Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.9\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 10) = \binom{12}{10} (0.9)^{10} (0.1)^2 = 0.230128\)

\(P(X = 11) = \binom{12}{11} (0.9)^{11} (0.1)^1 = 0.376573\)

\(P(X = 12) = \binom{12}{12} (0.9)^{12} (0.1)^0 = 0.282430\)

\(P(X \geq 10) = 0.230128 + 0.376573 + 0.282430 = 0.889131\)

Thus, \(P(X < 10) = 1 - 0.889131 = 0.110869\).

Therefore, the probability that fewer than 10 callers are connected immediately is approximately 0.111.