I have a couple of similar questions, which I think I know how to solve (but wish to see if I am right), and suspect can be solved using a Negative Binomial distribution, but not sure if this is the case and how to define the distribution and it's parameters.
We toss a dies until it falls on "6" three times. What is the probability we will need to toss it: a) 4 times b) 5 times
Somebody shoots a target until he hits it 4 times. The probability of hitting the target in a single shot is 0.7. What is the probability that he will need: a) 5 shots b) 6 shots
What I though to do, was (if to take the first question as an example), to say that needing 4 tosses means that I didn't get "6" either in my first toss, second, or third. So the probability is: 3*(5/6)(1/6)(1/6)*(1/6). Am I correct ?