Suppose you have a box containing 10 balls, of which $\theta$ are white and the rest are green. Suppose we take two balls for without replacement and let $X_i = 1$ if the i-th drawn ball is white and $0$ otherwise. Find the maximum likelihood estimator for $\theta$
I was wondering how to solve this problem, if there were replacements in question I could do it easily, since then the $X_i$ would be independent. But in this case I'm not sure how to proceed, I even thought of considering the sum of both outcomes, but didn't come to any solution as well
\begin{align} P(X_1=1, X_2=1) &= \frac{\theta}{10} \cdot \frac{\theta-1}{9} \\ P(X_1=1, X_2=0) &= \frac{\theta}{10} \cdot \frac{10-\theta}{9} \\ P(X_1=0, X_2=1) &= \frac{10-\theta}{10} \cdot \frac{\theta}{9} \\ P(X_1=0, X_2=0) &= \frac{10-\theta}{10} \cdot \frac{9-\theta}{9} \end{align}
This can be written in one expression as $$P(X_1=x_1, X_2 =x_2) = \frac{1}{90} [\theta(\theta-1)]^{x_1 x_2} [\theta(10-\theta)]^{x_1 (1-x_2)} [(10-\theta)\theta]^{(1-x_1)x_2} [(10-\theta)(9-\theta)]^{(1-x_1)(1-x_2)}. $$ I think you can then take the logarithm of this likelihood and then optimize for $\theta$.