Sum of three dice (six faced) throws is $15$. What is the probability that first throw was $4$?
The way I thought of solving this was... - Given - sum of second and third throw is $11$ - Probability of getting first throw = $4$ is $1$ out of $6$, that is $\frac{1}{6}$
Is this correct?
On
Joffan pretty much pointed how to find the answer.
$P(D_{1}=4\mid Sum=15) = \frac {ways\ to\ get\ 11\ with\ 2\ dices}{ways\ to\ get\ 15\ with\ 3\ dices}$ (Note that getting 11 with 2 dices is like getting 4 with the first one).
$ways\ to\ get\ 11\ with\ 2\ dices = 2$ (both $(5,6)$ and $(6,5)$)
$ways\ to\ get\ 15\ with\ 3\ dices =10$
You can have 15 with $(3,6,6), (4,5,6),(5,5,5)\ and\ its\ permutations: (6,3,6), (6,6,3), (4,6,5), (5,4,6), (5,6,4), (6,4,5)\ and\ (6,5,4)$
All together gives us that $P(D_{1}=4\mid Sum=15) =\frac {2}{10}=0.2$
$0.2$, is the probability you were looking for.
You can state this very nicely using conditional probabilities:
Following your start, I'd look at $$ \begin{split} P\left(\sum_{i=1}^3 x_i = 15, x_1 = 4\right) &= P\left(\sum_{i=1}^3 x_i = 15|x_1 = 4 \right)P\left(x_1 = 4\right) \\ &= P\left(\sum_{i=2}^3 x_i = 11\right)P\left(x_1 = 4\right), \end{split} $$ where by independence (unless a roll depends on previous rolls) the second line is derived from the first.
You have the second term on the right side, and now need to calculate the first term.