Let the probability that a customer shops online be \(p = 0.35\). The number of customers chosen is \(n = 6\). We need to find the probability that more than three customers shop online, i.e., \(P(X > 3)\) where \(X\) is a binomial random variable with parameters \(n = 6\) and \(p = 0.35\).

We calculate \(P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6)\).

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 4) = \binom{6}{4} (0.35)^4 (0.65)^2\)

\(P(X = 5) = \binom{6}{5} (0.35)^5 (0.65)^1\)

\(P(X = 6) = \binom{6}{6} (0.35)^6 (0.65)^0\)

Calculating each term:

\(P(X = 4) = 15 \times 0.35^4 \times 0.65^2\)

\(P(X = 5) = 6 \times 0.35^5 \times 0.65\)

\(P(X = 6) = 1 \times 0.35^6\)

Summing these probabilities gives \(P(X > 3) = 0.117\).