There are two possibilities: the soldier hits the target (probability p=0.3); or he misses (probability q=0.7). There are only two possibilities so the distribution is binomial.
For a finite number of shots n, (p+q)n=(pn+npn-1q+n(n-1)pn-2q2/2!+...(n!/(r!(n-r)!))pn-rqr+...n(n-1)p2qn-2/2!+npqn-1+qn
When n=10, the terms on the right mean probability of all hits+9 hits out of 10 shots+8 hits out of 10 shots+...+2 hits out of 10 shots+1 hit out of 10 shots+no hits.
Probability has two parameters: n and r where r is the exact number of hits.
So p(n,r)=(n!/(r!(n-r)!))0.3r0.7n-r where 0≤r≤n.
For example, if n=10 and r=3 (3 hits out of 10 shots)=p(10,3)=(10!/(3!7!))0.330.77=120(0.027)(0.08235)=0.2668=26.68% approx.