Let the random variable \(X\) represent the number of people who choose a mobile phone made by Company A. \(X\) follows a binomial distribution with parameters \(n = 13\) and \(p = 0.6\).

We need to find \(P(X < 11)\), which is equivalent to \(1 - P(X \geq 11)\).

Using the complement rule, \(P(X \geq 11) = P(X = 11) + P(X = 12) + P(X = 13)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 11) = \binom{13}{11} (0.6)^{11} (0.4)^2\)

\(P(X = 12) = \binom{13}{12} (0.6)^{12} (0.4)^1\)

\(P(X = 13) = \binom{13}{13} (0.6)^{13} (0.4)^0\)

Thus, \(P(X \geq 11) = \binom{13}{11} (0.6)^{11} (0.4)^2 + \binom{13}{12} (0.6)^{12} (0.4)^1 + \binom{13}{13} (0.6)^{13}\).

Calculate \(P(X < 11) = 1 - P(X \geq 11)\).

\(P(X < 11) = 1 - (\binom{13}{11} (0.6)^{11} (0.4)^2 + \binom{13}{12} (0.6)^{12} (0.4)^1 + \binom{13}{13} (0.6)^{13})\)

\(P(X < 11) = 0.942\).

Let S be the event that a vehicle goes straight, L be the event that a vehicle turns left, and R be the event that a vehicle turns right.

The probability that one vehicle goes straight and the other two turn left is given by:

\(P(SLL) = (0.3)(0.55)(0.55) = 0.09075\)

The probability that one vehicle goes straight and the other two turn right is given by:

\(P(SRR) = (0.3)(0.15)(0.15) = 0.00675\)

There are 3 ways to arrange SLL (SSL, SLL, LSL) and 3 ways to arrange SRR (SSR, SRR, RSR), so the total probability is:

\(\binom{3}{1} \times P(SLL) + \binom{3}{1} \times P(SRR) = 3 \times 0.09075 + 3 \times 0.00675 = 0.27225 + 0.02025 = 0.2925\)

Thus, the probability is approximately 0.293.

Let the probability of a family owning a dishwasher be \(p = 0.22\). The number of families is \(n = 15\).

We need to find \(P(4 \, \text{to} \, 6)\), which is the probability that between 4 and 6 families own a dishwasher.

Using the binomial probability formula:

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

Calculate for \(k = 4, 5, 6\):

\(P(4) = \binom{15}{4} (0.22)^4 (0.78)^{11}\)

\(P(5) = \binom{15}{5} (0.22)^5 (0.78)^{10}\)

\(P(6) = \binom{15}{6} (0.22)^6 (0.78)^9\)

Sum these probabilities:

\(P(4, 5, 6) = \binom{15}{4} (0.22)^4 (0.78)^{11} + \binom{15}{5} (0.22)^5 (0.78)^{10} + \binom{15}{6} (0.22)^6 (0.78)^9\)

\(= 0.398\)

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 \(0.111\).

Let \(X\) be the number of students who own a car out of the five chosen. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = \frac{1}{4}\).

We need to find \(P(X \geq 4) = P(X = 4) + P(X = 5)\).

Using the binomial probability formula:

\(P(X = 4) = \binom{5}{4} \left( \frac{1}{4} \right)^4 \left( \frac{3}{4} \right)^1\)

\(P(X = 5) = \binom{5}{5} \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0\)

Calculating each term:

\(P(X = 4) = 5 \times \left( \frac{1}{4} \right)^4 \times \frac{3}{4} = 0.014648\)

\(P(X = 5) = 1 \times \left( \frac{1}{4} \right)^5 = 0.00097656\)

Adding these probabilities gives:

\(P(X \geq 4) = 0.014648 + 0.00097656 = 0.0156\)

Thus, the probability that at least four of these students own a car is \(0.0156\) or \(\frac{1}{64}\).