Your calculated probability of observing 1 or fewer failures in the sample of 850 as equal to 0.86018 is correct. This calculation is based upon the probability of observing a failure in a single trial as p=1/1300 (taken from the original 1300 item sample).

One way that a confidence interval can be constructed for p is through the Clopper-Pearson (1) method – recommended over the normal distribution approximation in this case due to the observed small number of failures in the original 1300 item sample. This method calculates the upper and lower bounds for p by solving exact two-sided confidence interval equations for Pupper and Plower, respectively. (Although the equations can be tedious, a web search revealed a number of calculators that can accomplish this operation). In this case, the upper and lower bounds of a 90% confidence interval for p are 0.003646 and 0.0000396, respectively.

Substituting the upper and lower bounds for p in the binomial probability distribution equation results in the probability, with 90% confidence, of observing 1 or fewer failures in the 850 item sample as ranging from 0.184 to 0.999, with an expected value of 0.860.

Reference:
(1) C.J. Clopper and E.S. Pearson, “The use of confidence or fiducial limits in the case of the binomial”, Biometrika 26:404-413, 1934.