Let the random variable \(X\) represent the number of adults who travel to work by car in a sample of 12. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.6\).

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

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

Using the binomial probability formula:

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

\(P(X = 10) = \binom{12}{10} (0.6)^{10} (0.4)^2 = 0.063852\)

\(P(X = 11) = \binom{12}{11} (0.6)^{11} (0.4)^1 = 0.017414\)

\(P(X = 12) = \binom{12}{12} (0.6)^{12} (0.4)^0 = 0.0021768\)

Thus, \(P(X \geq 10) = 0.063852 + 0.017414 + 0.0021768 = 0.083443\).

Therefore, \(P(X < 10) = 1 - 0.083443 = 0.917\).

The problem involves a binomial distribution where the probability of obtaining a 5 on a single spin is \(p = \frac{1}{5} = 0.2\), and the probability of not obtaining a 5 is \(1 - p = 0.8\). George spins the spinner 10 times, so \(n = 10\).

We need to find the probability of obtaining a 5 more than 4 times but fewer than 8 times, i.e., \(P(5 \leq X \leq 7)\).

Using the binomial probability formula:

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

Calculate \(P(X = 5)\), \(P(X = 6)\), and \(P(X = 7)\):

\(P(X = 5) = \binom{10}{5} (0.2)^5 (0.8)^5 = 0.02642\)

\(P(X = 6) = \binom{10}{6} (0.2)^6 (0.8)^4 = 0.005505\)

\(P(X = 7) = \binom{10}{7} (0.2)^7 (0.8)^3 = 0.0007864\)

Sum these probabilities:

\(P(5 \leq X \leq 7) = 0.02642 + 0.005505 + 0.0007864 = 0.0327\)

Let the probability that a student does not like any sports be \(p = 0.3\). The number of students chosen is \(n = 10\). We need to find the probability that at least 3 students do not like any sports, which is \(P(X \geq 3)\).

This can be calculated as \(1 - P(X < 3)\), where \(X\) is a binomial random variable with parameters \(n = 10\) and \(p = 0.3\).

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

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

\(P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} = 0.028248\)

\(P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 = 0.121061\)

\(P(X = 2) = \binom{10}{2} (0.3)^2 (0.7)^8 = 0.233474\)

Thus, \(P(X < 3) = 0.028248 + 0.121061 + 0.233474 = 0.382985\).

Therefore, \(P(X \geq 3) = 1 - 0.382985 = 0.617015\).

Rounding to three decimal places, the probability is \(0.617\).

Let the random variable \(X\) represent the number of days the train is late out of 7 days. \(X\) follows a binomial distribution with parameters \(n = 7\) and \(p = 0.35\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 0) = \binom{7}{0} (0.35)^0 (0.65)^7 = 0.65^7 = 0.049022\)

\(P(X = 1) = \binom{7}{1} (0.35)^1 (0.65)^6 = 7 \times 0.35 \times 0.65^6 = 0.184776\)

\(P(X = 2) = \binom{7}{2} (0.35)^2 (0.65)^5 = 21 \times 0.35^2 \times 0.65^5 = 0.29848\)

Sum these probabilities:

\(P(X < 3) = 0.049022 + 0.184776 + 0.29848 = 0.532\)

(a) The probability of not getting a 4 in a single throw is \(\frac{5}{6}\). The probability of getting a 4 for the first time on the 6th throw is \(\left( \frac{5}{6} \right)^5 \times \frac{1}{6}\). Therefore, the probability of obtaining a 4 in fewer than 6 throws is:

\(1 - \left( \frac{5}{6} \right)^5\)

Calculating this gives:

\(1 - \left( \frac{5}{6} \right)^5 = 1 - \frac{3125}{7776} = \frac{4651}{7776}\)

(b) The probability of getting a 4 in a single throw is \(\frac{1}{6}\). We use the binomial distribution to find the probability of getting at least 3 fours in 10 throws:

\(1 - P(0, 1, 2) = 1 - \left( \binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} + \binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 + \binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 \right)\)

Calculating each term:

\(\binom{10}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{10} = 0.1615056\)

\(\binom{10}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^9 = 0.3230111\)

\(\binom{10}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^8 = 0.290710\)

Summing these gives:

\(0.1615056 + 0.3230111 + 0.290710 = 0.7752267\)

Thus, the probability of getting at least 3 fours is:

\(1 - 0.7752267 = 0.2247733\)

Rounding to three decimal places gives \(0.225\).