 class="math inline">\(0.5^P\)</span>. Let <span class="math inline">\(K_+\)</span> and <span class="math inline">\(K_-\)</span> denote the number of positive and negative observations contained in a random draw. Since our observations are independently drawn, we see that for any pair of natural numbers <span class="math inline">\((k_+,k_-)\)</span>, <span class="math display">\[\begin{equation*}
\mathbb{P}(K_+=k_+, K_-=k_-) = {N\choose k_+}{N-k_+ \choose k_-} \left(\frac{1}{2^P}\right)^{k_+ + k_-}.
\end{eq