To find \(r\), note that the only way to get a score of 6 is by landing on 3 on both spinners, so \(P(\text{score is } 6) = r^2 = \frac{1}{36}\). Solving for \(r\), we get \(r = \frac{1}{6}\).

For a score of 5, the possible outcomes are (2, 3) and (3, 2). Thus, \(P(\text{score is } 5) = P(2, 3) + P(3, 2) = qr + rq = 2qr = \frac{1}{9}\). Substituting \(r = \frac{1}{6}\), we have:

\(2q \times \frac{1}{6} = \frac{1}{9}\)

\(\frac{q}{3} = \frac{1}{9}\)

\(q = \frac{1}{3}\)

Since \(p + q + r = 1\), we have:

\(p + \frac{1}{3} + \frac{1}{6} = 1\)

\(p + \frac{1}{2} = 1\)

\(p = \frac{1}{2}\)