The probability that a mango is not small is 0.85 (since 15% are small).

The probability that none of the n mangoes is small is given by (0.85)^n.

We need to find the largest n such that (0.85)^n \geq 0.1.

Taking logarithms on both sides, we have:

\(n \leq \frac{\log(0.1)}{\log(0.85)} \approx 14.2\)

\(Since n must be an integer, the largest possible value is n = 14.\)

The probability that a student does not have blue eyes is given by:

\(P( ext{not blue eyes}) = 1 - \frac{1}{5} = \frac{4}{5} = 0.8\)

For n students, the probability that none of them has blue eyes is:

\((0.8)^n\)

We need this probability to be less than 0.001:

\((0.8)^n < 0.001\)

Taking logarithms on both sides:

\(n \log(0.8) < \log(0.001)\)

Solving for n:

\(n > \frac{\log(0.001)}{\log(0.8)}\)

Calculating the values:

\(n > \frac{-3}{-0.09691} \approx 30.9\)

Since n must be an integer, the least possible value of n is 31.

(i) The integers between 7 and 21 that are multiples of 5 are 10, 15, and 20. There are 3 such numbers out of 15 total numbers (7 to 21 inclusive), so the probability \(p\) that a randomly chosen integer is a multiple of 5 is \(\frac{3}{15} = 0.2\). Therefore, \(X \sim \text{Bin}(12, 0.2)\).

(ii) We need to calculate \(P(3 \leq X \leq 5)\). This is given by:

\(P(X = 3, 4, 5) = \binom{12}{3} (0.2)^3 (0.8)^9 + \binom{12}{4} (0.2)^4 (0.8)^8 + \binom{12}{5} (0.2)^5 (0.8)^7\)

Calculating each term:

\(\binom{12}{3} (0.2)^3 (0.8)^9 = 0.23622\)

\(\binom{12}{4} (0.2)^4 (0.8)^8 = 0.13287\)

\(\binom{12}{5} (0.2)^5 (0.8)^7 = 0.05315\)

Adding these probabilities gives \(0.23622 + 0.13287 + 0.05315 = 0.422\).

(iii) We want \(P(X = 0) < 0.01\). For \(X \sim \text{Bin}(n, 0.2)\), \(P(X = 0) = (0.8)^n\). We solve:

\((0.8)^n < 0.01\)

Taking logarithms, \(n \log(0.8) < \log(0.01)\).

\(n > \frac{\log(0.01)}{\log(0.8)} \approx 20.65\)

Thus, the smallest integer \(n\) is 21.

(i) Let the random variable \(X\) represent the number of customers who rate the food as ‘good’. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.65\). We need to find \(P(2 < X < 12)\).

\(P(2 < X < 12) = 1 - P(X \leq 2)\)

\(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(P(X = 0) = (0.35)^{12}\)

\(P(X = 1) = \binom{12}{1} (0.65)^1 (0.35)^{11}\)

\(P(X = 2) = \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = (0.35)^{12} + \binom{12}{1} (0.65)^1 (0.35)^{11} + \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = 0.0065359\)

\(P(2 < X < 12) = 1 - 0.0065359 = 0.993\)

(ii) Let \(Y\) be the number of customers who rate the food as ‘poor’. \(Y\) follows a binomial distribution with parameters \(n\) and \(p = 0.13\). We need \(P(Y \geq 1) > 0.95\).

\(P(Y \geq 1) = 1 - P(Y = 0)\)

\(P(Y = 0) = (0.87)^n\)

\(1 - (0.87)^n > 0.95\)

\((0.87)^n < 0.05\)

Solving \((0.87)^n < 0.05\) gives \(n = 22\).

The probability that Sue completes all n puzzles correctly is given by \((0.75)^n\).

We need to find the smallest n such that \((0.75)^n < 0.06\).

Taking logarithms on both sides, we have:

\(\log((0.75)^n) < \log(0.06)\)

\(n \cdot \log(0.75) < \log(0.06)\)

\(n > \frac{\log(0.06)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 9.78\).

Since n must be an integer, the smallest possible value is \(n = 10\).