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For part b, considering the case where X=1, the batch will only be accepted if another random sample of 9 has 0 defectives. The number of defective articles in this new sample is independent from the first sample, therefore we can multiply P(1 defective) by P(0 defective) to find P(first sample X=1 AND second sample X=0). Hope this helps.
Date Posted:
3 years ago