First, use the fact that the sum of probabilities must equal 1:

\(0.12 + p + q + 0.16 + 0.3 = 1\)

Simplifying gives:

\(p + q = 0.42\)

Next, use the given expected value \(E(X) = 0.28\):

\(-2(0.12) - 1(p) + 0.5(q) + 1(0.16) + 2(0.3) = 0.28\)

This simplifies to:

\(-0.24 - p + 0.5q + 0.16 + 0.6 = 0.28\)

\(-p + 0.5q = -0.24\)

Now solve the system of equations:

1. \(p + q = 0.42\)
2. \(-p + 0.5q = -0.24\)

Add the two equations to eliminate \(p\):

\((p + q) + (-p + 0.5q) = 0.42 - 0.24\)

\(1.5q = 0.18\)

\(q = 0.12\)

Substitute \(q = 0.12\) back into \(p + q = 0.42\):

\(p + 0.12 = 0.42\)

\(p = 0.3\)