489. Find the point estimate of the unknown parameter $p$ (the probability of an event occurring in a single trial) of the binomial distribution using the method of maximum likelihood:

$$
P_{m}\left(x_{i}\right)=C_{m}^{x_{i}} p^{x_{i}}(1-p)^{m-x_{i}}
$$

where $x_{i}$ is the number of occurrences of the event in the $i$-th experiment, $m$ is the number of trials in one experiment, and $n$ is the number of experiments.
Solution. Let's construct the likelihood function:

$$
L=p\left(x_{1} ; \theta\right) \cdot p\left(x_{2} ; \theta\right) \ldots p\left(x_{n} ; \theta\right)
$$

Considering that $\Theta=p$ and $P\left(X=x_{i}\right)=C_{m}^{x_{i}} p^{x_{i}}(1-p)^{m-x_{i}}$, we get

$L=\left[C_{m}^{x_{1}} p^{x_{1}}(1-p)^{m-x_{1}}\right] \cdot\left[C_{m}^{x_{2}} p^{x_{2}}(1-p)^{m-x_{2}}\right] \ldots\left[C_{m}^{x_{n}} p^{x_{n}}(1-p)^{m-x_{n}}\right]$, or

$$
L=\left(C_{m}^{x_{1}} C_{m}^{x_{2}} \ldots C_{m}^{x_{n}}\right) \cdot p^{x_{1}+x_{2}+\ldots+x_{n}} \cdot(1-p)^{n m-\left(x_{1}+x_{2}+\ldots+x_{n}\right)}
$$

Let's write the logarithmic likelihood function:

$\ln L=\ln \left[C_{m}^{x_{1}} C_{m}^{x_{2}} \ldots C_{m}^{x_{n}}\right]+\left(\sum x_{i}\right) \ln p+\cdot\left(n m-\sum x_{i}\right) \ln (1-p)$.

Let's find the first derivative with respect to $p$:

$$
\frac{\mathrm{d} \ln L}{\mathrm{~d} p}=\frac{\sum x_{i}}{p}-\left(n m-\sum x_{i}\right) \frac{1}{1-p}
$$

Setting the first derivative to zero and solving the resulting equation, we get the critical point

$$
\rho=\left(\sum x_{i}\right) /(n m)
$$

Let's find the second derivative with respect to $p$:

$$
\frac{\mathrm{d}^{2} \ln L}{\mathrm{~d} p^{2}}=-\frac{\sum x_{i}}{p^{2}}+\frac{n m-\sum x_{i}}{(1-p)^{2}}
$$

It is easy to verify that at $p=\left(\sum x_{i}\right) /(n m)$, the second derivative is negative; therefore, this point is a maximum and should be accepted as the maximum likelihood estimate of the unknown probability $p$ of the binomial distribution:

$$
\rho^{*}=\left(\sum x_{i}\right) /(n m)
$$

Obviously, if $x_{i}$ occurrences of the event were observed in $n_{i}$ trials, then

$$
p^{*}=\left(\sum n_{i} x_{i}\right) /(n m)
$$