(i) Let the probability of throwing a 4 be denoted by \(p\). The expected number of 4s in 30 throws is given by \(30p\). We are given that this average is 6.21, so:

\(30p = 6.21\)

Solving for \(p\), we get:

\(p = \frac{6.21}{30} = 0.207\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\). Here, \(n = 30\) and \(p = 0.207\), so:

\(\text{Variance} = 30 \times 0.207 \times (1 - 0.207)\)

\(= 30 \times 0.207 \times 0.793\)

\(= 4.92\)

(iii) For 15 throws, the probability of obtaining 2 or more 4s can be found using the complement rule. First, find the probability of obtaining fewer than 2 4s (i.e., 0 or 1 4s) using the binomial probability formula:

\(P(X = 0) = \binom{15}{0} (0.207)^0 (0.793)^{15} = (0.793)^{15}\)

\(P(X = 1) = \binom{15}{1} (0.207)^1 (0.793)^{14} = 15 \times 0.207 \times (0.793)^{14}\)

Calculate these probabilities and sum them:

\(P(X < 2) = (0.793)^{15} + 15 \times 0.207 \times (0.793)^{14}\)

Finally, the probability of obtaining 2 or more 4s is:

\(P(X \geq 2) = 1 - P(X < 2) = 0.848\)