Let the probability of landing on the blue side be \(p\). Then, the probability of landing on the red side is \(4p\), and the probability of landing on the green side is \(3p\).

Since the total probability must be 1, we have:

\(p + 4p + 3p = 1\)

\(8p = 1\)

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

Thus, the probability of landing on the blue side is \(\frac{1}{8}\).

For a binomial distribution with \(n = 136\) and \(p = \frac{1}{8}\), the mean \(\mu\) is:

\(\mu = np = 136 \times \frac{1}{8} = 17\)

The variance \(\sigma^2\) is:

\(\sigma^2 = np(1-p) = 136 \times \frac{1}{8} \times \frac{7}{8} = 14.875\)

Using a normal approximation, we standardize for \(P(X < 20)\) with continuity correction:

\(P(X < 20) = P\left( Z < \frac{19.5 - 17}{\sqrt{14.875}} \right)\)

\(= P(Z < 0.648)\)

Using standard normal distribution tables, \(\Phi(0.648) = 0.742\).

Therefore, the probability that the spinner lands on the blue side fewer than 20 times is 0.742.