Products produced by a machine all have a 3% defective rate. Suppose items are drawn in sequence from the machine for inspection. a) What is the probability that the first defective found is the fifth item inspected? Write the probability statement and then calculate your answer. b) What distribution does this follow? c) Suppose the first five items are not defective. What is the probability that at least 12 total items need to be inspected for the first defective to be found? Write the probability statement and then calculate your answer.

a) The probability that the first defective found is the fifth item insepcted is 0.0266.

b) The distribution of the total amount of items drawn until the first one defective is Geometric with parameter p = 0.03

c) The probability statement is

P(X≥12|X>5)

Where X is the geometric distribution described above. That probability is 0.833

Step-by-step explanation:

a) The probability of not picking a defect is 0.97. If we want the fifth item to be the first defective, we need the first fourth ones to not be defective. Also, each product being defective or not does not depend on the others, so we can multiply the probabilities. Therefore, the probability that the fifth item is the first defective is 0.97⁴ * 0.03 = 0.0266.

b) Lets call the distribution of total number of items picked until picking a defecitve one X. This distribution is finding out the first success for a sucesion of independent experiments with equally probability of success; as a consequence, the distribution is Geometric, with parameter p = 0.03.

c) The probability statement is P(X≥12 | X > 5). We need to note a few things here:

For independence on each individual outcome P(X≥12 | X > 5) = P(X≥12)/P(X>5)P(X ≥ 12) = P(X > 11)P(X>k) just computes the probability that the first k attemps of the experiment are a failure, for independance, this is 0.97^k.

With all this in mind, we conclude that

P(X≥12|X>5) = P(X>11)/P(X>5) = 0.97¹¹/0.97⁵ = 0.97⁶ = 0.833

50% of students voted for candidate b.