Question

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

Short answer

a. Yes, \(E_{1}\) and \(E_{2}\) are technically dependent events in this context, as selection of first board affects the composition of the boards that the second board is chosen from. But since number of defective is much less than total, the dependence is negligible.\n b. The probability of not \(E_{1}\) is 0.992.\n c. The two conditional probabilities, \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid E_{1}^{C}\right)\), are very similar.\n d. Given the similarity of the two conditional probabilities, \(E_{1}\) and \(E_{2}\) could be viewed as approximately independent.

Step-by-step solution

Dependence of events

Events \(E_{1}\) and \(E_{2}\) are dependent. This is because once a board is selected, the total number of boards reduces by 1. Hence, the probability of the second event depends on the outcome of the first event: if the first board is defective, there would be 39 defective boards left out of 4999, if not, there would still be 40 defective boards out of 4999.

Probability of not \(E_{1}\)

The event \(E_{1}^{C}\), or not \(E_{1}\), is that the first board selected is not defective. There are 5000-40=4960 such favourable cases. Hence, \(P\left(E_{1}^{C}\right) = \frac{4960}{5000} = 0.992\).

Comparison of the conditional probabilities

The conditional probability \(P\left(E_{2} \mid E_{1}\right) = \frac{39}{4999}\) because if the first selected board is defective (event \(E_{1}\)), we have 39 defective boards remaining out of 4999. The conditional probability \(P\left(E_{2} \mid E_{1}^{C}\right) = \frac{40}{4999}\) because if the first selected board is not defective, we still have 40 defective boards remaining out of 4999. These two probabilities are quite similar.

Independent or dependent

Even though the events are technically dependent, the two probabilities calculated in step 3 are very similar. This is due to the fact that the number of defective boards is very small compared to the total number of boards. Hence, it would be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent.