9. Solution. 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 $A \cap B: \quad \mathrm{P}(A \cap B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cup B)$. The probability that the Scientist will not hit a specific basket in one throw is $1-p$. Therefore, the probability that he will not hit it in six throws is $(1-p)^{6}$. Hence, $\mathrm{P}(A)=\mathrm{P}(B)=1-(1-p)^{6}$. It is impossible to hit both baskets at the same time, so the probability that the Scientist hits at least one basket is $2 p$. Therefore, the probability of not hitting any basket in one throw is $1-2 p$, and the probability of not hitting any basket in six throws is $(1-2 p)^{6}$. Consequently, $\mathrm{P}(A \cup B)=1-(1-2 p)^{6}$.

We get:

$$
\mathrm{P}(A \cap B)=2-2(1-p)^{6}-1+(1-2 p)^{6}=1-2(1-p)^{6}+(1-2 p)^{6}
$$
Answer: $1-2(1-p)^{6}+(1-2 p)^{6}$.

| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Solution is basically correct, but there is an error in transformations or in the transition to the probability of the opposite event. Possibly, the wrong answer | 2 |
| The formula for the probability of the intersection of events is clearly written, further steps are missing, insignificant or erroneous | 1 |
| Solution is missing or incorrect | 0 |