= Solution
For the same sample size $n$, the <geometric distribution> provides more <Fisher information>:
$$
\frac{I_n^{\rm geom}(\theta)}{I_n^{\rm Bern}(\theta)}=\frac1\theta>1
\qquad(0<\theta<1).
$$
The corresponding asymptotic variances of the <maximum-likelihood estimators> are
$$
\operatorname{Var}(\widehat\theta_{\rm geom})\approx\frac{\theta^2(1-\theta)}n,
\qquad
\operatorname{Var}(\widehat\theta_{\rm Bern})\approx\frac{\theta(1-\theta)}n.
$$
Hence \b[choose the geometric sample for more precise inference at equal numbers of observations]. Its asymptotic <variance> is $\theta$ times as large, and its <standard error> is $\sqrt\theta$ times as large. This illustrates <Fisher information per observation and per trial>. It is a per-observation comparison: one geometric observation records a waiting time containing, on average, $1/\theta$ Bernoulli trials. A constraint on the total number of underlying trials would be a different sampling-budget comparison.
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