If p=0.15, then q=1-p=0.85, the probability of failure, then we have the binomial distribution:
(p+q)n=pn+npn-1q+n(n-1)pn-2q2/2!+... where n=the number of independent trials=10. The sum of these terms =1 because (p+1-p)n=1 for all n.
Each term in this series represents an individual exact probability:
(10 successes)+(9 successes)+...+(4 successes)+(3 successes)+(2 successes)+(1 success)+(no successes). Alternatively, this is (10 failures)+(9 failures)+...+(1 failure)+(no failures).
x≤4 successes means at least 6 failures=P(q=10)+P(q=9)+P(q=8)+P(q=7)+P(q=6).
(q+p)n=qn+nqn-1p+n(n-1)qn-2p2/2!+...=
0.8510+10(0.859)(0.15)+45(0.858)(0.152)+120(0.857)(0.153)+210(0.856)(0.154)=0.9901.