Find the probability that you find 2 defective tires before 4 good ones. There is a chance of a tire being defective at a rate of 5%.
From my understanding with the negative binomial distribution we want to repeat the trial until (including) r successes are achieved. Each trial until we have achieved r successes can be composed of either a failure or up to r-1 successes.
P(X=?)=((4+2 - 1) choose (2))*(.95)^4 * (.05)^2
I don't really understand where the 4+2 -1 comes from and why is there no 2 - 1 to choose from instead of 2? If we want to find the probability of getting 2 defective tires before 4 good ones shouldn't we find the sum of the probability of getting both defective tires by the second pick, third pick, fourth, fifth pick since there can only be 3 good tires and 2 bad ones so as an example: GGGBB