Let the events be defined as follows:

• Event R: A red paper clip is taken from box A.
• Event W: A white paper clip is taken from box A.

Calculate the probabilities for each scenario:

1. If a red paper clip is taken from box A and then a white paper clip from box B:

\(P(R, W) = \frac{5}{6} \times \frac{2}{10} = \frac{10}{60}\)

2. If a white paper clip is taken from box A and then a white paper clip from box B:

\(P(W, W) = \frac{1}{6} \times \frac{3}{10} = \frac{3}{60}\)

Now, calculate the probability distribution of X:

• \(x = 0\): No red paper clips are taken. This occurs when a white paper clip is taken from box A and then from box B. Thus, \(P(X=0) = \frac{3}{60}\).
• \(x = 1\): One red paper clip is taken. This can occur in two ways: either a red paper clip is taken from box A and a white paper clip from box B, or a white paper clip is taken from box A and a red paper clip from box B. The probability is \(P(X=1) = \frac{10}{60} + \frac{7}{60} = \frac{17}{60}\).
• \(x = 2\): Two red paper clips are taken. This occurs when a red paper clip is taken from both boxes. Thus, \(P(X=2) = \frac{40}{60}\).

The probability distribution table is:

(a) The probability distribution table for X is:

(b) To find E(X), calculate:

E(X) = \frac{1 \times 2 + 4 \times 3 + 10 \times 4 + 12 \times 5 + 9 \times 6}{36} = \frac{2 + 12 + 40 + 60 + 54}{36} = \frac{168}{36} = \frac{14}{3}

To find Var(X), calculate:

\(\text{Var}(X) = \frac{1 \times 2^2 + 4 \times 3^2 + 10 \times 4^2 + 12 \times 5^2 + 9 \times 6^2}{36} - \left(\frac{14}{3}\right)^2\)

\(= \frac{4 + 36 + 160 + 300 + 324}{36} - \left(\frac{14}{3}\right)^2\)

\(= \frac{824}{36} - \frac{196}{9} = \frac{10}{9}\)

To find the probability distribution for the number of green peppers taken, we consider the possible values of the random variable X, which are 0, 1, 2, and 3.

1. Probability of taking 0 green peppers:

The number of ways to choose 3 peppers from the 8 non-green peppers (3 red + 5 yellow) is \(\binom{8}{3} = 56\).

The total number of ways to choose 3 peppers from all 12 is \(\binom{12}{3} = 220\).

Thus, \(P(X=0) = \frac{56}{220} = \frac{14}{55}\).

2. Probability of taking 1 green pepper:

Choose 1 green pepper from 4 and 2 non-green peppers from 8: \(\binom{4}{1} \times \binom{8}{2} = 4 \times 28 = 112\).

Thus, \(P(X=1) = \frac{112}{220} = \frac{28}{55}\).

3. Probability of taking 2 green peppers:

Choose 2 green peppers from 4 and 1 non-green pepper from 8: \(\binom{4}{2} \times \binom{8}{1} = 6 \times 8 = 48\).

Thus, \(P(X=2) = \frac{48}{220} = \frac{12}{55}\).

4. Probability of taking 3 green peppers:

Choose all 3 peppers from the 4 green peppers: \(\binom{4}{3} = 4\).

Thus, \(P(X=3) = \frac{4}{220} = \frac{1}{55}\).

The probability distribution table is:

(i) Mario pays $1 for each throw. After missing twice, he hits on the third throw and receives $3. His total cost is $3, so his profit is \(3 - 3 = 0\). However, the mark-scheme states $2, indicating a profit calculation of \(3 - 1 = 2\) for the third throw.

(ii) The probability that Mario’s profit is $0 is the probability that he misses all five throws or hits on the fourth or fifth throw. This is calculated as:

\(P(MMMMH) + P(MMMMM) = 0.8^3 \times 0.2 + 0.8^4 \times 0.2 = 0.184\)

(iii) The probability distribution table for Mario’s profit is:

(iv) The expected profit \(E(X)\) is calculated as:

\(E(X) = 4 \times 0.2 + 2 \times 0.288 + 0 \times 0.184 - 1 \times 0.328 = 0.8 + 0.576 - 0.328 = 1.05\)

(i) The probability of a single die landing on a green face is \(\frac{1}{4} = 0.25\). The probability of a die not landing on a green face is \(1 - 0.25 = 0.75\).

We use the binomial probability formula: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\).

For 4 dice landing on a green face, \(n = 5\), \(k = 4\), \(p = 0.25\):

\(P(X = 4) = \binom{5}{4} (0.25)^4 (0.75)^1\)

\(= 5 \times (0.25)^4 \times 0.75\)

\(= 5 \times 0.00390625 \times 0.75\)

\(= 0.0146484375\)

Rounding to 4 decimal places gives \(0.0146\).

(ii) The probability distribution of \(X\) is calculated using the binomial formula for each value of \(X\) from 0 to 5: