Every day, the dog Patrick gnaws one slipper from the available supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers need to be bought so that with a probability of at least 0.8, Patrick does not get upset for a whole week (7 days)?
Let in a week Patrick wants to eat $S$ left and $7-S$ right slippers. We need to find such $k$ that the inequality $\mathrm{P}(S \leq k \cap 7-S \leq k) \geq 0.8$ holds. Rewrite the event in parentheses: $P(7-k \leq S \leq k)$. It is clear that $7-k \leq k$, that is, $k \geq 4$.

The probability on the left side of the inequality is the sum

$C_{7}^{7-k} \frac{1}{2^{7}}+C_{7}^{8-k} \frac{1}{2^{7}}+\ldots+C_{7}^{k} \frac{1}{2^{7}}=\frac{1}{128} \cdot\left(C_{7}^{7-k}+C_{7}^{8-k}+\ldots+C_{7}^{k}\right)=\frac{1}{64} \cdot\left(C_{7}^{4}+C_{7}^{5}+\ldots+C_{7}^{k}\right)$. Therefore,

$C_{7}^{4}+C_{7}^{5}+\ldots+C_{7}^{k} \geq 64 \cdot 0.8=51.2$

Note that $C_{4}^{7}=35, \quad C_{5}^{7}=21$, so the sum $C_{7}^{4}+C_{7}^{5}=35+21>52$. Therefore, the smallest $k$ is 5.

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