Answer: This is a binomial distribution problem with parameters n = 17 and p = 0.3, where n is the sample size and p is the probability of a defective bulb. We need to find the probability that between 3 and 6 (both inclusive) bulbs from the sample are defective.
Let X be the number of defective bulbs in the sample. Then X has a binomial distribution with parameters n = 17 and p = 0.3, i.e., X ~ B(17, 0.3).
We want to find P(3 ≤ X ≤ 6). Using the binomial probability formula, we have:
P(3 ≤ X ≤ 6) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
= (17 choose 3)(0.3^3)(0.7^14) + (17 choose 4)(0.3^4)(0.7^13) + (17 choose 5)(0.3^5)(0.7^12) + (17 choose 6)(0.3^6)(0.7^11)
≈ 0.1566 (rounded to four decimal places)
Therefore, the probability that between 3 and 6 (both inclusive) bulbs from the sample are defective is approximately 0.1566.
Explanation: