(i) The probability that the die lands on 4 is \(\frac{1}{4}\). The probability of getting exactly 2 heads in 4 coin tosses is given by the binomial probability formula:

\(P(4, 2H) = \frac{1}{4} \times \binom{4}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^2\)

\(= \frac{1}{4} \times 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{2}{27}\)

(ii) The probability that the die lands on 3 is \(\frac{1}{4}\). The probability of getting exactly 3 heads in 3 coin tosses is:

\(P(3, 3H) = \frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)

(iii) We need to find the probability that the number the die lands on is the same as the number of heads. This includes the following cases:

• Die lands on 1 and 1 head: \(\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
• Die lands on 2 and 2 heads: \(\frac{1}{4} \times \binom{2}{2} \left( \frac{1}{3} \right)^2 = \frac{1}{36}\)
• Die lands on 3 and 3 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)
• Die lands on 4 and 4 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^4 = \frac{1}{324}\)

Summing these probabilities gives:

\(\frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \frac{1}{324} = \frac{10}{81}\)