(i) To find the probability that the two digits chosen are equal, consider the digits 1, 3, and 5 in the number 113 333 555. The probabilities are:

\(P(1, 1) = \frac{2}{9} \times \frac{1}{8}\)

\(P(3, 3) = \frac{4}{9} \times \frac{3}{8}\)

\(P(5, 5) = \frac{3}{9} \times \frac{2}{8}\)

Summing these probabilities gives:

\(\frac{2}{9} \times \frac{1}{8} + \frac{4}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{2}{8} = \frac{5}{18}\)

(ii) To find the probability that one digit is a 5 and one digit is not a 5, consider:

\(P(5, \overline{5}) + P(\overline{5}, 5)\)

\(= \frac{3}{9} \times \frac{6}{8} + \frac{6}{9} \times \frac{3}{8} = \frac{36}{72} = \frac{1}{2}\)

(a) The probability that it does not rain on Sunday is \(0.6\). If it does not rain on any day, the probability for each subsequent day is \(0.8\). Therefore, the probability that it does not rain on any of the 4 days is:

\(P(\text{no rain}) = 0.6 \times 0.8^3 = \frac{192}{625}\)

(b) The probability that it does not rain on Sunday is \(0.6\). The probability that it rains on Tuesday, given it did not rain on Monday, is \(0.2\). Therefore, the probability that the first day it rains is Tuesday is:

\(0.6 \times 0.8 \times 0.2 = \frac{12}{125}\)

(c) The probability that it rains on exactly one of the 4 days can be calculated by considering the scenarios where it rains on one specific day and not on the others. The scenarios are:

• Rains on Sunday: \(0.4 \times 0.3 \times 0.8 \times 0.8 = \frac{48}{625}\)
• Rains on Monday: \(0.6 \times 0.2 \times 0.3 \times 0.8 = \frac{18}{625}\)
• Rains on Tuesday: \(0.6 \times 0.8 \times 0.2 \times 0.3 = \frac{18}{625}\)
• Rains on Wednesday: \(0.6 \times 0.8 \times 0.8 \times 0.2 = \frac{48}{625}\)

Adding these probabilities gives:

\(\frac{48}{625} + \frac{18}{625} + \frac{18}{625} + \frac{48}{625} = \frac{132}{625}\)

The probability of picking a robot card from Jack's pack is \(\frac{10}{15} = \frac{2}{3}\).

Emma has 7 robot cards, so the total number of cards in Emma's pack is \(x + 4\) because \(x - 3\) cards are aeroplanes, making \(x - 3 + 7 = x + 4\) cards in total.

The probability of picking a robot card from Emma's pack is \(\frac{7}{x+4}\).

The probability that both cards have pictures of robots is \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Setting up the equation: \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Solving for \(x\):

\(\frac{14}{3(x+4)} = \frac{7}{18}\)

Cross-multiply: \(14 \times 18 = 7 \times 3(x+4)\)

\(252 = 21(x+4)\)

\(252 = 21x + 84\)

\(168 = 21x\)

\(x = 8\)

Let the houses with numbers greater than 14 be 16, 18, 20, 22, and 24. There are 5 such houses.

We need to find the probability that at least 2 newspapers are delivered to these 5 houses.

This can be calculated as \(P(2) + P(3)\) or \(1 - P(0, 1)\).

Calculate \(P(2)\):

\(P(2) = \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} \times \binom{3}{2}\)

Calculate \(P(3)\):

\(P(3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)

Thus, \(P(\text{at least 2}) = P(2) + P(3) = \frac{4}{11}\).

Alternatively, using combinations:

\(\frac{\binom{5}{3} + \binom{5}{2} \times \binom{7}{1}}{\binom{12}{3}} = \frac{4}{11}\).

Each tile can be black, white, or grey, giving a probability of \(\frac{1}{3}\) for each color. To ensure no two adjacent tiles are the same color, the first tile can be any color, but each subsequent tile must be a different color from the one before it.

For each of the 7 subsequent tiles, there are 2 choices (not the same as the previous tile), so the probability for each is \(\frac{2}{3}\).

The total probability is:

\(\left( \frac{2}{3} \right)^7 = \frac{128}{2187}\)

Thus, the probability that no two adjacent tiles are the same color is \(\frac{128}{2187}\) or approximately 0.0585.