Rolling dice probability

[S_100]

Two dice have been thrown, giving a total of at least 10. What is the probability that the throw of a third die will bring the total of the three numbers shown to 15 or higher?

My solution

if scoring a 10 then, 5 or 6 can be rolled
and since 3 combinations of two die can score a 10 : (4,6) (6,4) (5,5) , so (1/6)(1/6)(1/6)*2*3

if scoring a 11, then 4 ,5, 6 can be rolled

and since 2 combinations
(5,6) 1/6)(1/6)(1/6)* 3*2
(6,5)

if scoring a 12, then 3 ,4 5,6 can be rolled

(6,6) (1/6)(1/6)(1/6)* 4


This gives a total of 16 ways to score 15 or over which gives the probability 16*1/216 = 2/27

However the solution (image attached) gives 6/36 + 6/36 + 4/36 = 4/9 but I don't understand how this is, which 'solution' is correct?

 

[ksdhart2]

Your notation is unclear and quite confusing to me, and I don't think l understand any of your workings that went into it... but you did get the correct answer. The given answer is just plain wrong, no two ways about it. The theory behind their answer is fine but the execution is where they fail. Just taking the first case as an example, they've (tried) to write it in the form:

$\text{(Probability of rolling a 10 on the first two dice)} \cdot \text{(Probability of rolling a 5 or 6 on the third dice)}$.

But the actual numbers they wrote are not correct. That would correspond to saying the probability of rolling exactly 10 on the first two dice is $\displaystyle \frac{3}{6} = \frac{1}{2}$, and that's clearly false. There's no way that 18 of the 36 outcomes on 2d6 sum to exactly 10. What they should have written is:

$\displaystyle P(T \ge 15) = \left( \frac{3}{\mathbf{36}} \times \frac{2}{6} \right) + \left( \frac{2}{\mathbf{36}} \times \frac{3}{6} \right) + \left( \frac{1}{\mathbf{36}} \times \frac{4}{6} \right) = \frac{16}{216} = \frac{2}{27}$

 

[pka]

Two dice have been thrown, giving a total of at least 10. What is the probability that the throw of a third die will bring the total of the three numbers shown to 15 or higher?

Please look at this PAGE. That generating polynomial, $\displaystyle x^{12}+2x^{11}+3x^{10}+\cdots$ tells us that throwing two dice that there is one pair to get 12, two pairs to get 11 & three pairs to get a 10. Let $\displaystyle S$ be the sum and $\displaystyle D$ be the digit on the third die.
\(\displaystyle \begin{array}{*{20}{c}} S&D&{\mathcal{P}(S)}&{\mathcal{P}(D)}&{\mathcal{P}(S)\mathcal{P}(D)} \\ \hline
{10}&{\left\{ {5,6} \right\}}&{\frac{3}{{36}}}&{\frac{2}{6}}&{\frac{6}{{216}}} \\
{11}&{\left\{ {4,5,6} \right\}}&{\frac{2}{{36}}}&{\frac{3}{6}}&{\frac{6}{{216}}} \\
{12}&{\left\{ {3,4,5,6} \right\}}&{\frac{1}{{36}}}&{\frac{4}{6}}&{\frac{4}{{216}}}
\end{array}\)