Find an estimator for the probability of a given random variable 
Let $X$ be a random variable with a Binomial distribution $X \sim B(100,p)$. We want to find the probability $P\{X<4\}$ by estimating the parameter $p$ using the method of moments and maximum likelihood method.

I'm stuck. All I could do was conclude that using the method of moments, an estimator for $p$ would be $\bar X/100$. What am I supposed to do with this information and how can I use the ML method for the binomial distribution?
 A: If I understand correctly, the sample size is $1$.  The first sample moment is the mean of the observations in the sample, and since there is only one such observation, it is just the value of that observation, $X$.  The first population moment for $B(100,p)$ is $100p$.  In the method of moments one then equates the two and solves for $p$, thus:
$$
X=100p
$$
$$
p = \frac X {100}.
$$
That is the estimate of $p$ by the method of moments.  The likelihood function is
$$
L(p) = \binom {100} X p^X (1-p)^{100-X}.
$$
Finding the value of $p$ that maximizes this is the same as finding the value of $p$ that maximizes its logarithm because the logarithm function is increasing.  The logarithm is
$$
\ell(p) = \log\binom{100} X + \left( X\log p + (100-X)\log(1-p) \right).
$$
Then we have
$$
\ell\,'(p) = \frac X p - \frac{100-X}{1-p} = \frac{-100\left( p - \dfrac X {100} \right)}{p(1-p)} \quad \begin{cases} \ge 0 & \text{if }0<p\le 100/X, \\[8pt]  \le 0 & \text{if }100/X \le p < 1. \end{cases}
$$
Hence the maximizing value of $p$ is $\dfrac X {100}$.
Both methods yield the same estimate in this case.
If you use that estimate in place of $p$ in $\Pr(X<4)$ then you get
\begin{align}
& \Pr(X\in\{0,1,2,3\}) \\[10pt]
= {} & \binom {100} 0 \hat p^0(1-\hat p)^{100} + \binom{100} 1 \hat p^1(1-\hat p)^{99} + \binom{100} 2 p^2 (1-\hat p)^{98} + \binom{100} 3 \hat p^3(1-\hat p)^{97}
\end{align}
where $\hat p = X/100$ is the estimate.
