The question:
In order to determine the probability of success of a certain experiment, each of one hundred persons perform a series of these experiments until one results in success. A model to describe this: the number of attempts $N$ required until the first success follows a $Geo(p)$-distribution. We wish to estimate $p$ using the maximum likelihood method.
The data:
My reasoning:
Let $X_i$ with $i=1, ..., 100$ denote the number of attempts required until the first succes for person $i$. We now consider the follwing probability mass function
$$ P(X_1=x_1, ..., X_{100}=x_{100})=\prod_{i=1}^{100}P(X_i=x_i)=p^{100}\prod_{i=1}^{100}(1-p)^{x_i-1} \\ =p^{100}(1-p)^{30}[(1-p)^{2}]^{15}[(1-p)^{\sum_{i'}x_{i'}-1}]$$
where $x_{i'}\geq4$ and $i'$ the index for the corresponding group of people. The possible answers are
So the only possible answer would be d, is my reasoning correct?
No, your reasoning is not correct.
Recall that the PMF for a geometric distribution with probability $p$ of success that counts the total trials needed to observe the first success, is given by $$\Pr[X = x] = (1-p)^{x-1} p, \quad x \in \{1, 2, \ldots \}.$$ From this we conclude that $$\Pr[X \ge x] = \sum_{k=x}^\infty \Pr[X = x] = p (1-p)^{x-1} \sum_{k=0}^\infty (1-p)^k = p (1-p)^{x-1} \frac{1}{1-(1-p)} = (1-p)^{x-1}.$$ We can also see this intuitively since it takes at least $x$ tries to obtain the first success if and only if the first $x-1$ tries are all failures.
Therefore, the likelihood for $p$ given the sample is $$\begin{align} \mathcal L(p \mid \boldsymbol x) &= \Pr[X = 1]^{40} \Pr[X = 2]^{30} \Pr[X = 3]^{15} \Pr[X \ge 4]^{15} \\ &= ((1-p)^0 p)^{40}((1-p)^1 p)^{30} ((1-p)^2 p)^{15} ((1-p)^3)^{15} \\ &= (1-p)^{30 + 2(15) + 3(15)} p^{40 + 30 + 15} \\ &= p^{85} (1-p)^{105}. \end{align}$$ Your error lies in counting every person's experiment as contributing a factor of $p$ to the likelihood, when in fact only those people who reported an exact number of trials contribute $p$ to the likelihood. Those $15$ people who reported at least four trials were needed because their first three failed, do not contribute a factor of $p$. If instead they all reported exactly four trials were needed, then you would get $p^{100}(1-p)^{105}$ as you claim.