(a) The probability that Kayla takes more than 6 throws to achieve a success is given by the geometric distribution formula:

\(P(X > 6) = (0.75)^6\)

Calculating this gives:

\(P(X > 6) = 0.178\)

(b) To find the probability that Kayla achieves at least 3 successes in 10 throws, we use the binomial distribution:

\(1 - P(0, 1, 2) = 1 - \left( (0.75)^{10} + \binom{10}{1} (0.25)^1 (0.75)^9 + \binom{10}{2} (0.25)^2 (0.75)^8 \right)\)

Calculating each term:

\((0.75)^{10} = 0.0563135\)

\(\binom{10}{1} (0.25)^1 (0.75)^9 = 0.1877117\)

\(\binom{10}{2} (0.25)^2 (0.75)^8 = 0.2815676\)

Summing these gives:

\(0.0563135 + 0.1877117 + 0.2815676 = 0.5256\)

Thus,

\(1 - 0.5256 = 0.474\)

(a) The probability that a student owns a car is 0.22. The probability that all 3 students own a car is given by:

\(0.22^3 = 0.0106\)

(b) We use the binomial probability formula to find the probability that the number of students who own a car is at least 2 and at most 4 out of 16 students. The probability that a student does not own a car is 0.78.

The probability is given by:

\(P(2, 3, 4) = \binom{16}{2} (0.22)^2 (0.78)^{14} + \binom{16}{3} (0.22)^3 (0.78)^{13} + \binom{16}{4} (0.22)^4 (0.78)^{12}\)

Calculating each term:

\(\binom{16}{2} (0.22)^2 (0.78)^{14} = 0.179205\)

\(\binom{16}{3} (0.22)^3 (0.78)^{13} = 0.235877\)

\(\binom{16}{4} (0.22)^4 (0.78)^{12} = 0.216221\)

Adding these probabilities gives:

\(0.179205 + 0.235877 + 0.216221 = 0.631\)

(a) We use the binomial distribution with parameters \(n = 12\) and \(p = 0.72\). We need to find \(P(X \leq 9)\), which is \(1 - P(X \geq 10)\).

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

\(P(X = 10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(X = 11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(X = 12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Calculating these probabilities:

\(P(X = 10) = 0.19372\)

\(P(X = 11) = 0.09057\)

\(P(X = 12) = 0.01941\)

\(P(X \geq 10) = 0.19372 + 0.09057 + 0.01941 = 0.30304\)

\(P(X \leq 9) = 1 - 0.30304 = 0.696\)

(b) The probability that Moena does not message Pasha for 3 consecutive days and then messages on the 4th day is:

\((0.28)^3 \times 0.72 = 0.0158\)

Let the random variable \(X\) represent the number of boxes containing more jellies than chocolates. \(X\) follows a binomial distribution with parameters \(n = 10\) and \(p = 0.64\).

We need to find \(P(X \leq 7)\), which is equivalent to \(1 - P(X \geq 8)\).

Calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

Using the binomial probability formula:

\(P(X = 8) = \binom{10}{8} (0.64)^8 (0.36)^2\)

\(P(X = 9) = \binom{10}{9} (0.64)^9 (0.36)^1\)

\(P(X = 10) = \binom{10}{10} (0.64)^{10}\)

Calculate each probability:

\(P(X = 8) = 0.164156\)

\(P(X = 9) = 0.064852\)

\(P(X = 10) = 0.11529\)

Sum these probabilities: \(P(X \geq 8) = 0.164156 + 0.064852 + 0.11529 = 0.344298\)

Therefore, \(P(X \leq 7) = 1 - 0.344298 = 0.759\).

Let the random variable \(X\) represent the number of adults who own a car in a sample of 8. \(X\) follows a binomial distribution with parameters \(n = 8\) and \(p = 0.7\).

We need to find \(P(X < 6)\), which is equivalent to \(1 - P(X \geq 6)\).

Calculate \(P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8)\).

\(P(X = 6) = \binom{8}{6} (0.7)^6 (0.3)^2\)

\(P(X = 7) = \binom{8}{7} (0.7)^7 (0.3)^1\)

\(P(X = 8) = \binom{8}{8} (0.7)^8 (0.3)^0\)

\(P(X \geq 6) = \binom{8}{6} (0.7)^6 (0.3)^2 + \binom{8}{7} (0.7)^7 (0.3)^1 + \binom{8}{8} (0.7)^8 (0.3)^0\)

\(P(X \geq 6) = 0.55177\)

Thus, \(P(X < 6) = 1 - 0.55177 = 0.448\).