The probability of seeing the first 61 in a batch good and the last 3 defective is (.96)^61 * (.04)^3 = X
Now then, count the number of ways a batch of 64 could have exactly 3 defectives. The formula for n-choose-c or 64-choose-3 is (n!/c!(n-c)!) (where ! means factorial).
64!/3!*61! = 41664.
Finally 41664*X = .221 or 22.1%
QED
A company manufactures calculators in batches of 64 and there is a 4% rate of defects. Find the probability of getting exactly three defects in a batch.
1 answer