We may as well suppose the game continues until all targets have been hit (which will happen eventually if all $p_j > 0$; we may as well remove any targets that have $p_j = 0$).
For each subset $S$ of $\{1,\ldots, N\}$, let $p_S = \sum_{s \in S} p_s$ be the probability that a shot hits a member of $S$, and let
$a_{i,t}(S)$ be the probability that target $i$ is one of the first $t$ targets in set $S$ to be hit. You want $1 - a_{i,X}(\{1,\ldots,N\})$.
Of course $a_{i,t}(S) = 0$ if $i \notin S$, and we also take it to be $0$ if
$t = 0$.
Otherwise, conditioning on the first target in $S$ to be hit,
$$ a_{i,t}(S) = \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}} \dfrac{p_j}{p_S} a_{i,t-1}(S \backslash \{j\}) $$
Now I claim that
$$ a_{i,t}(S) = \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,|S|) \frac{p_i}{p_{T \cup \{i\}}} $$
for some constants $c(t,m,n)$, $0 \le m \le n-1$.
I will prove this by induction on $t$.
In the case $t=1$ we have $a_{i,1}(S) = p_i/p_S$, so $c(1,m,n) = 1$ if
$m = n-1$, $0$ otherwise.
If $t >1$, we have (with $|S|=n$):
$$ \eqalign{a_{i,t}(S) &= \dfrac{p_i}{p_S} + \sum_{j \in S \backslash \{i\}}
\dfrac{p_j}{p_S}
\sum_{T \subseteq S \backslash \{i,j\}} c(t-1, |T|,n-1) \dfrac{p_i}{p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \sum_{j \in S \backslash (T \cup \{i\})} \dfrac{p_j p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{p_{S \backslash (T \cup \{i\})} p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \dfrac{(p_S - p_{T \cup \{i\}}) p_i}{p_S p_{T \cup \{i\}}}\cr
&= \dfrac{p_i}{p_S} + \sum_{m=0}^{n-2} c(t-1,m,n-1) \sum_{T \subseteq S \backslash \{i\}: |T| = m} \left(\dfrac{p_i}{p_{T \cup \{i\}}} - \dfrac{p_i}{p_S}\right)\cr
&= \sum_{T \subseteq S \backslash \{i\}} c(t,|T|,n) \dfrac{p_i}{p_{T \cup \{i\}}}
}$$
where $c(t, m,n) = c(t-1, m,n-1)$ if $m < n-1$ while
$$c(t, n-1, n) = 1 - \sum_{m=0}^{n-2} {n-1 \choose m} c(t-1,m,n-1)$$
Hmm: it looks like
$$ c(t, m, n) = \cases{1 & for $t=n,m=0$\cr
(-1)^{n+m+t} {m-1 \choose {t+m-n}} & $ n-m \le t \le n$\cr
0 & otherwise\cr}$$
There ought to be an inclusion-exclusion proof for this.
Your solution to the first problem is correct.
Let $A$ be the event that the first shooter hits the target; let $B$ be the event that the second shooter hits the target; let $C$ be the event that the third shooter hits the target. If at least two shooters hit the target, then either exactly two of them hit the target or all three do. Thus, we must calculate
$$\Pr(A)\Pr(B)\Pr(C^C) + \Pr(A)\Pr(B^C)\Pr(C) + \Pr(A^C)\Pr(B)\Pr(C) + \Pr(A)\Pr(B)\Pr(C)$$
given $\Pr(A) = 0.70$, $\Pr(B) = 0.80$, $\Pr(C) = 0.90$. Can you proceed?
Best Answer
It is given that the arrow lands on the target uniformly. This means the probability of landing on a circle is proportional to the area of the circle. In this case we have $10$ circles of equal area, so the probability that the arrow lands on any circle is $1/10$. The expected value of the points is$$\frac1{10}[1+2+\cdot\cdot\cdot+10]=\frac{11}2.$$