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

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

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