I am revising and came upon these questions, but I am not sure if the answers are right especially for part B
Suppose the probability of defective is p(D) = 0.009 , p(d^c)=0.991 and mean 7 and std = 1.5
A) Suppose you select one battery at a time from the manufacturer, what is the probability that the first defective batery is found in the fourth selection?
Answer: 3(0.0099) * (0.991)^3 = 0.026
B)Suppose you select one battery at a time from the manufacturer, what is the probability that the third defective battery is found in the tenth selection?
Answer: will be 9(0.0099)^3 * (0.991)^7 =0.000
Let $p = \text{Pr}(D)$.
A) First 3 should not be defective and fourth should be defective, by product rule the probability of this happening is $(1-p)^{3} p$.
B) First there should be 2 out of 9 defective which can happen in ${9}\choose{2}$ ways and 10th one defective, so probability is ${9}\choose{2}$ $p^{3} (1-p)^{7}$.