Let the probability of a bag being underweight be \(p = 0.1\). The number of underweight bags follows a binomial distribution \(X \sim B(10, 0.1)\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.1)^0 (0.9)^{10} = 0.348678\)

\(P(X = 1) = \binom{10}{1} (0.1)^1 (0.9)^9 = 0.38742\)

\(P(X = 2) = \binom{10}{2} (0.1)^2 (0.9)^8 = 0.19371\)

Sum these probabilities:

\(P(X < 3) = 0.348678 + 0.38742 + 0.19371 = 0.929808\)

Rounding to three decimal places, the probability is \(0.930\).

Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.9\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 10) = \binom{12}{10} (0.9)^{10} (0.1)^2 = 0.230128\)

\(P(X = 11) = \binom{12}{11} (0.9)^{11} (0.1)^1 = 0.376573\)

\(P(X = 12) = \binom{12}{12} (0.9)^{12} (0.1)^0 = 0.282430\)

\(P(X \geq 10) = 0.230128 + 0.376573 + 0.282430 = 0.889131\)

Thus, \(P(X < 10) = 1 - 0.889131 = 0.110869\).

Therefore, the probability that fewer than 10 callers are connected immediately is approximately 0.111.

(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:

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

(i) The probability of selecting a non-orange disc is \(\frac{200}{300} = \frac{2}{3}\). Since the selection is with replacement, the probability that no orange discs are selected in 5 draws is \(\left( \frac{2}{3} \right)^5\).

(ii) The probability of selecting a disc with a number ending in 6 is \(\frac{10}{100} = \frac{1}{10}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3\).

(iii) The probability of selecting an orange disc with a number ending in 6 is \(\frac{10}{300} = \frac{1}{30}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{30} \right)^2 \left( \frac{29}{30} \right)^3\).

(iv) The probability of selecting a pink disc is \(\frac{100}{300} = \frac{1}{3}\). For a binomial distribution with \(n = 5\) and \(p = \frac{1}{3}\), the mean is \(np = 5 \times \frac{1}{3} = \frac{5}{3}\) and the variance is \(np(1-p) = 5 \times \frac{1}{3} \times \frac{2}{3} = \frac{10}{9}\).

Let the number of plants per box be denoted by \(n\), and the probability of a plant producing yellow flowers be \(p\). The number of plants producing yellow flowers follows a binomial distribution \(B(n, p)\).

The mean of a binomial distribution is given by \(np = 11\) and the variance is \(np(1-p) = 4.95\).

From \(np = 11\), we have \(p = \frac{11}{n}\).

Substitute \(p\) into the variance equation:

\(np(1-p) = 4.95\)

\(n \left(\frac{11}{n}\right) \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 - \frac{121}{n} = 4.95\)

\(11 - 4.95 = \frac{121}{n}\)

\(6.05 = \frac{121}{n}\)

\(n = \frac{121}{6.05} \approx 20\)

Thus, there are 20 plants per box.

For part (b), we need to find the probability that exactly 12 plants produce yellow flowers, i.e., \(P(X = 12)\).

Using \(n = 20\) and \(p = \frac{11}{20} = 0.55\), the probability mass function of a binomial distribution is:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

\(P(X = 12) = \binom{20}{12} (0.55)^{12} (0.45)^{8}\)

\(P(X = 12) \approx 0.162\)