The probability that a phone is not made by Company B is 0.65 (since 35% are made by Company B).

The probability that none of the n phones is made by Company B is given by \((0.65)^n\).

We want the probability that at least one phone is made by Company B to be more than 0.98, so:

\(1 - (0.65)^n > 0.98\)

\((0.65)^n < 0.02\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.02)\)

\(n > \frac{\log(0.02)}{\log(0.65)}\)

Calculating the right-hand side gives \(n > 9.08\).

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

The probability that a student does not own a car is \(1 - \frac{1}{4} = \frac{3}{4} = 0.75\).

The probability that none of the \(n\) students owns a car is \(0.75^n\).

We need the probability that at least one student owns a car to be greater than 0.995:

\(1 - 0.75^n > 0.995\)

\(0.75^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.75) < \log(0.005)\)

\(n > \frac{\log(0.005)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 18.4\).

Therefore, the least possible integer value of \(n\) is 19.

(i) The mean number of cracked eggs is 1.4, so the probability of a cracked egg, p, is given by:

\(p = \frac{1.4}{20} = 0.07\)

The probability of exactly 2 cracked eggs in a box of 20 is calculated using the binomial distribution:

\(P(2) = \binom{20}{2} (0.07)^2 (0.93)^{18}\)

Calculating this gives:

\(P(2) = 0.252\)

(ii) The probability of at least 1 cracked egg is:

\(P(\text{at least 1 cracked egg}) = 1 - P(0)\)

Where:

\(P(0) = (0.93)^{20}\)

Thus:

\(P(\text{at least 1 cracked egg}) = 1 - (0.93)^{20} = 0.766\)

(iii) We need to find the smallest n such that:

\((0.766)^n < 0.01\)

Taking logarithms:

\(n \log(0.766) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.766)} \approx 17.8\)

Thus, the smallest integer n is 18.

The probability that a person takes more than t minutes is 0.12. Therefore, the probability that a person takes t minutes or less is 0.88.

The probability that none of the n people takes more than t minutes is given by:

\((0.88)^n < 0.003\)

Taking logarithms on both sides, we have:

\(n \log(0.88) < \log(0.003)\)

Solving for n gives:

\(n > \frac{\log(0.003)}{\log(0.88)}\)

Calculating the right-hand side:

\(n > 45.4\)

Since n must be an integer, the smallest possible value is:

\(n = 46\)

(i) The probability that a person buys something is 0.66 (since 34% do not buy anything). We model this situation with a binomial distribution: \(X \sim B(15, 0.66)\).

The probability that at least 14 people buy something is \(P(X \geq 14) = P(X = 14) + P(X = 15)\).

Using the binomial probability formula:

\(P(X = 14) = \binom{15}{14} (0.66)^{14} (0.34)^1\)

\(P(X = 15) = \binom{15}{15} (0.66)^{15}\)

Calculating these gives:

\(P(X = 14) = 15 \times (0.66)^{14} \times 0.34\)

\(P(X = 15) = (0.66)^{15}\)

Adding these probabilities: \(P(X \geq 14) = 0.0171\).

(ii) The probability that a person spends less than $50 is 0.87. We want \((0.87)^n < 0.04\).

Taking logarithms: \(n \log(0.87) < \log(0.04)\).

Solving for \(n\):

\(n > \frac{\log(0.04)}{\log(0.87)} \approx 23.5\).

The smallest integer \(n\) is 24.