Final answer:
a. The probability of no defective calculators in a sample of three is (295/300) * (295/300) * (295/300). b. The probability of all defective calculators in a sample of three is (5/300) * (5/300) * (5/300). c. The probability of at least 1 defective calculator in a sample of three is 1 - [(295/300) * (295/300) * (295/300)]. d. The probability of at least 1 non-defective calculator in a sample of three is 1 - [(5/300) * (5/300) * (5/300)].
Step-by-step explanation:
a. To find the probability of no defective calculators in a sample of three, we need to calculate the probability of selecting a non-defective calculator three times in a row. Since there are 5 defective units out of 300, there are 295 non-defective calculators. The probability of selecting a non-defective calculator on the first draw is 295/300. Since each draw is independent, the probability on the second draw is also 295/300, and the probability on the third draw is also 295/300. To find the overall probability, we multiply these probabilities together: (295/300) * (295/300) * (295/300).
b. To find the probability of all defective calculators in a sample of three, we need to calculate the probability of selecting a defective calculator three times in a row. Since there are 5 defective units out of 300, the probability of selecting a defective calculator on the first draw is 5/300. Since each draw is independent, the probability on the second draw is also 5/300, and the probability on the third draw is also 5/300. To find the overall probability, we multiply these probabilities together: (5/300) * (5/300) * (5/300).
c. To find the probability of at least 1 defective calculator in a sample of three, we can use the complement rule. The complement of having no defective calculators is having at least one defective calculator. Therefore, the probability of at least one defective calculator is 1 - probability of no defective calculators. Using the probabilities from part a, we find the probability of at least one defective calculator is 1 - [(295/300) * (295/300) * (295/300)].
d. To find the probability of at least 1 non-defective calculator in a sample of three, we can use the complement rule. The complement of having all defective calculators is having at least one non-defective calculator. Therefore, the probability of at least one non-defective calculator is 1 - probability of all defective calculators. Using the probabilities from part b, we find the probability of at least one non-defective calculator is 1 - [(5/300) * (5/300) * (5/300)].