statistics
posted by Jennifer .
The probability any one widget is free of defects is 0.93. For quality control purposes the company is concerned about bad batches of widgets. If the widgets are shipped in batches of 20, find the probability of at most 2 defective widgets in one batch.

Prob 0 defective = (.93)^20
prob 1 defective = C(20,1) (.93)^19 (.07)
prob 2 defective  C(20,2) (.93)^18 (.07)^2
prob (at most 2 defective) = 1  (sum of the above 3 probs)