Let the probability that a person does not support either choir be denoted by \(p\). Given that 30% support Notes and 45% support Classics, the probability of not supporting either is:

\(p = 1 - 0.30 - 0.45 = 0.25\)

The mean number of people in the sample who do not support either choir is:

\(\text{Mean} = 240 \times 0.25 = 60\)

The variance is:

\(\text{Variance} = 240 \times 0.25 \times 0.75 = 45\)

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{45}\)

We apply the continuity correction for \(X < 50\):

\(P(X < 50) = P\left( Z < \frac{49.5 - 60}{\sqrt{45}} \right) = P(Z < -1.565)\)

Using standard normal distribution tables, \(P(Z < -1.565)\) is approximately:

\(1 - 0.9412 = 0.0588\)

Thus, the probability that fewer than 50 do not support either choir is \(0.0588\).