The third success occurs on the $30$th trial if and only if there are exactly $2$ successes in the first $29$ trials, and then a success on the $30$th.
If $p$ is the probability of success on any trial, here $0.1$, then the probability of $2$ successes in $29$ trials is $\binom{29}{2}p^2(1-p)^{27}$. Now multiply by $p$.
Yes.
The key to the question is that "skillfulness is constant" means that the three distances are independent and identically distributed. It is also almost impossible for them to be tied, since they are continuous RV.
This means that even without knowing what the distribution of distances may be, we do know that the $6$ order arrangements are equally probable.
$$\boxed{\boxed{D_1{<}D_2{<}D_3 \;,\; D_1{<}D_3{<}D_2} \;,\; D_3{<}D_1{<}D_2}\\ D_2{<}D_1{<}D_3 \;,\; D_2{<}D_3{<}D_1 \;,\; D_3{<}D_2{<}D_1$$
Now we were asked to evaluate the probability that the third throw is placed further than the first throw, when given that the first throw is placed closer than the second. That is: $\mathsf P(D_1<D_3\mid D_1<D_2)$. Clearly this is: $2/3$.
If doing it the easy way isn't convincing, we shall do it the hard way:
Let $f_1(r), f_2(r), f_3(r)$ be the probability density functions of the distance from the center of the three darts, respectively, and $F_1(r), F_2(r), F_3(r)$ be their cumulative distribution functions. (These of course will be co-equal because the darts are i.i.d., but the indices are a readability guide.)
Without knowing the skill of the dart thrower we may as well assume the darts are uniformly distributed over the board (why? why not?). Then ...,
$$\newcommand{\dint}{{\displaystyle \int}}
\begin{align}
f_1(r) & = \dfrac{2r}{R^2} & =f_2(r)=f_3(r)
\\[2ex]
F_1 (r) & = \dfrac{r^2}{R^2} & =F_2(r)=F_3(r)
\\[2ex]
\mathsf P(D_1<D_2)
& = \int_0^R f_1(r)\,\big(1-F_2(r)\big) \operatorname d r
\\[1ex] & = \frac 2{R^4}\int_0^R rR^2-r^3\operatorname d r
\\[1ex] & = \dfrac 1 2
\\[2ex]\mathsf P(D_1<D_3\mid D_1<D_2)
& = \int_0^R f_1(r \mid D_1<D_2)\; \big(1-F_3(r)\big) \operatorname d r
\\[1ex] ~ & = \dfrac{
\dint_0^R f_1(r) \, \big(1-F_2(r)\big) \, \big(1-F_3(r)\big) \operatorname d r
}{
\dint_0^R f_1(r) \, \big(1-F_2(r)\big)\operatorname d r
}
\\[1ex] ~ & = \dfrac{
\frac{2}{R^6} \dint_0^R r \, (R^2-r^2)^2 \operatorname d r
}{
\frac 1 2
}
\\[1ex] ~ & = \dfrac 2 3
\end{align}$$
As an exercise for the student, try this with some other assumption of the dart thrower's skill, such as $f_1(r)=\frac 1 R$.
Finally examine the situation when you make no assumption, and convince yourself of the symmetry argument.
$\mathsf P(D_1<D_2) = \dint_0^R f(x) \dint_x^R f(y)\operatorname d y\operatorname d x \\ \mathsf P(D_1<D_3\mid D_1< D_2) = \frac{\int_0^R f(x) \left(\int_x^R f(y)\operatorname d y\right)^2\operatorname d x}{\int_0^R f(x) \int_x^R f(y)\operatorname d y\operatorname d x}$
Best Answer
Only positive scores are relevant. So it is enough to look at them. Under condition that the score is positive by aiming on certain positive score ($2$ or $3$) there is probability of $\frac{56}{80}=\frac{7}{10}$ that this score is reached and a probability of $\frac{24}{80}=\frac{3}{10}$ that the other positive score is reached.
Let $p_{n}$ denote the probability that score $n$ is reached by applying a strategy that is optimal in the sense that it leads to a maximal probability to achieve score $n$.
Then $p_{1}=0$, $p_{2}=p_{3}=p_{5}=\frac{7}{10}$ and $p_{4}=p_{6}=\frac{7}{10}\frac{7}{10}=\frac{49}{100}$.
Choosing a strategy (at the start) to come to a score of $7$ we have the following options:
Aiming at $2$: there is a probability of $\frac{7}{10}p_{5}+\frac{3}{10}p_{4}=\frac{7}{10}\frac{7}{10}+\frac{3}{10}\frac{49}{100}=0.637$.
Aiming at $3$: we have an optimal probability of $\frac{7}{10}p_{4}+\frac{3}{10}p_{5}=\frac{7}{10}\frac{49}{100}+\frac{3}{10}\frac{7}{10}=0.553$.
Any 'other' strategy will give a probability $p\times0.637+\left(1-p\right)\times0.553\in\left[0.553,0.637\right]$ for some $p\in\left[0,1\right]$
So at the start we choose for aiming at $2$ and $p_{7}=0.637$ is the probability that score $7$ will be reached.