Say you have a manufacturer who manufactures a product and the process historically averages 5% defective products.
Now suppose the products are randomly selected and inspected for defects one after another. What is the probability that the 3rd defective product will be detected at the 7th inspection (Answer: 0.001527)
I think it needs a binomial distribution but I don't know how to handle a case where they are chosen one after another and the 3rd defective product is detected exactly at the 7th inspection.
The binomial distribution applies to a finite and fixed number of independent Bernoulli trials, each with the same probability of "success" $p$. Under these criteria, a binomially distributed random variable $X$ then counts the total number of successes observed out of $n$ trials. As such, it does not model your stated event.
What you want is a negative binomial random variable, one parametrization of which counts the total number of trials needed to obtain $r$ successes. You can see how this differs from the regular binomial scenario, which counts the successes when the number of trials is fixed: in the negative binomial case, we count the total number of trials when the number of successes is fixed.
In your case, the number of successes is fixed at $r = 3$, and we wish to find the probability that it takes $X = 7$ trials to observe three successes (where "success" is defined here as observing a defective product). The probability distribution for $X$ is given by $$\Pr[X = x] = \binom{x - 1}{x - r} p^r (1-p)^{x-r}, \quad x = r, r+1, r+2, \ldots.$$ So for $x = 7$, $r = 3$, $p = 0.05$, you will get your desired probability of observing the third defect on the seventh trial.
Where does this formula come from? We reason as follows. Clearly, in order for the $r^{\rm th}$ success to occur on the $x^{\rm th}$ trial, the last trial must be successful and that's when we stop sampling. Now there are $r-1$ successes to be distributed among the $x-1$ trials and there are exactly $\dbinom{x-1}{r-1} = \dbinom{x-1}{x-r}$ ways to do this. There are $r$ successes that occur with probability $p$ and $x-r$ failures that occur with probability $1-p$, so the probability of the desired outcome is what is claimed above.