(i) We know the total probability of viewing photos fewer than 3 times is 0.801. This can be expressed as:

\((1-x) \times 0.9 + x \times 0.24 = 0.801\)

Solving for \(x\):

\(0.9 - 0.9x + 0.24x = 0.801\)

\(0.9 - 0.66x = 0.801\)

\(0.66x = 0.9 - 0.801\)

\(0.66x = 0.099\)

\(x = \frac{0.099}{0.66} = 0.15\)

(ii) We need to find \(P(\text{at least 100 photos} \mid \text{at least 3 views})\).

Using Bayes' theorem:

\(P(\text{at least 100 photos} \mid \text{at least 3 views}) = \frac{P(\text{at least 100 photos} \cap \text{at least 3 views})}{P(\text{at least 3 views})}\)

\(P(\text{at least 100 photos} \cap \text{at least 3 views}) = 0.85 \times 0.1\)

\(P(\text{at least 3 views}) = 1 - 0.801 = 0.199\)

\(P(\text{at least 100 photos} \mid \text{at least 3 views}) = \frac{0.85 \times 0.1}{0.85 \times 0.1 + 0.15 \times 0.76}\)

\(= \frac{0.085}{0.085 + 0.114}\)

\(= \frac{0.085}{0.199} \approx 0.427\)