Let p=0.75 (probability that an Ewok is startled) and q=0.25 (probability that an Ewok is not startled).
p+q=1 and (p+q)¹⁰=1=p¹⁰+10p⁹q+45p⁸q²+... contains all probabilities. Each term represents a unique probability: first term: all Ewoks startled; 2nd term: exactly 9 startled; 3rd term: exactly 8 startled; ...
Probability of 5 startled=252p⁵q⁵; probability of 6 startled=210p⁶q⁴; ...
Probability of at least 5 startled is the sum of the last 6 terms (5, 6, 7, 8, 9, 10).
C(10,r) or ¹⁰Cᵣ where r=(5, 6, ..., 10) is the coefficient=(10×...×(10-r+1))/r!.
The coefficients we need (if we don’t use tables) are 252, 210, 120, 45, 10, 1 (10th row of Pascal’s Triangle). The sum is 0.98027 approx or about 98.03%. We could have used q¹⁰+10q⁹p+...+210q⁶p⁴ (5 terms) and subtracted the sum from 1 to give us the same answer.
The probability of all startled is p¹⁰=0.05631 or about 5.63%.