The sum of all probabilities must equal 1:

\(p + p + 0.1 + q + q = 1\)

Simplifying, we get:

\(2p + 0.1 + 2q = 1\)

Given that \(P(X \geq 0) = 3P(X < 0)\), we have:

\(0.1 + 2q = 3(2p)\)

Now, solve the system of equations:

1. \(2p + 0.1 + 2q = 1\)
2. \(0.1 + 2q = 6p\)

From equation (2), express \(q\) in terms of \(p\):

\(2q = 6p - 0.1\)

\(q = 3p - 0.05\)

Substitute \(q = 3p - 0.05\) into equation (1):

\(2p + 0.1 + 2(3p - 0.05) = 1\)

\(2p + 0.1 + 6p - 0.1 = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Substitute \(p = \frac{1}{8}\) back to find \(q\):

\(q = 3 \times \frac{1}{8} - 0.05\)

\(q = \frac{3}{8} - \frac{1}{20}\)

\(q = \frac{15}{40} - \frac{2}{40}\)

\(q = \frac{13}{40}\)