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.
Best Answer
Hint :
1) The probability that the target is missed by BOTH shooters is : $p=0.3\times 0.2$
2) Your answer is correct.