5. Solution. Let the probabilities of heads and tails be $p$ and $q=1-p$ respectively. We form the equation

$$
C_{10}^{7} p^{7} q^{3}=C_{10}^{6} p^{6} q^{4}
$$

from which it follows: $120 p=210 q ; \frac{p}{q}=\frac{7}{4} ; p=\frac{7}{11}$.
Answer: $\frac{7}{11}$.

Grading Criteria

| Correct and justified solution | 2 points |
| :--- | :---: |
| Correct equation is set up, but an error is made or the solution is not completed | 1 point |
| Solution is incorrect or missing (including only providing the answer) | 0 points |