Let's say we are throwing a ball until we hit a target exactly $3$ times. Hits are indipedent and the probability of hitting a target is $p=0.25$. What's the probabability of that we need exactly 20 trials to hit the target exactly $3$ times?
My teacher gave this hint: "What is happening in the first 19 tries". What i have thought taking into account this hint is calculating the binomial probability $b(19, 0.25)$: $$P(X = 2) = \binom {19}{2}(0.25)^2(0.75)^{17} = 0.08$$.
However, i also thought using negative binomial probability $Nb(3, 0.25)$: $$P(X = 20) = \binom {20-1}{3-1}(0.25)^3(0.75)^{17} = 0.02$$
What is correct? I guess the first one is and i also guess i have not fully comprehend Negative binom distribution.
$\binom {19}{2}(0.25)^2(0.75)^{17} = 0.08$ is the conditional probability, given that the 20th attempt was a hit, that it was the third hit
$\binom {20-1}{3-1}(0.25)^3(0.75)^{17} = 0.02$ is the overall probability that the 20th attempt was the third hit
It looks to me as if the question is asking for the second of these, which you can either state directly from the negative binomial, or by multiplying the first (ordinary binomial) by the probability that the 20th attempt was a hit, i.e. multiply the first probability by $0.25$