1{20}, \quad s = \frac 1{10} $$ So, the overall probability of getting a square side on one roll of the die is $$ 6 \cdot s = \frac 3{5} $$ Perhaps you could take it from there.

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  • $\begingroup$ That is the procedure that the solution describes, but I don't understand why I have to put the number of sides in the equations, since they are flat out telling me the ratio of the probability of the events. $\endgroup$ – Nick Cassol Jul 24 '17 at 1:05
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    $\begingroup$ They're telling you that the probability of getting any particular square face is twice the probability of getting any particular triangular face. They are not telling you that the overall probability of getting a square face is twice that of getting a triangular face. If that were the case, then you're right: the number of sides wouldn't matter. $\endgroup$ – Ben Grossmann Jul 24 '17 at 1:15
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    $\begingroup$ There should be a distinction for that. I read the question and I didn't know which interpretation they wanted. $\endgroup$ – tarashir342 Jul 24 '17 at 1:21
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    $\begingroup$ @tarashir342 of course, given the correct answer, it's clear which one they meant. But I agree that the question could have been phrased more clearly. $\endgroup$ – Ben Grossmann Jul 24 '17 at 1:29
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    $\begingroup$ @tarashir342 I'm translating the statement of the exercise, and I can't translate it directly, but after asking this question, reading the answers and then rereading the problem, now I can see that the interpretation used in this answer is the one wanted (and the most obvious) from what the exercise states. $\endgroup$ – Nick Cassol Jul 24 '17 at 1:31
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There are two possibilities to understand the sentence:

The probability of getting a squared face is twice the probability of getting a triangular one

The first possibility is the one you understood is that the probability of getting any square is double of getting any triangle:
$P(x\in\{S_1,S_2,S_3,S_4,S_5,S_6\}) = 2\cdot P(x\in\{T_1,T_2,T_3,T_4,T_5,T_6,T_7,T_8\}))$
This leads to your result of $\frac{4}{9}$.

The second possibility is that the probability of getting a specific square is twice the probability of getting a specific triangle:
$P(x=S_1) = ... = P(x=S_6) = 2\cdot P(x=T_1)=...= 2\cdot P(x=T_8)$
With this basis the result is $\frac{9}{25}$

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