9. Solution. Let's introduce the notations for the events:

$A=\{$ the first basket has an idea $\}, B=\{$ the second basket has an idea $\}$.

We need to find the probability of the event $A \cap B: \mathrm{P}(A \cap B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A \cup \dot{\boldsymbol{\phi}}$. From the condition, we find: $\mathrm{P}(A)=\mathrm{P}(B)=1-p^{5}$. The probability that the Scientist will hit one specific basket in one throw is $1-p$. It is impossible to hit both baskets at once, so the probability that the Scientist hits 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 $(2 p-1)^{5}$. Thus,

$$
\mathrm{P}(A \cup B)=1-(2 p-1)^{5}
$$

We get:

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

| Evaluation Criterion | Score |
| :--- | :---: |
| Solution is complete and correct | 3 |
| Solution is generally correct, but an error has been made in transformations or in the transition to the probability of the opposite event. Possibly, an incorrect 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 |

Variant 2

## Short Answer Questions

| Question | Answer | Question | Answer |
| :---: | :---: | :---: | :---: |
| 1 | $15 \%$ | 4 | 3 |
| 2 | $3340 g$ | 5 | $A$ |
| 3 | 7 | 6 | 0.5 |

Note. Questions 1 - 6 are considered completed if the correct answer is given. In question 2, the answer can be given in any units; for example, 3.34 kg.

## Extended Answer Questions