Chapter 2 - Section 2-6 - Independence - Exercises - Page 52: 2-145

a. Events A and B are dependent. b. Events A and B are independent.

a. Let 𝐴: be the event that the first container is defective. Let 𝐵: be the event that the second container is defective. The probability of A is given by: $P(A)=$number of defective first containers / total number of containers $ = 5/500=0.01$ Now, to determine if 𝐴 and 𝐵 are independent, we check if 𝑃(𝐴∣𝐵)=𝑃(𝐴) If 𝑃(𝐴∣𝐵) is not equal to 𝑃(𝐴), then the events 𝐴 and 𝐵 are dependent, meaning the occurrence of one event affects the probability of the other event occurring. Conditional Probability Formula The conditional probability 𝑃(𝐴∣𝐵) is calculated as: $𝑃(𝐴∣𝐵)=𝑃(𝐴∩𝐵)/𝑃(𝐵)$ Law of Total Probability If 𝐴1, 𝐴2,…,𝐴𝑛 are mutually exclusive and ⋃𝑖^n=1A𝑖=Ω, then for any event 𝐵: $𝑃(𝐵)=𝑃(𝐵∣𝐴1)𝑃(𝐴1)+𝑃(𝐵∣𝐴2)𝑃(𝐴2)+⋯+𝑃(𝐵∣𝐴𝑛)𝑃(𝐴𝑛)$ or more compactly: $𝑃(𝐵)=n∑𝑖=1𝑃(𝐵∣𝐴𝑖)𝑃(𝐴𝑖)$ Using the Law of Total Probability, we have: $𝑃(𝐵)=𝑃(𝐵∣𝐴)⋅𝑃(𝐴)+𝑃(𝐵∣𝐴′)⋅𝑃(𝐴′)$ where 𝐴′ is the complement of 𝐴. In this problem: 𝑃(𝐵∣𝐴)=4/499 because 4 defective containers remain out of 499 after taking one defective container. 𝑃(𝐵∣𝐴')=5/499 because all 5 defective containers remain out of 499 after taking a non-defective container. 𝑃(𝐴′)=495/500 So, 𝑃(𝐵)=(4/499)(5/500)+(5/499)(495/500)=0.01 Thus, we confirm that 𝑃(𝐵)=0.01. Finally, we check the independence condition: $𝑃(𝐴∣𝐵)=𝑃(𝐴∩𝐵)/𝑃(𝐵)$ 𝑃(𝐴∩𝐵)=4/499⋅5/500=20/249500 𝑃(𝐴∣𝐵)=20/249500/0.01=0.008≠0.01=𝑃(𝐴) This shows that events 𝐴 and 𝐵 are not independent. b. Probabilities of A and B With sampling done with replacement, the probabilities are: 𝑃(𝐴)=5/500=0.01 𝑃(𝐵)=5/500=0.01 The number of favorable outcomes in both cases is 5 because sampling is done with replacement. Intersection of Events A and B The number of favorable outcomes in the intersection 𝐴∩𝐵 is:5×5=25 This is because we can select the first defective container in 5 ways and the second defective container in 5 ways (since sampling is done with replacement, there are still 5 defective containers each time). The number of outcomes in the sample space is: 500×500=250,000 So, the probability of the intersection is: 𝑃(𝐴∩𝐵)=5×5/500×500=25/250,000=0.0001 Conditional Probability P(A ∣ B) The conditional probability 𝑃(𝐴∣𝐵) is calculated as: $𝑃(𝐴∣𝐵)=𝑃(𝐴∩𝐵)/𝑃(𝐵)$=0.0001/0.01=0.01 Independence Check: We know that: 𝑃(𝐴)=0.01 Since: 𝑃(𝐴∣𝐵)=0.01=𝑃(𝐴) This means that events 𝐴 and 𝐵 are independent.