I have done this by the following way :

the probability of drawing a black ball is either 0.25 or 0.75.

If n balls are drawn from the urn, the distribution of X, the number of black balls, is given by the *Binomial Distribution*

$f(x;p)=\binom{n}{x}p^x(1-p)^{n-x}\quad ; x=0,1,2,...,n$

where $p$ is the probability of drawing a black ball. Here $p=0.25$ or $0.75$

We shall draw a sample of three balls, that is, n=3, with replacement and attempt to estimate the unknown parameter $p$ of the distribution.

Let us anticipate the results of the drawing of the sample. The possible outcomes and their probabilities are given below :

$$

\begin{array}{c|lcr}

Outcome:x & 0 & 1 & 2 & 3 \\

\hline

f(x;\frac{3}{4}) & \frac{1}{64} & \frac{9}{64} & \frac{27}{64} & \frac{27}{64} \\

f(x;\frac{1}{4}) & \frac{27}{64} & \frac{27}{64} & \frac{9}{64} & \frac{1}{64} \\

\end{array}

$$

In the present example if we found x=0 or 1 in a sample of 3, the estimate for $p$ would be preferred over 0.75.

The estimator may be defined as :

$\hat p$ =

\begin{cases}

0.25, &\text{for $x=0,1$} \\

0.75, &\text{for $x=2,3$} \\

\end{cases}

I want to estimate $p$ for the above situation in R. I tried to do that in the following procedure which seems totally incorrect to me because if several alternative values of $p$ were possible, i can't proceed in the same manner. But i have no more idea also :

x=2

ifelse(x==0 & 1, "hat_p=0.25", "hat_p=0.75")

x=0

ifelse(x==0 & 1, "hat_p=0.25", "hat_p=0.75")

*"How can i compute the above situation in R?"*