We model this situation using a binomial distribution with parameters: number of trials \(n = 10\) and probability of success \(p = 0.35\).

The probability that Janice buys at most 7 items is given by:

\(P(X \leq 7) = 1 - P(X = 8) - P(X = 9) - P(X = 10)\)

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

\(P(X = 8) = \binom{10}{8} (0.35)^8 (0.65)^2\)

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

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

Substituting these into the equation:

\(P(X \leq 7) = 1 - (0.004281 + 0.0005123 + 0.00002759)\)

\(P(X \leq 7) = 1 - 0.00482089\)

\(P(X \leq 7) = 0.995\)