Let the probability of selling less than 16 kg of grapes on any day be denoted by \(p = 0.1\). We are interested in finding the probability that less than 16 kg of grapes are sold on more than 2 out of 12 days. This is equivalent to finding \(1 - P(0, 1, 2)\), where \(P(0, 1, 2)\) is the probability of selling less than 16 kg on 0, 1, or 2 days.
Using the binomial distribution formula, we have:
\(P(0, 1, 2) = \binom{12}{0} (0.1)^0 (0.9)^{12} + \binom{12}{1} (0.1)^1 (0.9)^{11} + \binom{12}{2} (0.1)^2 (0.9)^{10}\)
Calculating each term:
\(\binom{12}{0} (0.1)^0 (0.9)^{12} = 1 \times 1 \times 0.2824 = 0.2824\)
\(\binom{12}{1} (0.1)^1 (0.9)^{11} = 12 \times 0.1 \times 0.3138 = 0.3766\)
\(\binom{12}{2} (0.1)^2 (0.9)^{10} = 66 \times 0.01 \times 0.3487 = 0.2301\)
Thus, \(P(0, 1, 2) = 0.2824 + 0.3766 + 0.2301 = 0.8891\).
Therefore, the probability that less than 16 kg of grapes are sold on more than 2 days is:
\(1 - 0.8891 = 0.1109\)
Rounding to three significant figures, the probability is \(0.111\).
