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

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

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

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

\(= 7.3 - 6.25\)

\(= 1.05\)