Let the probability that a household has no broadband service be 0.35. We are choosing 10 households, so this is a binomial distribution with parameters \(n = 10\) and \(p = 0.35\).

We need to find the probability that fewer than 3 households have no broadband service, i.e., \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.35)^0 (0.65)^{10} = 1 \times 1 \times 0.013463 = 0.013463\)

\(P(X = 1) = \binom{10}{1} (0.35)^1 (0.65)^9 = 10 \times 0.35 \times 0.072492 = 0.072492\)

\(P(X = 2) = \binom{10}{2} (0.35)^2 (0.65)^8 = 45 \times 0.1225 \times 0.178565 = 0.17565\)

Add these probabilities:

\(P(X < 3) = 0.013463 + 0.072492 + 0.17565 = 0.261605\)

Rounding to three decimal places, the probability is \(0.262\).