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+#import "@preview/catppuccin:1.1.0": catppuccin, flavors
+#show: catppuccin.with(flavors.mocha)
+
+#set text(size: 16pt)
+
+== task 3
+
+the following article explains an algorithm that solves our problem:
+#link("https://en.wikipedia.org/wiki/Reservoir_sampling#Simple:_Algorithm_R")
+
+the main idea is to keep a reservoir of $k$ items, deciding randomly if the $i$-th item you revealed should be added to the reservoir (replacing any item already in there at that slot), otherwise discarding the item.
+
+the random variable determining the inclusion of the item in the reservoir is given as a random integer between 1 and $i$; if it is less than or equal to $k$, it will fit in the reservoir and then occupies that drawn slot in the reservoir. if it is greater than $k$, it will be discarded.
+
+this ensures a sampling with a uniform probability distribution of $n$
+
+proof: since an item $i$ is included in the reservoir with a probability of $k/i$, the next item $i + 1$ has probability $k/(i + 1)$ of being included in the reservoir. but since processing the $(i + 1)$-th input affects the probability of the previous elements, the probability of $i$ is now $k/i times (1 - 1/(i + 1)) = k/(i + 1)$. but then it isn't actually the case that the probability is different for the previous elements, i.e. the sampling was uniform and independent. 
+
+as the wiki mentions, the algorithm is needlessly slow, but plenty apt for a simple solution to our problem.