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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.

Book answer:

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

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"There are (exactly) 2 defective tires before the 4th good one" is the same thing as "of the first 5 tires, 2 are defective and 3 are good, and then the 6th tire is good". The $5 \choose 2$ is because any 2 of those 5 tires could be the defective ones.

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Your approach is correct (and the book's answer is wrong). Let $p$ denote the probability that a tire is defective. The probability to find $2$ defective tires before $4$ good ones is $$ s=p^2+2p^2(1-p)+3p^2(1-p)^2+4p^2(1-p)^3. $$ Plugging $p=\frac1{20}$ into this yields $s=0.0225925$, that is, approximately $2.26\%$.

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