When a Brilliant Idea comes to the Scattered Scholar, he writes it down on a piece of paper, but immediately realizes that the idea is not brilliant, crumples the paper, and throws it under the table, where there are two wastebaskets. The Scholar misses the first wastebasket with a probability of $p>0.5$, and with the same probability, he misses the second one. In the morning, the Scholar threw five crumpled brilliant ideas under the table. Find the probability that each wastebasket ended up with at least one of the morning ideas.
Let's introduce the notations for the events: $A=\{$ An idea is in the first basket $\}, B=\{$ An idea is in the second basket $\}$. We need to find the probability of the event $A \cap B$. From the condition, it follows that $\mathrm{P}(A)=\mathrm{P}(B)=1-p^{5}$. The probability that the Scientist will hit one specific basket in one throw is $1-p$. It is impossible to hit both at the same time, so the probability that the Scientist hit at least one basket is 2 - $2 p$. Therefore, the probability of not hitting any basket in one throw is $1-(2-2 p)=2 p-1$, and the probability of not hitting any basket in five throws is $(2 p-1)^{5}$. Hence,

$\mathrm{P}(A \cup B)=1-(2 p-1)^{5}$, and $\mathrm{P}(A \cap B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cup B)=2-2 p^{5}-1+(2 p-1)^{5}=1-2 p^{5}+(2 p-1)^{5}$.

## Answer

$1-2 p^{5}+(2 p-1)^{5}$

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