(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:

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

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

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

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

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

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

(i) The probability that the person chosen is from country X is given by the ratio of the population of country X to the total population:

\(P(X) = \frac{20}{28} = \frac{5}{7} \approx 0.714\)

(ii) The probability that the person chosen has fair hair is calculated by considering both countries:

In country X, 25% have fair hair, so \(0.25 \times 20 = 5\) million people.

In country Y, 60% have fair hair, so \(0.60 \times 8 = 4.8\) million people.

Total with fair hair = \(5 + 4.8 = 9.8\) million people.

\(P(F) = \frac{9.8}{28} = \frac{7}{20} = 0.35\)

(iii) The probability that the person is from country X given they have fair hair is:

\(P(X|F) = \frac{P(X \cap F)}{P(F)}\)

\(P(X \cap F) = \frac{5}{28}\) (people with fair hair from X)

\(P(X|F) = \frac{\frac{5}{28}}{\frac{7}{20}} = \frac{5}{28} \times \frac{20}{7} = \frac{25}{49} \approx 0.51\)

To find the probability that the coin shows a head given that Jodie’s score is 8, we use conditional probability:

\(P(H | 8) = \frac{P(H \cap 8)}{P(8)}\)

First, calculate \(P(8)\):

\(P(8) = P(H \text{ and } 4+4) + P(T \text{ and } 2 \times 4) + P(T \text{ and } 4 \times 2)\)

\(= \frac{1}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16}\)

\(= \frac{1}{48} + \frac{2}{48} + \frac{2}{48}\)

\(= \frac{5}{48}\)

Next, calculate \(P(H \cap 8)\):

\(P(H \cap 8) = P(H \text{ and } 4+4) = \frac{1}{3} \times \frac{1}{16} = \frac{1}{48}\)

Now, substitute these into the conditional probability formula:

\(P(H | 8) = \frac{\frac{1}{48}}{\frac{5}{48}} = \frac{1}{5}\)

(i) The probability that Ben becomes the champion after playing exactly 2 games is when Ben wins both games. The probability of Ben winning a game is 0.7. Therefore, the probability is:

\(P(B \text{ champ}) = 0.7 \times 0.7 = 0.49\)

(ii) The probability that Ben becomes the champion can occur in the following sequences: WW, WLW, LWW.

\(P(B \text{ champ}) = P(WW) + P(WLW) + P(LWW)\)

\(= (0.7 \times 0.7) + (0.7 \times 0.3 \times 0.7) + (0.3 \times 0.7 \times 0.7)\)

\(= 0.49 + 0.147 + 0.147\)

\(= 0.784\)

(iii) Given that Tom becomes the champion, we need to find the probability that he won the 2nd game. This can occur in the sequences: TT or TBT.

\(P(T2 \cap T) = (0.3 \times 0.3) + (0.7 \times 0.3 \times 0.3) = 0.216\)

The probability that Tom becomes the champion is \(1 - 0.784 = 0.216\).

Thus, the probability is:

\(P(T2 | T) = \frac{P(T2 \cap T)}{P(T)} = \frac{0.216}{0.216} = 0.708\)

To find the probability that Nur chose playground Y given that she chose a climbing frame, we use Bayes' theorem:

\(P(Y | C) = \frac{P(Y \cap C)}{P(C)}\)

First, calculate \(P(Y \cap C)\):

Nur chooses playground Y with probability \(\frac{1}{4}\) and then chooses a climbing frame with probability \(\frac{1}{12}\) (since there is 1 climbing frame out of 12 total pieces of equipment in playground Y).

\(P(Y \cap C) = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48}\)

Next, calculate \(P(C)\), the total probability of choosing a climbing frame:

\(P(C) = P(X \cap C) + P(Y \cap C) + P(Z \cap C)\)

\(P(X \cap C) = 0\) since there are no climbing frames in playground X.

\(P(Y \cap C) = \frac{1}{48}\) as calculated above.

\(P(Z \cap C) = \frac{1}{2} \times \frac{4}{16} = \frac{1}{8}\) since there are 4 climbing frames out of 16 total pieces of equipment in playground Z.

\(P(C) = 0 + \frac{1}{48} + \frac{1}{8} = \frac{1}{48} + \frac{6}{48} = \frac{7}{48}\)

Finally, substitute back into Bayes' theorem:

\(P(Y | C) = \frac{\frac{1}{48}}{\frac{7}{48}} = \frac{1}{7}\)