(b) The probability of getting a leopard in one magazine is \(p = 0.2\). We need to find the probability of getting more than two leopards in 12 magazines. This is given by:

\(1 - P(0, 1, 2) = 1 - \left( \binom{12}{0} (0.2)^0 (0.8)^{12} + \binom{12}{1} (0.2)^1 (0.8)^{11} + \binom{12}{2} (0.2)^2 (0.8)^{10} \right)\)

\(= 1 - (0.06872 + 0.20615 + 0.28347)\)

\(= 0.442\)

(c) The probability that after 5 weeks Sahim has exactly one of each animal is given by:

\((0.2)^5 \times 5!\)

\(= 0.0384 = \frac{24}{625}\)

The problem involves a binomial distribution where the probability of being an office worker is given as 0.22. The probability of not being an office worker is therefore 1 - 0.22 = 0.78.

We need to find the probability that between 4 and 6 people inclusive are office workers out of 8 people. This is calculated using the binomial probability formula:

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

where \(n = 8\), \(p = 0.22\), and \(r\) is the number of office workers.

Calculate for \(r = 4, 5, 6\):

\(P(4) = \binom{8}{4} (0.22)^4 (0.78)^4\)

\(P(5) = \binom{8}{5} (0.22)^5 (0.78)^3\)

\(P(6) = \binom{8}{6} (0.22)^6 (0.78)^2\)

Sum these probabilities:

\(P(4, 5, 6) = P(4) + P(5) + P(6)\)

\(= \binom{8}{4} (0.22)^4 (0.78)^4 + \binom{8}{5} (0.22)^5 (0.78)^3 + \binom{8}{6} (0.22)^6 (0.78)^2\)

\(= 0.0763\)

(i) The number of fireworks that fail follows a binomial distribution with parameters \(n = 20\) and \(p = 0.05\). We need to find \(P(X > 1)\).

\(P(X > 1) = 1 - P(X = 0) - P(X = 1)\)

\(P(X = 0) = (0.95)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.95)^{19} (0.05)^1\)

\(P(X > 1) = 1 - (0.95)^{20} - \binom{20}{1} (0.95)^{19} (0.05)\)

\(P(X > 1) = 1 - 0.3585 - 0.3770 = 0.264\)

(ii) The profit if 19 or 20 fireworks work is \(450 \times 10 - 480 = 4020\).

If more than 1 firework fails, the profit is \(-480\).

The expected profit is:

\(4020 \times (1 - 0.264) - 480 \times 0.264\)

\(= 4020 \times 0.736 - 480 \times 0.264\)

\(= 2960.64 - 126.72 = 2833.92\)

Rounding gives \(\$2830\).

The total number of pairs of shoes is 20, and the number of pairs with designer labels is 8. Therefore, the probability of choosing a pair with designer labels is \(p = \frac{8}{20} = 0.4\).

The probability of choosing a pair without designer labels is \(1 - p = 0.6\).

We need to find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of 7. This is a binomial probability problem where \(n = 7\), \(p = 0.4\), and we want \(P(X \leq 4)\).

Using the complement rule, \(P(X \leq 4) = 1 - P(X = 5) - P(X = 6) - P(X = 7)\).

Calculate each probability:

\(P(X = 5) = \binom{7}{5} (0.4)^5 (0.6)^2 = 21 \times 0.01024 \times 0.36 = 0.0774144\).

\(P(X = 6) = \binom{7}{6} (0.4)^6 (0.6)^1 = 7 \times 0.004096 \times 0.6 = 0.0172032\).

\(P(X = 7) = \binom{7}{7} (0.4)^7 (0.6)^0 = 1 \times 0.0016384 \times 1 = 0.0016384\).

Thus, \(P(X \leq 4) = 1 - (0.0774144 + 0.0172032 + 0.0016384) = 1 - 0.096256 = 0.903744\).

Rounding to three decimal places, the probability is \(0.904\).

(i) Let the random variable \(X\) represent the number of days with more than 20 cm of snow in a 7-day period. \(X\) follows a binomial distribution with parameters \(n = 7\) and \(p = 0.21\).

We need to find \(P(X < 5)\), which is \(1 - P(X \geq 5)\).

Calculate \(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\).

\(P(X = 5) = \binom{7}{5} (0.21)^5 (0.79)^2\)

\(P(X = 6) = \binom{7}{6} (0.21)^6 (0.79)^1\)

\(P(X = 7) = \binom{7}{7} (0.21)^7\)

\(P(X \geq 5) = 21(0.21)^5(0.79)^2 + 7(0.21)^6(0.79) + (0.21)^7\)

\(P(X \geq 5) = 0.006 + 0.0003 + 0.0000008 \approx 0.0063\)

Thus, \(P(X < 5) = 1 - 0.0063 = 0.994\).

(ii) Let \(Y\) be the number of 7-day periods with at least 1 day of more than 20 cm of snow. The probability of at least 1 day with more than 20 cm of snow in a 7-day period is \(1 - (0.79)^7 = 0.808\).

\(Y\) follows a binomial distribution with parameters \(n = 4\) and \(p = 0.808\).

We need to find \(P(Y = 3)\).

\(P(Y = 3) = \binom{4}{3} (0.808)^3 (0.192)^1\)

\(P(Y = 3) = 4(0.808)^3(0.192)\)

\(P(Y = 3) = 4(0.527)(0.192) \approx 0.405\).