Answer:
the probability to be accepted is 0.302 (30.2%) (many will be rejected)
Step-by-step explanation:
assuming that the rate of 4% applies to the 60 tablets then since each tablet behaves independently, the random variable X= number of tablets with defects out of 60 tablets has a binomial distribution , where:
p(X)=n!/((n-x)!*x!)*p^x*(1-p)^(n-x)
where
n= total number of tablets tested = 60
x = number of defective tablets
p= probability to be defective = 0.04
then in order to be accepted x≤0 , then the probability that the batch is accepted Pa is
Pa=P(x≤1) = P(0) + P(1) = (1-p)^n + n*p*(1-p)^(n-1)
replacing values
Pa= (1-p)^n + n*p*(1-p)^(n-1) = 0.96^60 + 60*0.04*0.96^59 = 0.302 (30.2%)
then the probability to be accepted is 0.302 (30.2%) (many will be rejected)