 Solution

Assume each sampled response is truthful, card selection is independent of the student's status, and responses follow the usual independent sampling model. In randomized response, a yes answer occurs either for a user receiving the first question or for a nonuser receiving the complementary question. Its probability isConsequently $Y\sim \mathrm{Bin}(n,q)$, and the likelihood isFor $x\ne 1/2$, first maximize the binomial likelihood in the unrestricted probability $q$: its maximum occurs at $Y/n$. Inverting the affine relation givesFor an interior value $0<Y<n$, the same result follows from the scoreThe endpoint cases follow by monotonicity. This estimator has expected value $p$ and varianceThese identities follow immediately from $\mathbb{E}Y=nq$ and $\mathrm{Var}(Y)=nq(1-q)$.

The printed claim needs a parameter-space qualification. The admissible yes probabilities lie in $[\min (x,1-x),\max (x,1-x)]$, which may exclude $Y/n$. The constrained maximum-likelihood estimator isThe displayed untrimmed estimator is the maximum-likelihood estimator only when it already lies in $[0,1]$. Its unbiased estimator property does not carry over to the clipped estimator in general. For example, if $n=1$, $x=3/4$ and $p=0$, the clipped estimate equals 1 exactly when the answer is yes, so its expectation is $1/4$, not zero. At $x=1/2$, $q=1/2$ for every $p$: the parameter is unidentifiable and the printed quotient is undefined. Thus the unbiased formula requires $x\ne 1/2$, and the constrained likelihood assertion requires the stated clipping.