(a) To find \(k\), use the fact that the sum of probabilities must equal 1:

\(2k + 6k + 12k + 20k = 1\)

\(40k = 1\)

\(k = \frac{1}{40}\)

Now, calculate the probabilities:

(b) Calculate \(E(X)\):

\(E(X) = \frac{1 \times 2 + 2 \times 6 + 3 \times 12 + 4 \times 20}{40}\)

\(E(X) = \frac{2 + 12 + 36 + 80}{40}\)

\(E(X) = \frac{130}{40} = 3.25\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1^2 \times 2 + 2^2 \times 6 + 3^2 \times 12 + 4^2 \times 20}{40} - (E(X))^2\)

\(\text{Var}(X) = \frac{2 + 24 + 108 + 320}{40} - (3.25)^2\)

\(\text{Var}(X) = \frac{454}{40} - 10.5625\)

\(\text{Var}(X) = 11.35 - 10.5625 = 0.7875\)

(a) To find the probability distribution of \(X\), we consider all possible sums of the numbers from the two spinners:

• Spinner 1: 0, 1, 2, 2
• Spinner 2: -1, 0, 1

Possible sums \(X\) are: -1, 0, 1, 2, 3.

Calculate probabilities:

• \(P(X = -1) = \frac{1}{12}\) (only one combination: 0 + (-1))
• \(P(X = 0) = \frac{2}{12}\) (combinations: 0 + 0, 1 + (-1))
• \(P(X = 1) = \frac{4}{12}\) (combinations: 0 + 1, 1 + 0, 2 + (-1), 2 + (-1))
• \(P(X = 2) = \frac{3}{12}\) (combinations: 1 + 1, 2 + 0, 2 + 0)
• \(P(X = 3) = \frac{2}{12}\) (combinations: 2 + 1, 2 + 1)

(b) To find \(\text{Var}(X)\), first calculate \(E(X)\):

\(E(X) = \sum x p(x) = -1 \times \frac{1}{12} + 0 \times \frac{2}{12} + 1 \times \frac{4}{12} + 2 \times \frac{3}{12} + 3 \times \frac{2}{12} = \frac{15}{12}\)

Then, calculate \(E(X^2)\):

\(E(X^2) = (-1)^2 \times \frac{1}{12} + 0^2 \times \frac{2}{12} + 1^2 \times \frac{4}{12} + 2^2 \times \frac{3}{12} + 3^2 \times \frac{2}{12} = \frac{35}{12}\)

Finally, \(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{35}{12} - \left( \frac{15}{12} \right)^2 = \frac{65}{48} = 1.35\).

(a) To find the probability distribution of X, we consider all possible outcomes from spinning the two spinners:

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

\(E(X) = \frac{-1 \times 1 + 0 \times 2 + 1 \times 1 + 2 \times 3 + 3 \times 2}{9} = \frac{-1 + 0 + 1 + 6 + 6}{9} = \frac{12}{9} = \frac{4}{3}\)

To find Var(X), we calculate:

\(\text{Var}(X) = \frac{(-1)^2 \times 1 + (0)^2 \times 2 + (1)^2 \times 1 + (2)^2 \times 3 + (3)^2 \times 2}{9} - (E(X))^2\)

\(= \frac{1 + 0 + 1 + 12 + 18}{9} - \left(\frac{4}{3}\right)^2 = \frac{32}{9} - \frac{16}{9} = \frac{16}{9}\)

(a) To find \(P(X = 3)\), we calculate the probability of opening two non-carrot tins followed by a carrot tin. The probability of the first tin not being carrots is \(\frac{4}{7}\) (since there are 4 non-carrot tins out of 7 total). The probability of the second tin not being carrots is \(\frac{3}{6}\) (since there are now 3 non-carrot tins out of 6 remaining). The probability of the third tin being carrots is \(\frac{3}{5}\) (since there are 3 carrot tins out of 5 remaining). Therefore, \(P(X = 3) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{6}{35}\).

(b) The probability distribution table for \(X\) is:

(c) To find \(\text{Var}(X)\), we first calculate \(E(X)\):

\(E(X) = 1 \times \frac{15}{35} + 2 \times \frac{10}{35} + 3 \times \frac{6}{35} + 4 \times \frac{3}{35} + 5 \times \frac{1}{35} = \frac{70}{35} = 2\)

Then, \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = \frac{1^2 \times 15 + 2^2 \times 10 + 3^2 \times 6 + 4^2 \times 3 + 5^2 \times 1}{35} - 2^2\)

\(= \frac{15 + 40 + 54 + 48 + 25}{35} - 4 = \frac{182}{35} - 4 = \frac{6}{5}\)

(a) The probability distribution table for X is:

To find k, use the fact that the sum of probabilities is 1:

\(4k + 6k + 6k + 4k = 1\)

\(20k = 1\)

\(k = \frac{1}{20}\)

(b) Calculate the expected value \(E(X)\):

\(E(X) = 1 \times \frac{4}{20} + 2 \times \frac{6}{20} + 3 \times \frac{6}{20} + 4 \times \frac{4}{20}\)

\(= \frac{4}{20} + \frac{12}{20} + \frac{18}{20} + \frac{16}{20}\)

\(= \frac{50}{20} = 2.5\)

Calculate the variance \(\text{Var}(X)\):

\(\text{Var}(X) = 1^2 \times \frac{4}{20} + 2^2 \times \frac{6}{20} + 3^2 \times \frac{6}{20} + 4^2 \times \frac{4}{20} - (2.5)^2\)

\(= \left(\frac{4}{20} + \frac{24}{20} + \frac{54}{20} + \frac{64}{20}\right) - 6.25\)

\(= \frac{146}{20} - 6.25\)

\(= 7.3 - 6.25\)

\(= 1.05\)

(a) To find the probability of obtaining exactly 2 heads and 1 tail, consider the scenarios: HHT, HTH, and THH.

For HHT: \(\frac{2}{3} \times \frac{2}{3} \times \frac{1}{5} = \frac{4}{45}\)

For HTH: \(\frac{2}{3} \times \frac{1}{3} \times \frac{4}{5} = \frac{8}{45}\)

For THH: \(\frac{1}{3} \times \frac{2}{3} \times \frac{4}{5} = \frac{8}{45}\)

Total probability = \(\frac{4}{45} + \frac{8}{45} + \frac{8}{45} = \frac{20}{45} = \frac{4}{9}\)

(b) The probability distribution table for \(X\) is:

(c) Given \(\text{E}(X) = \frac{32}{15}\), find \(\text{Var}(X)\):

\(\text{Var}(X) = 0^2 \times \frac{1}{45} + 1^2 \times \frac{8}{45} + 2^2 \times \frac{20}{45} + 3^2 \times \frac{16}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{8}{45} + \frac{80}{45} + \frac{144}{45} - \left( \frac{32}{15} \right)^2\)

\(= \frac{136}{225}\) or 0.604

(a) To find the probability that Sadie takes exactly 1 red ball, we calculate the probability of choosing 1 red ball and 2 blue balls. The probability is given by:

\(P(1 \text{ red}) = \frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{15}{56}\)

(b) The probability distribution table for X is:

(c) To find \(\text{Var}(X)\), we use the formula:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

First, calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times \frac{1}{56} + 1^2 \times \frac{15}{56} + 2^2 \times \frac{30}{56} + 3^2 \times \frac{10}{56}\)

\(E(X^2) = \frac{15}{56} + \frac{120}{56} + \frac{90}{56} = \frac{225}{56}\)

Now, substitute into the variance formula:

\(\text{Var}(X) = \frac{225}{56} - \left( \frac{15}{8} \right)^2\)

\(\text{Var}(X) = \frac{225}{56} - \frac{225}{64}\)

\(\text{Var}(X) = 0.502\)

(a) To find the probability distribution of Y, consider all possible pairs of X values:

• \((1,1)\), \((2,2)\), \((3,3)\), \((4,4)\): Y takes the value 1, 2, 3, 4 respectively.
• \((1,2)\), \((2,1)\): Y = 1.
• \((1,3)\), \((3,1)\): Y = 2.
• \((1,4)\), \((4,1)\): Y = 3.
• \((2,3)\), \((3,2)\): Y = 1.
• \((2,4)\), \((4,2)\): Y = 2.
• \((3,4)\), \((4,3)\): Y = 1.

Calculate probabilities:

• \(P(Y=1) = \frac{7}{16}\)
• \(P(Y=2) = \frac{5}{16}\)
• \(P(Y=3) = \frac{3}{16}\)
• \(P(Y=4) = \frac{1}{16}\)

(b) Given Y is even, possible values are 2 and 4. Calculate:

\(P(Y=2 \mid Y \text{ is even}) = \frac{P(Y=2)}{P(Y=2) + P(Y=4)} = \frac{\frac{5}{16}}{\frac{5}{16} + \frac{1}{16}} = \frac{5}{6}\)

(a) To find the probability distribution table for X, we first list all possible outcomes of the sums of the numbers from the two spinners:

• Four-sided spinner outcomes: 1, 2, 2, 3
• Three-sided spinner outcomes: -2, -1, 1

Possible sums (X):

• 1 + (-2) = -1
• 1 + (-1) = 0
• 1 + 1 = 2
• 2 + (-2) = 0
• 2 + (-1) = 1
• 2 + 1 = 3
• 3 + (-2) = 1
• 3 + (-1) = 2
• 3 + 1 = 4

Count the occurrences of each sum and divide by the total number of outcomes (12) to get the probabilities:

• \(P(X = -1) = \frac{1}{12}\)
• \(P(X = 0) = \frac{3}{12}\)
• \(P(X = 1) = \frac{3}{12}\)
• \(P(X = 2) = \frac{2}{12}\)
• \(P(X = 3) = \frac{2}{12}\)
• \(P(X = 4) = \frac{1}{12}\)

(b) To find \(\text{Var}(X)\), we first calculate \(\text{E}(X)\):

\(\text{E}(X) = \frac{-1 + 0 + 3 + 4 + 6 + 4}{12} = \frac{16}{12} = \frac{4}{3}\)

Then, calculate \(\text{E}(X^2)\):

\(\text{E}(X^2) = \frac{1 + 0 + 3 + 8 + 18 + 16}{12} = \frac{46}{12}\)

Finally, \(\text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2\):

\(\text{Var}(X) = \frac{46}{12} - \left(\frac{4}{3}\right)^2 = \frac{37}{18} \approx 2.06\)

(a) To find \(P(X = 3)\), consider the outcomes where the maximum value is 3. The three-sided spinner can land on 3, and the five-sided spinner can land on 3, or the five-sided spinner can land on 3 while the three-sided spinner lands on 1 or 2. The possible outcomes are (3,3), (3,1), (3,2), (1,3), (2,3). Counting these outcomes gives 7 favorable outcomes out of 15 total possible outcomes, so \(P(X = 3) = \frac{7}{15}\).

(b) The probability distribution table for X is:

(c) To find \(E(X)\), use the formula \(E(X) = \sum x_i P(x_i)\):

\(E(X) = 1 \times \frac{2}{15} + 2 \times \frac{6}{15} + 3 \times \frac{7}{15} = \frac{2 + 12 + 21}{15} = \frac{35}{15} = \frac{7}{3}\).

To find \(Var(X)\), use the formula \(Var(X) = E(X^2) - (E(X))^2\):

\(E(X^2) = 1^2 \times \frac{2}{15} + 2^2 \times \frac{6}{15} + 3^2 \times \frac{7}{15} = \frac{2 + 24 + 63}{15} = \frac{89}{15}\).

\(Var(X) = \frac{89}{15} - \left(\frac{7}{3}\right)^2 = \frac{89}{15} - \frac{49}{9} = \frac{22}{45}\).

To find the probability distribution for the number of jellies Jemeel chooses, we calculate the probability for each possible number of jellies (0, 1, 2, 3) out of 3 sweets chosen.

There are a total of 8 sweets (5 jellies and 3 chocolates), and Jemeel chooses 3 sweets. The total number of ways to choose 3 sweets from 8 is \(\binom{8}{3} = 56\).

Probability of choosing 0 jellies:

Choose 3 chocolates from 3: \(\binom{3}{3} = 1\)

Probability: \(\frac{1}{56}\)

Probability of choosing 1 jelly:

Choose 1 jelly from 5 and 2 chocolates from 3: \(\binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15\)

Probability: \(\frac{15}{56}\)

Probability of choosing 2 jellies:

Choose 2 jellies from 5 and 1 chocolate from 3: \(\binom{5}{2} \times \binom{3}{1} = 10 \times 3 = 30\)

Probability: \(\frac{30}{56}\)

Probability of choosing 3 jellies:

Choose 3 jellies from 5: \(\binom{5}{3} = 10\)

Probability: \(\frac{10}{56}\)

The probability distribution table is:

(a) Given \(P(X = 4) = 3P(X = 2)\), we have:

\(4k(4 + a) = 3 \times 2k(2 + a)\)

\(16k + 4ak = 12k + 6ak\)

\(4k = 2ak\)

\(a = 2\)

Using the sum of probabilities:

\(k(1 + a) + 2k(2 + a) + 3k(3 + a) + 4k(4 + a) = 1\)

\(3k + 8k + 15k + 24k = 1\)

\(50k = 1\)

\(k = \frac{1}{50}\)

(b) The probability distribution table is:

(c) To find \(\text{Var}(X)\), use:

\(\text{Var}(X) = \sum (x^2 \cdot P(X = x)) - (E(X))^2\)

\(\text{Var}(X) = \frac{3}{50} \times 1^2 + \frac{8}{50} \times 2^2 + \frac{15}{50} \times 3^2 + \frac{24}{50} \times 4^2 - 3.2^2\)

\(\text{Var}(X) = \frac{3 + 32 + 135 + 384}{50} - 10.24\)

\(\text{Var}(X) = \frac{554}{50} - 10.24\)

\(\text{Var}(X) = 11.08 - 10.24\)

\(\text{Var}(X) = 0.84\)

\(\text{Var}(X) = \frac{21}{25}\)

To solve part (b), we need to find the probability distribution for the number of times a 1 or a 6 is rolled in 3 throws of a die. The probability of rolling a 1 or a 6 in a single throw is \(\frac{1}{3}\), and the probability of not rolling a 1 or a 6 is \(\frac{2}{3}\).

Calculating each probability:

\(P(0) = \left( \frac{2}{3} \right)^3 = \frac{8}{27}\)

\(P(1) = \binom{3}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^2 = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27}\)

\(P(2) = \binom{3}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^1 = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27}\)

\(P(3) = \left( \frac{1}{3} \right)^3 = \frac{1}{27}\)

For part (c), the expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{8}{27} + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}\)

\(E(X) = \frac{0}{27} + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}\)

\(E(X) = \frac{27}{27} = 1\)

(i) The probability that the first ball is red is \(\frac{3}{8}\). After removing one red ball, the probability that the second ball is red is \(\frac{2}{7}\). Thus, the probability that both balls are red is \(\frac{3}{8} \times \frac{2}{7} = \frac{3}{28}\).

(ii) The probability of choosing a red then a white ball is \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\). The probability of choosing a white then a red ball is \(\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\). Therefore, the probability of different colors is \(\frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\).

(iii) Using conditional probability, \(P(\text{first red} | \text{second red}) = \frac{P(\text{first red and second red})}{P(\text{second red})} = \frac{\frac{3}{28}}{\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}} = \frac{2}{7}\).

(iv) The probability distribution table for \(X\) is:

(v) The expected value \(E(X) = \frac{30}{56} = \frac{15}{28}\). The variance \(\text{Var}(X) = \frac{30}{56} - \left(\frac{15}{28}\right)^2 = \frac{45}{112}\) or 0.402.

(i) To find the probability distribution of \(X\), calculate \(X = \text{Red} - \text{Blue}\) for all combinations:

(ii) Calculate \(\text{E}(X)\):

\(\text{E}(X) = \frac{-1 + 0 + 3 + 4 + 9 + 8}{12} = \frac{23}{12}\)

Calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{1 + 0 + 3 + 8 + 27 + 32}{12} - \left(\frac{23}{12}\right)^2 = \frac{71}{12} - \left(\frac{23}{12}\right)^2\)

\(\text{Var}(X) = \frac{323}{144} \text{ or } \frac{35}{144}\)

(i) The possible scores are calculated by multiplying the outcomes of the two spinners. The five-sided spinner outcomes are \(1, 1, 1, 2, 3\) and the three-sided spinner outcomes are \(1, 2, 3\). The scores are:

(ii) To find the mean, use the formula \(\mu = \sum x_i p_i\):

\(\mu = \frac{3 \times 1 + 4 \times 2 + 4 \times 3 + 1 \times 4 + 2 \times 6 + 1 \times 9}{15} = \frac{48}{15} = 3.2\)

To find the variance, use the formula \(\sigma^2 = \sum x_i^2 p_i - \mu^2\):

\(\sigma^2 = \frac{3 \times 1^2 + 4 \times 2^2 + 4 \times 3^2 + 1 \times 4^2 + 2 \times 6^2 + 1 \times 9^2}{15} - 3.2^2\)

\(\sigma^2 = \frac{224}{15} - 3.2^2 = 4.69\)

(iii) The scores greater than the mean (3.2) are 4, 6, and 9. The probability is:

\(P(X > 3.2) = \frac{1}{15} + \frac{2}{15} + \frac{1}{15} = \frac{4}{15}\)

(i) The tree diagram is as follows:

First draw two branches for the first sweet: Toffee (T) with probability \(\frac{6}{7}\) and Chocolate (C) with probability \(\frac{1}{7}\).

For the second sweet, if the first was a Toffee (T), draw branches for Toffee (T) with probability \(\frac{5}{9}\) and Chocolate (C) with probability \(\frac{4}{9}\).

If the first was a Chocolate (C), draw branches for Toffee (T) with probability \(\frac{6}{9}\) and Chocolate (C) with probability \(\frac{3}{9}\).

(ii) The probability distribution table for the number of toffees taken is:

(iii) The mean number of toffees taken is calculated as:

\(E(X) = \frac{90}{63} \times \left( \frac{10}{7} \right) = 1.43\)

(iv) The probability that the first sweet taken is a chocolate, given that the second sweet taken is a toffee, is:

\(P(\text{1st C | 2nd T}) = \frac{P(C \cap T)}{P(T)} = \frac{\frac{1}{7} \times \frac{6}{9}}{\frac{1}{7} \times \frac{6}{9} + \frac{6}{7} \times \frac{5}{9}} = \frac{6}{63} \div \frac{36}{63} = \frac{1}{6}\)

(i) The probability that Amy loses her original $1 is the probability that she fails both attempts. The probability of failing one attempt is given by:

\(P(F) = 1 - 0.2 = 0.8\)

Therefore, the probability of failing both attempts is:

\(P(\text{loses } \$1) = P(F \text{ and } F) = 0.8 \times 0.8 = 0.64\)

(ii) The probability distribution table for the amount that Amy gains is:

The probability of gaining $0.50 is the probability of failing the first attempt and succeeding on the second:

\(P(0.50) = P(F \text{ and } S) = 0.8 \times 0.2 = 0.16\)

The probability of gaining $2 is the probability of succeeding on the first attempt:

\(P(2) = 0.2\)

(iii) The expected gain is calculated as follows:

\(E(\text{winnings}) = (-1 \times 0.64) + (0.5 \times 0.16) + (2 \times 0.2)\)

\(E(\text{winnings}) = -0.64 + 0.08 + 0.4 = -0.16\)

Therefore, Amy's expected gain is -$0.16, indicating a loss of 16 cents.

(i) The probability distribution table is:

Since the sum of probabilities must equal 1, we have:

\(k + k + 4k + 9k = 15k = 1\)

Solving for \(k\), we get \(k = \frac{1}{15}\).

(ii) To find \(E(X)\), we calculate:

\(E(X) = (-1)k + 1k + 2(4k) + 3(9k) = -k + k + 8k + 27k = 35k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 35 \times \frac{1}{15} = \frac{35}{15} = \frac{7}{3}\)

To find \(Var(X)\), we use:

\(Var(X) = E(X^2) - (E(X))^2\)

\(E(X^2) = (-1)^2k + 1^2k + 2^2(4k) + 3^2(9k) = k + k + 16k + 81k = 99k\)

\(Var(X) = 99k - \left(\frac{7}{3}\right)^2\)

Substituting \(k = \frac{1}{15}\), we get:

\(Var(X) = 99 \times \frac{1}{15} - \left(\frac{7}{3}\right)^2 = \frac{99}{15} - \frac{49}{9}\)

\(Var(X) = \frac{52}{45}\)

To solve this problem, we first need to determine the possible values of \(X\) and their probabilities.

The possible outcomes for the die are \(-1, -1, 0, 0, 1, 2\) and for the spinner are \(-1, 0, 1\). The possible values of \(X\) are the sums of these outcomes.

Possible values of \(X\) are: \(-2, -1, 0, 1, 2, 3\).

Next, we calculate the expected value \(E(X)\):

\(E(X) = \frac{-4 - 4 + 0 + 4 + 4 + 3}{18} = \frac{1}{6}\)

Now, we calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{8 + 4 + 0 + 4 + 8 + 9}{18} - \left(\frac{1}{6}\right)^2\)

\(= \frac{11}{6} - \frac{1}{36}\)

\(= \frac{65}{36}\)

Thus, \(\text{Var}(X) = \frac{65}{36}\) or approximately 1.81.

(i) To find the probability distribution of \(X\), calculate \(X = R - B\) for each combination of red (R) and blue (B) spinner outcomes. The possible values of \(X\) are -2, -1, 0, 1, 2, 3. The probability distribution table is:

(ii) The variance \(\text{Var}(X)\) is calculated using:

\(E(X) = \frac{-2 \times 1 + (-1) \times 2 + 0 \times 3 + 1 \times 3 + 2 \times 2 + 3 \times 1}{12} = 0.5\)

\(\text{Var}(X) = \frac{(-2)^2 \times 1 + (-1)^2 \times 2 + 0^2 \times 3 + 1^2 \times 3 + 2^2 \times 2 + 3^2 \times 1}{12} - (0.5)^2\)

\(= \frac{26}{12} - \frac{1}{4} = \frac{23}{12}\)

(iii) The probability that \(X = 1\) given \(X\) is non-zero is:

\(P(X \neq 0) = \frac{9}{12}\)

\(P(X = 1 \mid X \neq 0) = \frac{P(X = 1 \cap X \neq 0)}{P(X \neq 0)} = \frac{\frac{3}{12}}{\frac{9}{12}} = \frac{1}{3}\)

(i) The sum of all probabilities must equal 1:

\(6p + 0.1 = 1\)

Solving for \(p\):

\(6p = 0.9\)

\(p = 0.15\)

(ii) The variance \(\text{Var}(X)\) is calculated using:

\(\text{Var}(X) = (-1)^2 \times p + 0^2 \times p + 1^2 \times 2p + 2^2 \times 2p + 4^2 \times 0.1 - (1.15)^2\)

Substitute \(p = 0.15\):

\(\text{Var}(X) = 1 \times 0.15 + 0 + 1 \times 0.3 + 4 \times 0.3 + 16 \times 0.1 - 1.15^2\)

\(= 0.15 + 0 + 0.3 + 1.2 + 1.6 - 1.3225\)

\(= 1.9275\)

Rounded to three significant figures:

\(\text{Var}(X) = 1.93\)

(a) To find the probabilities, we use \(P(X = x) = k(x^2 - 1)\) for each value of \(x\):

• For \(x = -2\), \(P(X = -2) = k((-2)^2 - 1) = 3k\).
• For \(x = 2\), \(P(X = 2) = k(2^2 - 1) = 3k\).
• For \(x = 3\), \(P(X = 3) = k(3^2 - 1) = 8k\).

Since the total probability must be 1, we have:

\(3k + 3k + 8k = 1\)

\(14k = 1\)

\(k = \frac{1}{14}\)

Thus, the probability distribution table is:

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

\(E(X) = (-2) \times \frac{3}{14} + 2 \times \frac{3}{14} + 3 \times \frac{8}{14}\)

\(= \frac{-6}{14} + \frac{6}{14} + \frac{24}{14}\)

\(= \frac{12}{14} = \frac{12}{7}\)

To find \(\text{Var}(X)\), we use:

\(\text{Var}(X) = (-2)^2 \times \frac{3}{14} + 2^2 \times \frac{3}{14} + 3^2 \times \frac{8}{14} - (E(X))^2\)

\(= 4 \times \frac{3}{14} + 4 \times \frac{3}{14} + 9 \times \frac{8}{14} - \left(\frac{12}{7}\right)^2\)

\(= \frac{12 + 12 + 72}{14} - \frac{144}{49}\)

\(= \frac{96}{14} - \frac{144}{49}\)

\(= \frac{192}{49}\)

To solve this problem, we need to calculate the probability distribution for the number of heads obtained when the three coins are tossed.

Let the random variable X represent the number of heads obtained. The possible values of X are 0, 1, 2, and 3.

1. Probability of 0 heads, P(0):
\(P(0) = (1 - 0.4) \times (1 - 0.75) \times (1 - 0.5) = 0.6 \times 0.25 \times 0.5 = 0.075\)

2. Probability of 1 head, P(1):
\(P(1) = 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 + 0.6 \times 0.25 \times 0.5 = 0.35\)

3. Probability of 2 heads, P(2):
\(P(2) = 0.4 \times 0.75 \times 0.5 + 0.4 \times 0.25 \times 0.5 + 0.6 \times 0.75 \times 0.5 = 0.425\)

4. Probability of 3 heads, P(3):
\(P(3) = 0.4 \times 0.75 \times 0.5 = 0.15\)

The probability distribution table is:

To calculate the mean, E(X):
\(E(X) = 0 \times 0.075 + 1 \times 0.35 + 2 \times 0.425 + 3 \times 0.15 = 1.65\)

To calculate the variance, Var(X):
\(Var(X) = (0^2 \times 0.075) + (1^2 \times 0.35) + (2^2 \times 0.425) + (3^2 \times 0.15) - (1.65)^2 = 0.678\)

To solve this problem, we first need to determine the probability distribution of the random variable \(X\), which represents the number of cats chosen when 3 animals are selected from 5 dogs and 2 cats.

We can have 0, 1, or 2 cats chosen. The probabilities are calculated as follows:

• \(P(X = 0) = \frac{\binom{5}{3} \cdot \binom{2}{0}}{\binom{7}{3}} = \frac{\frac{5 \times 4 \times 3}{3 \times 2 \times 1}}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{2}{7}\)
• \(P(X = 1) = \frac{\binom{5}{2} \cdot \binom{2}{1}}{\binom{7}{3}} = \frac{\frac{5 \times 4}{2 \times 1} \cdot 2}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{4}{7}\)
• \(P(X = 2) = \frac{\binom{5}{1} \cdot \binom{2}{2}}{\binom{7}{3}} = \frac{5}{\frac{7 \times 6 \times 5}{3 \times 2 \times 1}} = \frac{1}{7}\)

The probability distribution table is:

Given \(E(X) = \frac{6}{7}\), we find \(\text{Var}(X)\) using the formula:

\(\text{Var}(X) = E(X^2) - [E(X)]^2\)

Calculate \(E(X^2)\):

\(E(X^2) = 0^2 \cdot \frac{2}{7} + 1^2 \cdot \frac{4}{7} + 2^2 \cdot \frac{1}{7} = 0 + \frac{4}{7} + \frac{4}{7} = \frac{8}{7}\)

Then, \(\text{Var}(X) = \frac{8}{7} - \left(\frac{6}{7}\right)^2 = \frac{8}{7} - \frac{36}{49} = \frac{56}{49} - \frac{36}{49} = \frac{20}{49}\)

Thus, \(\text{Var}(X) = \frac{20}{49}\) or approximately 0.408.

(i) To find the probability that the socks taken are of different colours, we can use the formula:

\(P(\text{different colours}) = P(RB) + P(BR)\)

Where:

• \(P(RB) = \frac{4}{12} \times \frac{8}{11}\)
• \(P(BR) = \frac{8}{12} \times \frac{4}{11}\)

Thus,

\(P(\text{different colours}) = \frac{4}{12} \times \frac{8}{11} + \frac{8}{12} \times \frac{4}{11} = \frac{16}{33}\)

(ii) The probability distribution table for \(X\), the number of red socks taken, is:

(iii) To find \(E(X)\), the expected value of \(X\), we use:

\(E(X) = 0 \times \frac{14}{33} + 1 \times \frac{16}{33} + 2 \times \frac{3}{33}\)

\(E(X) = \frac{16}{33} + \frac{6}{33} = \frac{22}{33} = \frac{2}{3}\)

First, calculate the possible values of X by squaring each face value and subtracting 4:

• For face 1: \(1^2 - 4 = -3\)
• For face 2: \(2^2 - 4 = 0\)
• For face 3: \(3^2 - 4 = 5\)
• For face 6: \(6^2 - 4 = 32\)

The probability distribution table is:

To find \(E(X)\):

\(E(X) = (-3) \cdot \frac{1}{6} + 0 \cdot \frac{1}{2} + 5 \cdot \frac{1}{6} + 32 \cdot \frac{1}{6} = \frac{-3}{6} + \frac{5}{6} + \frac{32}{6} = \frac{34}{6} = \frac{17}{3} \approx 5.67\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{(-3)^2}{6} + \frac{0^2}{2} + \frac{5^2}{6} + \frac{32^2}{6} - \left(\frac{34}{6}\right)^2\)

\(= \frac{9}{6} + \frac{25}{6} + \frac{1024}{6} - \left(\frac{34}{6}\right)^2\)

\(= 144 \left( \frac{1298}{9} \right)\)

(i) To find \(P(X = 3)\), we need the probability of drawing two red discs followed by a blue disc, i.e., \(P(RRB)\).

The probability of drawing the first red disc is \(\frac{2}{6}\), the second red disc is \(\frac{1}{5}\), and the first blue disc is \(\frac{4}{4}\).

Thus, \(P(X = 3) = \frac{2}{6} \times \frac{1}{5} \times \frac{4}{4} = \frac{1}{15}\).

(ii) The probability distribution table for \(X\) is as follows:

\(P(X = 1)\) is the probability of drawing a blue disc first, which is \(\frac{4}{6} = \frac{2}{3}\).

\(P(X = 2)\) is the probability of drawing one red disc followed by a blue disc, i.e., \(P(RB) = \frac{2}{6} \times \frac{4}{5} = \frac{4}{15}\).

\(P(X = 3)\) is already calculated as \(\frac{1}{15}\).

(i) The probability \(P(X = -2) = k(-2)^2 = 4k\) and \(P(X = 2) = k(2)^2 = 4k\). Thus, \(P(X = -2) = P(X = 2)\).

(ii) The probability distribution table is:

Sum of probabilities: \(4k + k + 4k + 16k = 1\)

\(25k = 1\)

\(k = \frac{1}{25}\)

(iii) \(E(X) = (-2)(4k) + (-1)(k) + (2)(4k) + (4)(16k)\)

\(E(X) = -8k - k + 8k + 64k = 63k\)

Substitute \(k = \frac{1}{25}\):

\(E(X) = 63 \times \frac{1}{25} = \frac{63}{25}\)

(i) To find \(P(X = 2)\), we need the combinations where the sum of the numbers from pack A and pack B equals 2. The possible combinations are:

• Card from A is 0 and card from B is 2: Probability = \(\frac{2}{10} \times \frac{4}{6} = \frac{8}{60} = \frac{2}{15}\)

Thus, \(P(X = 2) = \frac{2}{15}\).

(ii) The probability distribution table for \(X\) is constructed by considering all possible sums:

(iii) Given \(X = 3\), we find the probability that the card from pack A is 1. The combinations for \(X = 3\) are:

• Card from A is 1 and card from B is 2: Probability = \(\frac{5}{10} \times \frac{4}{6} = \frac{20}{60} = \frac{1}{3}\)
• Card from A is 3 and card from B is 0: Probability = \(\frac{4}{10} \times \frac{2}{6} = \frac{8}{60} = \frac{2}{15}\)

Total probability for \(X = 3\) is \(\frac{13}{30}\). The probability that the card from A is 1 given \(X = 3\) is:

\(P(A1 | \text{Sum 3}) = \frac{\frac{5}{10} \times \frac{4}{6}}{\frac{13}{30}} = \frac{10}{13}\)

(i) To find the probability that Noor chooses exactly one T-shirt, we use combinations. The total number of ways to choose 3 items from 12 is given by \(\binom{12}{3}\).

The number of ways to choose 1 T-shirt from 3 is \(\binom{3}{1}\), and the number of ways to choose 2 items from the remaining 9 (blouses and jumpers) is \(\binom{9}{2}\).

Thus, the probability is:

\(\frac{\binom{3}{1} \times \binom{9}{2}}{\binom{12}{3}} = \frac{3 \times 36}{220} = \frac{108}{220} = \frac{27}{55}\)

(ii) The probability distribution table for X is constructed by considering the possible values of X (0, 1, 2, 3) and their respective probabilities:

To find the expectation of the difference between the two scores, we first determine the probability distribution of the differences.

The expectation is calculated as:

\(E(X) = \sum (\text{difference} \times \text{probability})\)

\(E(X) = (0 \times \frac{6}{36}) + (1 \times \frac{10}{36}) + (2 \times \frac{8}{36}) + (3 \times \frac{6}{36}) + (4 \times \frac{4}{36}) + (5 \times \frac{2}{36})\)

\(E(X) = \frac{0 + 10 + 16 + 18 + 16 + 10}{36}\)

\(E(X) = \frac{70}{36}\)

\(E(X) \approx 1.94\)

(i) To find the probability distribution, consider all possible outcomes when two dice are thrown. There are 36 possible outcomes (6 sides on the first die and 6 sides on the second die).

- The score is 0 when both dice show the same number. There are 6 such outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, P(0) = \(\frac{6}{36}\).

- For scores 1 to 5, count the outcomes where the smaller number is 1, 2, 3, 4, or 5, respectively:

• Score 1: (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1) - 10 outcomes, so P(1) = \(\frac{10}{36}\).
• Score 2: (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2) - 8 outcomes, so P(2) = \(\frac{8}{36}\).
• Score 3: (3,4), (3,5), (3,6), (4,3), (5,3), (6,3) - 6 outcomes, so P(3) = \(\frac{6}{36}\).
• Score 4: (4,5), (4,6), (5,4), (6,4) - 4 outcomes, so P(4) = \(\frac{4}{36}\).
• Score 5: (5,6), (6,5) - 2 outcomes, so P(5) = \(\frac{2}{36}\).

(ii) The expected score is calculated using the formula for expected value: \(E(X) = \sum x_i p_i\).

\(E(X) = 0 \times \frac{6}{36} + 1 \times \frac{10}{36} + 2 \times \frac{8}{36} + 3 \times \frac{6}{36} + 4 \times \frac{4}{36} + 5 \times \frac{2}{36}\)

\(E(X) = \frac{0 + 10 + 16 + 18 + 16 + 10}{36} = \frac{70}{36} \approx 1.94\)

(a) To find \(P(X = 2)\), we consider the scenarios where the highest number is 2. The possible outcomes are (2,2,2), (2,2,1), (2,1,2), (1,2,2), (2,1,1), (1,2,1), and (1,1,2). There are 7 such outcomes.

The total number of outcomes when spinning three 4-sided spinners is \(4 \times 4 \times 4 = 64\).

Thus, \(P(X = 2) = \frac{7}{64}\).

(b) To complete the probability distribution table, we need to find \(P(X = 1)\) and \(P(X = 4)\).

\(P(X = 1)\) occurs only when all spinners show 1, which is (1,1,1). Thus, \(P(X = 1) = \frac{1}{64}\).

For \(P(X = 4)\), we use the fact that the probabilities must sum to 1:

\(P(X = 4) = 1 - \left( \frac{1}{64} + \frac{7}{64} + \frac{19}{64} \right) = \frac{37}{64}\).

(i) The probability distribution table is:

Since the total probability must equal 1, we have:

\(k + 2k + 3k + 4k = 1\)

\(10k = 1\)

Thus, \(k = \frac{1}{10}\).

(ii) The mean \(E(X)\) is calculated as:

\(E(X) = 1 \times \frac{1}{10} + 2 \times \frac{2}{10} + 3 \times \frac{3}{10} + 4 \times \frac{4}{10}\)

\(E(X) = \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{16}{10} = 3\)

The variance \(\text{Var}(X)\) is calculated as:

\(\text{Var}(X) = \left(1^2 \times \frac{1}{10} + 2^2 \times \frac{2}{10} + 3^2 \times \frac{3}{10} + 4^2 \times \frac{4}{10}\right) - E(X)^2\)

\(\text{Var}(X) = \left(\frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10}\right) - 3^2\)

\(\text{Var}(X) = \frac{100}{10} - 9 = 1\)

Let X be the random variable representing the number of green sweets taken. The possible values of X are 0, 1, and 2.

Calculate the probabilities:

\(P(X = 0) = P(B, B) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}\)

\(P(X = 1) = P(G, B) + P(B, G) = \frac{2}{7} \times \frac{5}{6} + \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}\)

\(P(X = 2) = P(G, G) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21}\)

Calculate the expected value:

\(E(X) = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21} = \frac{4}{7}\)

Calculate the variance:

\(\text{Var}(X) = 0^2 \times \frac{10}{21} + 1^2 \times \frac{10}{21} + 2^2 \times \frac{1}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\)

To find the probability distribution for the number of white roses chosen, we calculate the probabilities for choosing 0, 1, or 2 white roses.

There are a total of 10 roses (5 yellow, 3 red, 2 white).

Probability of choosing 0 white roses:

The probability of choosing 3 non-white roses is:

\(P(0) = \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{42}{90}\)

Probability of choosing 1 white rose:

The probability of choosing 1 white and 2 non-white roses is:

\(P(1) = \left( \frac{2}{10} \times \frac{8}{9} \times \frac{7}{8} \right) \times 3 = \frac{42}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Probability of choosing 2 white roses:

The probability of choosing 2 white and 1 non-white rose is:

\(P(2) = \left( \frac{2}{10} \times \frac{1}{9} \times \frac{8}{8} \right) \times 3 = \frac{6}{90}\)

The factor of 3 accounts for the different orders in which the roses can be chosen.

Thus, the probability distribution table is:

(i) To complete the table, calculate the sum of the numbers from Spinner A and Spinner B for each combination. For example, if Spinner A shows 1 and Spinner B shows -3, then \(X = 1 + (-3) = -2\). Complete the table similarly for all combinations.

(ii) To find the probability distribution of \(X\), count the frequency of each sum from the completed table and divide by the total number of outcomes (16). The probabilities are: \(-2: \frac{1}{16}, -1: \frac{2}{16}, 0: \frac{4}{16}, 1: \frac{3}{16}, 2: \frac{2}{16}, 3: \frac{3}{16}, 4: \frac{1}{16}\).

(iii) Calculate \(\text{Var}(X)\) using the formula \(\text{Var}(X) = E(X^2) - [E(X)]^2\). First, find \(E(X) = \sum x \cdot P(x) = 1\). Then, \(E(X^2) = \sum x^2 \cdot P(x) = \frac{62}{16}\). Thus, \(\text{Var}(X) = \frac{62}{16} - 1^2 = \frac{23}{8} = 2.875\).

(iv) To find the probability that \(X\) is even given \(X\) is positive, use conditional probability: \(P(\text{even} \mid \text{positive}) = \frac{P(\text{even and positive})}{P(\text{positive})}\). The even positive values are 2 and 4, with probabilities \(\frac{2}{16} + \frac{1}{16} = \frac{3}{16}\). The positive values are 1, 2, 3, and 4, with probabilities \(\frac{3}{16} + \frac{2}{16} + \frac{3}{16} + \frac{1}{16} = \frac{9}{16}\). Thus, \(P(\text{even} \mid \text{positive}) = \frac{3/16}{9/16} = \frac{5}{9}\).

(i) The tree diagram shows three stages, each with a probability of success \(S = 0.4\) and failure \(F = 0.6\). Each branch represents a login attempt.

(ii) The probability distribution is calculated as follows:

• \(P(X = 0) = 0.4\)
• \(P(X = 1) = 0.6 \times 0.4 = 0.24\)
• \(P(X = 2) = 0.6^2 \times 0.4 = 0.144\)
• \(P(X = 3) = 0.6^3 = 0.216\)

(iii) The expected number of unsuccessful attempts is calculated using \(E(X) = \sum x P(X = x)\):

\(E(X) = 0 \times 0.4 + 1 \times 0.24 + 2 \times 0.144 + 3 \times 0.216 = 1.176\)

(i) The probability that exactly one rabbit is white can be calculated by considering two cases: one white and one non-white rabbit.

Case 1: First rabbit is white, second is non-white: \(\frac{6}{9} \times \frac{3}{8}\).

Case 2: First rabbit is non-white, second is white: \(\frac{3}{9} \times \frac{6}{8}\).

Adding these probabilities gives: \(\frac{6}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{6}{8} = \frac{1}{2}\).

(ii) The probability distribution table is constructed as follows:

(iii) The expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{1}{12} + 1 \times \frac{1}{2} + 2 \times \frac{5}{12} = \frac{16}{12} = \frac{4}{3}\).

To find the probability distribution for \(S\), we list all possible combinations of 3 discs from the set \{1, 2, 4, 6, 7\}.

The possible combinations are: 124, 126, 127, 146, 147, 167, 246, 247, 267, 467.

For each combination, we determine the smallest number, \(S\).

Counting the occurrences of each smallest number:

• \(S = 1\): 124, 126, 127, 146, 147, 167 (6 combinations)
• \(S = 2\): 246, 247, 267 (3 combinations)
• \(S = 4\): 467 (1 combination)

The total number of combinations is 10.

The probability distribution is calculated as follows:

• \(P(S=1) = \frac{6}{10}\)
• \(P(S=2) = \frac{3}{10}\)
• \(P(S=4) = \frac{1}{10}\)

(i) To find \(P(X = 3)\), we consider the sequences where the process stops after 3 apples. The sequences are GRR and RGR. The probability for GRR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). The probability for RGR is \(\frac{2}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}\). Therefore, \(P(X = 3) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\).

(ii) The probability distribution table for \(X\) is:

(iii) To find the probability that all 3 peppers are orange given that at least 2 are orange, we use conditional probability:

\(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{P(3 \text{ orange })}{P(\text{at least 2 orange})}\).

\(P(3 \text{ orange }) = \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} = \frac{2}{7}\).

\(P(\text{at least 2 orange}) = P(2 \text{ orange, 1 yellow}) + P(3 \text{ orange})\).

\(P(2 \text{ orange, 1 yellow}) = \frac{5}{7} \times \frac{4}{6} \times \frac{2}{5} + \frac{5}{7} \times \frac{2}{6} \times \frac{4}{5} + \frac{2}{7} \times \frac{5}{6} \times \frac{4}{5} = \frac{6}{7}\).

Thus, \(P(3 \text{ orange } | \text{ at least 2 orange }) = \frac{\frac{2}{7}}{\frac{6}{7}} = \frac{1}{3}\).

Let \(X\) be the number of attempts Sharik makes until he gets the correct answer.

The possible values of \(X\) are 1, 2, and 3.

Probability that Sharik gets the correct answer on the first attempt (\(X = 1\)):

\(P(1) = \frac{1}{3}\)

Probability that Sharik gets the correct answer on the second attempt (\(X = 2\)):

First attempt is wrong and second attempt is correct:

\(P(2) = P(WC) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}\)

Probability that Sharik gets the correct answer on the third attempt (\(X = 3\)):

First two attempts are wrong and third attempt is correct:

\(P(3) = P(WWC) = \frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6}\)

Total probability for \(X = 2\) and \(X = 3\):

\(P(2) + P(3) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}\)

Probability distribution table:

Expected value \(E(X)\):

\(E(X) = 1 \times \frac{1}{3} + 2 \times \frac{1}{3} + 3 \times \frac{1}{3} = 2\)

(i) To find the probability of choosing exactly 2 hamsters, we use combinations. The total number of ways to choose 5 pets from 9 is \(\binom{9}{5}\). The number of ways to choose 2 hamsters from 3 is \(\binom{3}{2}\), and the number of ways to choose 3 rabbits from 6 is \(\binom{6}{3}\).

The probability is given by:

\(P(2) = \frac{\binom{3}{2} \times \binom{6}{3}}{\binom{9}{5}} = \frac{3 \times 20}{126} = \frac{60}{126} = \frac{10}{21}\)

(ii) The probability distribution table for X is constructed by calculating the probabilities for 0, 1, 2, and 3 hamsters:

• \(P(0) = \frac{\binom{6}{5} \times \binom{3}{0}}{\binom{9}{5}} = \frac{6}{126} = \frac{2}{42}\)
• \(P(1) = \frac{\binom{6}{4} \times \binom{3}{1}}{\binom{9}{5}} = \frac{45}{126} = \frac{15}{42}\)
• \(P(2) = \frac{\binom{6}{3} \times \binom{3}{2}}{\binom{9}{5}} = \frac{60}{126} = \frac{20}{42}\)
• \(P(3) = \frac{\binom{6}{2} \times \binom{3}{3}}{\binom{9}{5}} = \frac{15}{126} = \frac{5}{42}\)

The table is:

(a) To find \(P(B)\), consider the scenarios where at least one biased coin shows a head. The probability that a biased coin shows a tail is \(\frac{3}{4}\). The probability that both biased coins show tails is \(\left(\frac{3}{4}\right)^2 = \frac{9}{16}\). Therefore, \(P(B) = 1 - \frac{9}{16} = \frac{7}{16}\).

(b) To find \(P(A \mid B)\), use the formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). The event \(A \cap B\) occurs when all coins show heads or all show tails. The probability of all heads is \(\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32}\). The probability of all tails is \(\frac{1}{2} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{32}\). Thus, \(P(A \cap B) = \frac{1}{32}\). Therefore, \(P(A \mid B) = \frac{\frac{1}{32}}{\frac{7}{16}} = \frac{1}{14}\).

(c) The probability distribution for \(X\) is calculated as follows:

• \(P(X = 0) = \frac{9}{32}\)
• \(P(X = 1) = \frac{15}{32}\)
• \(P(X = 2) = \frac{7}{32}\)
• \(P(X = 3) = \frac{1}{32}\)

(i) To find the probability of obtaining exactly 1 head and 2 tails, consider the scenarios: HTT, THT, and TTH.

For HTT: \(P(H) = \frac{2}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(HTT) = \frac{2}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For THT: \(P(T) = \frac{1}{3}, P(H) = \frac{2}{3}\) for Coin A and \(P(T) = \frac{3}{4}\) for Coin B.

\(P(THT) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{2}{12} = \frac{1}{6}\).

For TTH: \(P(T) = \frac{1}{3}, P(T) = \frac{1}{3}\) for Coin A and \(P(H) = \frac{1}{4}\) for Coin B.

\(P(TTH) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\).

Total probability: \(\frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{12}{36} + \frac{1}{36} = \frac{13}{36}\).

(ii) The probability distribution table is:

(iii) The expectation \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{3}{36} + 1 \times \frac{13}{36} + 2 \times \frac{16}{36} + 3 \times \frac{4}{36}\).

\(E(X) = \frac{0}{36} + \frac{13}{36} + \frac{32}{36} + \frac{12}{36} = \frac{57}{36} = \frac{19}{12} \approx 1.58\).

(i) The probability of choosing exactly 2 paperback books is calculated using combinations. The number of ways to choose 2 paperback books from 6 is \(\binom{6}{2}\), and the number of ways to choose 2 hardback books from 2 is \(\binom{2}{2}\). The total number of ways to choose 4 books from 8 is \(\binom{8}{4}\).

Thus, the probability is:

\(P(\text{exactly 2}) = \frac{\binom{6}{2} \cdot \binom{2}{2}}{\binom{8}{4}} = \frac{15}{70} = \frac{3}{14}\)

(ii) The probability distribution table for X is:

(iii) Given \(E(X) = 3\), we find \(\text{Var}(X)\) using the formula \(\text{Var}(X) = \sum x^2 p - (E(X))^2\).

\(\text{Var}(X) = \frac{12}{14} + \frac{72}{14} + \frac{48}{14} - 3^2\)

\(= \frac{3}{7}\)

To find the probability distribution of \(X\), we consider the possible values of \(X\): 0, 1, or 2.

Case 1: \(X = 0\)

This means no 5s are chosen. There are 6 non-5 digits (1s and 3s) and 3 5s. The probability of choosing two non-5s is:

\(P(0) = \frac{6}{9} \times \frac{5}{8} = \frac{30}{72} = \frac{5}{12}\)

Case 2: \(X = 1\)

This means one 5 is chosen. From part (ii), \(P(1) = 0.5\).

Case 3: \(X = 2\)

This means two 5s are chosen. From part (i), \(P(2) = \frac{6}{72} = \frac{1}{12}\).

The probability distribution table is:

(i) To find the probability that the sum of the numbers on the three cards is 11, consider the possible combinations: (3, 4, 4), (4, 3, 4), or (4, 4, 3). The probability for each combination is calculated as follows:

\(\text{Prob} = \left( \frac{4}{10} \times \frac{6}{9} \times \frac{5}{8} \right) \times 3C1 = \frac{360}{720} = \frac{1}{2}\)

Alternatively, using combinations:

\(\frac{\binom{6}{2} \times \binom{4}{1}}{\binom{10}{3}} = \frac{1}{2}\)

(ii) The probability distribution table for the sum of the numbers on the three cards is:

Calculations for each sum:

• \(P(3, 3, 3) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{24}{720}\)
• \(P(3, 3, 4) = \frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} \times 3C1 = \frac{216}{720}\)
• \(P(4, 4, 4) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720}\)

(i) If the coin shows a head, the smallest sum of two dice throws is 2 (1+1). Therefore, X = 1 can only occur if the coin shows a tail and the first die shows 1. The probability of a tail is \(\frac{1}{2}\) and the probability of rolling a 1 is \(\frac{1}{4}\). Thus, \(P(X = 1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

(ii) For X = 3, if the coin shows a head, possible outcomes are (1,2) and (2,1), each with probability \(\frac{1}{16}\). If the coin shows a tail, the first die must show 3, with probability \(\frac{1}{8}\). Therefore, \(P(X = 3) = \frac{2}{32} + \frac{1}{8} = \frac{3}{16}\).

(iii) Completing the table:

(iv) Events Q and R are exclusive because if a tail is thrown, X cannot be 7. Thus, \(P(Q \cap R) = 0\).

The possible values of the random variable X are 0, 1, and 2.

To find the probability distribution, consider the following cases:

1. P(X = 0): No chocolates are taken. This occurs if a toffee or boiled sweet is taken from Susan's bag and then a toffee or boiled sweet is taken from Ahmad's bag.

\(P(T, B) + P(T, T) = \frac{5}{12} \times \frac{2}{10} + \frac{5}{12} \times \frac{5}{10} = \frac{7}{24} \approx 0.292\)

\(2. P(X = 2): Two chocolates are taken. This occurs if a chocolate is taken from Susan's bag and then a chocolate is taken from Ahmad's bag.\)

\(P(C, C) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120} = \frac{7}{30} \approx 0.233\)

\(3. P(X = 1): One chocolate is taken. This is the complement of the other two probabilities.\)

\(P(X = 1) = 1 - \frac{7}{24} - \frac{7}{30} = \frac{19}{40} \approx 0.475\)

Thus, the probability distribution is:

(i) To find \(P(X = 9)\), we consider the combinations of scores that sum to 9: (1, 4, 4), (2, 3, 4), and (3, 3, 3). The probabilities are calculated as follows:

\(P(1, 4, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 3 = \frac{3}{64}\)

\(P(2, 3, 4) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 6 = \frac{6}{64}\)

\(P(3, 3, 3) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\)

Adding these probabilities gives \(P(X = 9) = \frac{3}{64} + \frac{6}{64} + \frac{1}{64} = \frac{10}{64}\).

(ii) The completed probability distribution table is:

(iii) To determine if \(R\) and \(S\) are independent, we calculate:

\(P(S) = \frac{6}{64}\)

\(P(R \cap S) = \frac{3}{64}\)

Since \(P(R \cap S) \neq P(R) \times P(S)\), events \(R\) and \(S\) are not independent.

To find the probability distribution for the number of green pens Rod takes, we calculate the probabilities for 0, 1, 2, and 3 green pens.

Probability of 0 green pens:

\(P(0) = \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} = \frac{210}{720}\)

Probability of 1 green pen:

\(P(1) = \frac{3}{10} \times \frac{7}{9} \times \frac{6}{8} \times \binom{3}{1} = \frac{378}{720}\)

Probability of 2 green pens:

\(P(2) = \frac{3}{10} \times \frac{2}{9} \times \frac{7}{8} \times \binom{3}{2} = \frac{126}{720}\)

Probability of 3 green pens:

\(P(3) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720}\)

These probabilities sum to 1, confirming the distribution is correct.

(i) To find the mean of W, calculate the expected value:

Mean = \(\frac{1+1+1+2+3+3}{6} = \frac{11}{6} \approx 1.83\)

For the standard deviation, use:

SD = \(\sqrt{\frac{(1-1.83)^2 + (1-1.83)^2 + (1-1.83)^2 + (2-1.83)^2 + (3-1.83)^2 + (3-1.83)^2}{6}}\)

SD = \(\sqrt{\frac{29}{6}} \approx 0.898\)

(ii) The probability distribution table for X is calculated by considering all possible sums of two throws:

(iii) Given \(E(Y) = 8\), and \(Y\) follows a binomial distribution with \(p = \frac{1}{3}\), we have:

\(np = 8 \Rightarrow n = 24\)

Variance is given by \(Var(Y) = np(1-p) = 24 \times \frac{1}{3} \times \frac{2}{3} = \frac{16}{3} \approx 5.33\)

(i) To find the probability distribution of Y, consider the possible outcomes when two values of X are chosen:

• Y = 0: This occurs when both values are the same. The probabilities are:
  • P(2, 2) = 0.5 × 0.5 = 0.25
  • P(4, 4) = 0.4 × 0.4 = 0.16
  • P(6, 6) = 0.1 × 0.1 = 0.01
  Total probability for Y = 0 is 0.25 + 0.16 + 0.01 = 0.42.
• Y = 2: This occurs when the values are (2, 4), (4, 2), (4, 6), or (6, 4). The probabilities are:
  • P(2, 4) = 0.5 × 0.4 = 0.2
  • P(4, 2) = 0.4 × 0.5 = 0.2
  • P(4, 6) = 0.4 × 0.1 = 0.04
  • P(6, 4) = 0.1 × 0.4 = 0.04
  Total probability for Y = 2 is 0.2 + 0.2 + 0.04 + 0.04 = 0.48.
• Y = 4: This occurs when the values are (2, 6) or (6, 2). The probabilities are:
  • P(2, 6) = 0.5 × 0.1 = 0.05
  • P(6, 2) = 0.1 × 0.5 = 0.05
  Total probability for Y = 4 is 0.05 + 0.05 = 0.1.

The probability distribution table for Y is:

(ii) The expected value of Y is calculated as:

\(E(Y) = (0 imes 0.42) + (2 imes 0.48) + (4 imes 0.1) = 0 + 0.96 + 0.4 = 1.36\)

To find the probability distribution, we first determine \(k\) by ensuring the probabilities sum to 1:

\(k(-2)^2 + k(1)^2 + k(2)^2 + k(3)^2 = 1\)

\(4k + k + 4k + 9k = 18k = 1\)

\(k = \frac{1}{18}\)

Thus, the probability distribution table is:

To find \(E(X)\):

\(E(X) = \frac{4 \times (-2) + 1 \times 1 + 4 \times 2 + 9 \times 3}{18}\)

\(= \frac{-8 + 1 + 8 + 27}{18}\)

\(= \frac{28}{18} = \frac{14}{9}\)

To find \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{4 \times (-2)^2 + 1 \times 1^2 + 4 \times 2^2 + 9 \times 3^2}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{16 + 1 + 16 + 81}{18} - \left(\frac{14}{9}\right)^2\)

\(= \frac{114}{18} - \frac{196}{81}\)

\(= \frac{317}{81}\)

Let the probability of choosing a rope of length 3 m be denoted as \(P(3)\) and the probability of choosing a rope of length 5 m be \(P(5)\).

Given that there are four times as many ropes of length 3 m as there are ropes of length 5 m, we have:

\(P(3) = \frac{4}{5} \quad \text{and} \quad P(5) = \frac{1}{5}\)

The expectation \(E(X)\) is calculated as:

\(E(X) = 3 \times P(3) + 5 \times P(5) = 3 \times \frac{4}{5} + 5 \times \frac{1}{5} = \frac{12}{5} + \frac{5}{5} = \frac{17}{5} = 3.4\)

The variance \(\text{Var}(X)\) is calculated as:

First, find \(E(X^2)\):

\(E(X^2) = 3^2 \times P(3) + 5^2 \times P(5) = 9 \times \frac{4}{5} + 25 \times \frac{1}{5} = \frac{36}{5} + \frac{25}{5} = \frac{61}{5}\)

Then, use the formula for variance:

\(\text{Var}(X) = E(X^2) - (E(X))^2 = \frac{61}{5} - \left(\frac{17}{5}\right)^2 = \frac{61}{5} - \frac{289}{25}\)

Convert \(\frac{61}{5}\) to \(\frac{305}{25}\) and subtract:

\(\text{Var}(X) = \frac{305}{25} - \frac{289}{25} = \frac{16}{25} = 0.64\)

To solve this problem, we first need to determine the probability distribution of \(X\), the number of girls in the team.

We have 3 boys and 5 girls, and we are choosing a team of 4. The possible values for \(X\) are 1, 2, 3, and 4.

For \(X = 1\):

We choose 1 girl and 3 boys. The probability is given by:

\(P(X=1) = \frac{\binom{5}{1} \cdot \binom{3}{3}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

For \(X = 2\):

We choose 2 girls and 2 boys. The probability is given by:

\(P(X=2) = \frac{\binom{5}{2} \cdot \binom{3}{2}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 3\):

We choose 3 girls and 1 boy. The probability is given by:

\(P(X=3) = \frac{\binom{5}{3} \cdot \binom{3}{1}}{\binom{8}{4}} = \frac{10 \cdot 3}{70} = \frac{3}{7}\)

For \(X = 4\):

We choose 4 girls. The probability is given by:

\(P(X=4) = \frac{\binom{5}{4} \cdot \binom{3}{0}}{\binom{8}{4}} = \frac{5 \cdot 1}{70} = \frac{1}{14}\)

Thus, the probability distribution table is:

Given \(E(X) = \frac{5}{2}\), we calculate \(\text{Var}(X)\) using the formula:

\(\text{Var}(X) = E(X^2) - (E(X))^2\)

First, calculate \(E(X^2)\):

\(E(X^2) = 1^2 \cdot \frac{1}{14} + 2^2 \cdot \frac{3}{7} + 3^2 \cdot \frac{3}{7} + 4^2 \cdot \frac{1}{14}\)

\(= \frac{1}{14} + \frac{12}{7} + \frac{27}{7} + \frac{16}{14}\)

\(= \frac{1}{14} + \frac{24}{14} + \frac{54}{14} + \frac{16}{14}\)

\(= \frac{95}{14}\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = \frac{95}{14} - \left(\frac{5}{2}\right)^2\)

\(= \frac{95}{14} - \frac{25}{4}\)

\(= \frac{95}{14} - \frac{87.5}{14}\)

\(= \frac{15}{28}\)

Thus, \(\text{Var}(X) = \frac{15}{28}\) (0.536).

(i) The score is 6 if the cards drawn are (3, 9) or (9, 3). The probability of each pair is \(\frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\). Therefore, the probability of a score of 6 is \(2 \times \frac{1}{25} = \frac{2}{25} = 0.08\).

(ii) The probability distribution table is:

(iii) The mean score is calculated as \(\sum x p = 0 \times 0.2 + 1 \times 0.24 + 2 \times 0.08 + 3 \times 0.08 + 4 \times 0.16 + 5 \times 0.16 + 6 \times 0.08 = 2.56\).

(iv) Judy wins if the score is 4, 5, or 6. The probability of these scores is \(0.16 + 0.16 + 0.08 = 0.4\). The probability of a draw (score 0) is 0.2. Therefore, the probability that Judy wins with the second choice is \(0.2 \times 0.4 = 0.08\).

(v) The probability that Judy wins with the \(n\)th choice is \((0.2)^{n-1} \times 0.4\).

(i) The probability distribution table for \(X\) is:

Since the sum of all probabilities must equal 1, we have:

\(k + 2k + 3k + 4k + 5k = 1\)

\(15k = 1\)

\(k = \frac{1}{15}\) (0.0667)

(ii) To find \(E(X)\), we calculate:

\(E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 3k + 4 \cdot 4k + 5 \cdot 5k\)

\(= k + 4k + 9k + 16k + 25k\)

\(= 55k\)

Substituting \(k = \frac{1}{15}\), we get:

\(E(X) = 55 \times \frac{1}{15} = \frac{55}{15} = \frac{11}{3}\) (3.67)

(i) To find \(P(X = 2)\), consider the combinations: \((0, 2)\) and \((2, 0)\).

\(P(0, 2) = \frac{6}{10} \times \frac{3}{7}\)

\(P(2, 0) = \frac{3}{10} \times \frac{4}{7}\)

\(P(X = 2) = \frac{6}{10} \times \frac{3}{7} + \frac{3}{10} \times \frac{4}{7} = \frac{30}{70} = \frac{3}{7}\)

(ii) The probability distribution of X is:

(iii) To find \(E(X)\) and \(\text{Var}(X)\):

\(E(X) = \sum x \cdot P(X = x) = \frac{13}{7}\)

\(\text{Var}(X) = \sum x^2 \cdot P(X = x) - (E(X))^2\)

\(\text{Var}(X) = \frac{120}{70} + \frac{208}{70} + \frac{108}{70} - \left(\frac{13}{7}\right)^2 = 2.78\)

(iv) Given \(X = 2\), find \(P(A = 2)\):

\(P(A = 2 \mid X = 2) = \frac{P(A = 2 \cap X = 2)}{P(X = 2)} = \frac{\frac{3}{10} \times \frac{4}{7}}{\frac{30}{70}} = 0.4\)

(i) The possible values of \(X\) are 0, 1, and 2. The probability distribution is:

(ii) To find \(E(X)\), use the formula:

\(E(X) = \sum x P(X = x) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = \frac{8}{7}\).

To find \(\text{Var}(X)\), use the formula:

\(\text{Var}(X) = E(X^2) - (E(X))^2\).

First, calculate \(E(X^2)\):

\(E(X^2) = 0^2 \times \frac{1}{7} + 1^2 \times \frac{4}{7} + 2^2 \times \frac{2}{7} = \frac{12}{7}\).

Then, \(\text{Var}(X) = \frac{12}{7} - \left(\frac{8}{7}\right)^2 = \frac{12}{7} - \frac{64}{49} = \frac{20}{49}\).

(i) To find \(P(X = 40)\), we need to find the value of \(r\) such that \(\frac{120}{r} = 40\). Solving for \(r\), we get:

\(40 = \frac{120}{r}\)

\(r = \frac{120}{40} = 3\)

Now, substitute \(r = 3\) into the probability formula:

\(P(X = 40) = \frac{3}{45} = \frac{1}{15}\)

(ii) Construct the probability distribution table for \(X\):

(iii) The modal value of \(X\) is the value with the highest probability. From the table, \(X = 40\) has the highest probability of \(\frac{3}{45}\).

(iv) To find the probability that \(X\) lies between 18 and 100, sum the probabilities for \(X = 20, 24, 30, 40, 60\):

\(P(18 < X < 100) = \frac{6}{45} + \frac{5}{45} + \frac{4}{45} + \frac{3}{45} + \frac{2}{45}\)

\(= \frac{20}{45} = \frac{4}{9}\)

(ii) To find the probability distribution of \(X\), count the occurrences of each sum in the table and divide by the total number of outcomes (36). The probabilities are:

(iii) The expected value \(E(X)\) is calculated as:

\(E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + 6 \times \frac{5}{36} + 7 \times \frac{4}{36} + 8 \times \frac{4}{36} + 9 \times \frac{4}{36} + 10 \times \frac{4}{36} + 11 \times \frac{8}{36} + 12 \times \frac{4}{36}\)

\(= \frac{312}{36} = \frac{26}{3} \approx 8.67\)

(iv) The probability that \(X\) is greater than \(E(X)\) is:

\(P(X > 8.67) = P(X = 9, 10, 11, 12) = \frac{4}{36} + \frac{4}{36} + \frac{8}{36} + \frac{4}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.556\)

(i) The tree diagram should have branches for each call attempt with probabilities 0.5 for Answered (A) and 0.5 for Unanswered (U). There should be 4 levels of branches, representing up to 4 attempts.

(ii) The probability distribution table is completed as follows:

(iii) The expected number of unanswered calls \(E(X)\) is calculated as:

\(E(X) = \sum x \cdot P(X = x) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{16} + 4 \cdot \frac{1}{16}\)

\(E(X) = 0 + \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{16}\)

\(E(X) = \frac{1}{4} + \frac{1}{4} + \frac{3}{16} + \frac{4}{16}\)

\(E(X) = \frac{1}{2} + \frac{7}{16} = \frac{15}{16}\)

Thus, \(E(X) = \frac{15}{16}\) or approximately 0.938 or 0.9375.

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:

To find the probability distribution of L, we consider all possible combinations of drawing 3 balls from the set {1, 2, 3, 4, 5} and determine the largest number in each combination.

The possible values for L are 3, 4, and 5. The probabilities are calculated as follows:

For the expectation, we use the formula:

\(E(L) = \sum l \cdot P(L = l) = 3 \times 0.1 + 4 \times 0.3 + 5 \times 0.6 = 4.5\)

For the variance, we use the formula:

\(\text{Var}(L) = \sum l^2 \cdot P(L = l) - (E(L))^2 = 3^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.6 - 4.5^2 = 0.45\)

(i) To find the probability distribution of \(X\), we consider all possible outcomes of throwing two dice. There are a total of 36 outcomes.

- For \(x = 1\), the outcomes are \((1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1)\). There are 11 outcomes, so \(P(X = 1) = \frac{11}{36}\).

- For \(x = 2\), the outcomes are \((2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (4,2), (5,2), (6,2)\). There are 9 outcomes, so \(P(X = 2) = \frac{9}{36}\).

- For \(x = 3\), the outcomes are \((3,3), (3,4), (3,5), (3,6), (4,3), (5,3), (6,3)\). There are 7 outcomes, so \(P(X = 3) = \frac{7}{36}\).

- For \(x = 4\), the outcomes are \((4,4), (4,5), (4,6), (5,4), (6,4)\). There are 5 outcomes, so \(P(X = 4) = \frac{5}{36}\).

- For \(x = 5\), the outcomes are \((5,5), (5,6), (6,5)\). There are 3 outcomes, so \(P(X = 5) = \frac{3}{36}\).

- For \(x = 6\), the outcome is \((6,6)\). There is 1 outcome, so \(P(X = 6) = \frac{1}{36}\).

(ii) To find \(\mathbb{E}(X)\), we calculate:

\(\mathbb{E}(X) = 1 \times \frac{11}{36} + 2 \times \frac{9}{36} + 3 \times \frac{7}{36} + 4 \times \frac{5}{36} + 5 \times \frac{3}{36} + 6 \times \frac{1}{36}\)

\(= \frac{11}{36} + \frac{18}{36} + \frac{21}{36} + \frac{20}{36} + \frac{15}{36} + \frac{6}{36}\)

\(= \frac{91}{36}\)

To solve part (ii), we need to find the probability distribution for the number of new pens in a sample of two pens.

Let \(X\) be the random variable representing the number of new pens in the sample. \(X\) can take values 0, 1, or 2.

1. Probability of 0 new pens: \(P(X = 0) = \frac{7}{10} \times \frac{6}{9} = \frac{7}{15}\).

2. Probability of 1 new pen: \(P(X = 1) = \left( \frac{3}{10} \times \frac{7}{9} \right) + \left( \frac{7}{10} \times \frac{3}{9} \right) = \frac{7}{15}\).

3. Probability of 2 new pens: \(P(X = 2) = \frac{3}{10} \times \frac{2}{9} = \frac{1}{15}\).

For part (iii), calculate the expected value \(E(X)\):

\(E(X) = 0 \times \frac{7}{15} + 1 \times \frac{7}{15} + 2 \times \frac{1}{15} = \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}\).

(i) The possible scores on the die are 1, 2, 3, and 4. The corresponding areas A are 1, 4, 9, and 16. The probability distribution is:

(ii) To find the expected value E(A), calculate:

\(E(A) = 1 \times \frac{1}{2} + 4 \times \frac{1}{6} + 9 \times \frac{1}{6} + 16 \times \frac{1}{6} = 5.33.\)

To find the variance Var(A), calculate:

\(Var(A) = (1^2 \times \frac{1}{2} + 4^2 \times \frac{1}{6} + 9^2 \times \frac{1}{6} + 16^2 \times \frac{1}{6}) - (5.33)^2 = 30.9.\)

(a) To find the probability distribution of \(X\), we calculate all possible sums of the numbers from the red and blue spinners and their probabilities:

(b) Given \(E(X) = 0.25\), we use the formula for variance:

\(\text{Var}(X) = E(X^2) - [E(X)]^2\)

Calculate \(E(X^2)\):

\(E(X^2) = \frac{1}{16}(-2)^2 + \frac{3}{16}(-1)^2 + \frac{5}{16}(0)^2 + \frac{5}{16}(1)^2 + \frac{2}{16}(2)^2\)

\(= \frac{1}{16} \times 4 + \frac{3}{16} \times 1 + \frac{5}{16} \times 0 + \frac{5}{16} \times 1 + \frac{2}{16} \times 4\)

\(= \frac{4}{16} + \frac{3}{16} + 0 + \frac{5}{16} + \frac{8}{16} = \frac{20}{16} = 1.25\)

Now, calculate \(\text{Var}(X)\):

\(\text{Var}(X) = 1.25 - (0.25)^2 = 1.25 - 0.0625 = 1.1875\)

(a) To find the probability that exactly one marble is yellow, consider the arrangements: YGG, GYG, GGY. The probability for each arrangement is:

\(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7}\)

Multiply by 3 for the three arrangements:

\(3 \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} = \frac{180}{504} = \frac{5}{14}\)

(b) The probability distribution table for \(X\) is:

(c) The expected value \(E(X)\) is calculated as:

\(E(X) = 0 \times \frac{1}{21} + 1 \times \frac{5}{14} + 2 \times \frac{10}{21} + 3 \times \frac{5}{42} = 1.67\)