Since the coin is fair, the probability of getting heads on each toss is 1/2. The probability of getting two heads is the product of these probabilities, which is (1/2) * (1/2) = 1/4.

The probability of selecting a blue ball is the number of blue balls divided by the total number of balls. So, the probability is 3 / (5 + 3 + 2) = 3/10.

The sum of probabilities for all possible values of X must be equal to 1. So, we have: P(X = 1) + P(X = 2) + P(X = 3) = k * (1^2) + k * (2^2) + k * (3^2) = k * (1 + 4 + 9) = k * 14 = 1. Therefore, k = 1/14.

Since A and B are independent, the probability of both occurring is the product of their individual probabilities. So, P(A and B) = P(A) * P(B) = 0.4 * 0.6 = 0.24.

The probability of selecting a defective item on the first draw is 10/100. After the first draw, there are 9 defective items and 89 non-defective items remaining. The probability of selecting a defective item on the second draw is 9/99. Therefore, the probability of selecting two defective items is (10/100) * (9/99) = 1/100.

The probability of preferring coffee or tea is the sum of the probabilities of preferring coffee and preferring tea, minus the probability of preferring both. So, P(coffee or tea) = P(coffee) + P(tea) - P(coffee and tea) = 0.6 + 0.3 - 0.1 = 0.9.

The probability that X is less than or equal to 0.5 is given by the integral of f(x) from 0 to 0.5: P(X ≤ 0.5) = ∫[0, 0.5] 2x dx = [x^2]_[0, 0.5] = 0.5^2 - 0^2 = 0.25.

The probability that X is between 6 and 14 is given by the area under the normal distribution curve between these values. We can use the standard normal distribution (Z) to calculate this probability: P(6 ≤ X ≤ 14) = P((6 - 10)/2 ≤ Z ≤ (14 - 10)/2) = P(-2 ≤ Z ≤ 2) = 0.6827.

We can use the standard normal distribution (Z) to calculate this probability: P(X > 100) = P((X - 95)/10 > (100 - 95)/10) = P(Z > 0.5) = 1 - P(Z ≤ 0.5) = 1 - 0.6915 = 0.1587.

The probability of selecting a red ball on the first draw is 4/9. After the first draw, there are 3 red balls, 3 blue balls, and 2 green balls remaining. The probability of selecting a blue ball on the second draw is 3/8. Therefore, the probability of selecting a red ball on the first draw and a blue ball on the second draw is (4/9) * (3/8) = 3/35.

There are 36 possible outcomes when rolling a fair six-sided die twice. The outcomes that sum to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). So, the probability of getting a sum of 7 is 6/36 = 1/6.

The probability of preferring only coffee is the probability of preferring coffee minus the probability of preferring both coffee and tea. So, P(coffee only) = P(coffee) - P(coffee and tea) = 0.4 - 0.2 = 0.2.

The expected value of X is given by E(X) = Σ[x * P(X = x)] for all possible values of X. So, E(X) = 1 * (1/6) + 2 * (1/6) + 3 * (1/6) + 4 * (1/6) + 5 * (1/6) + 6 * (1/6) = 3.5.

The variance of X is given by Var(X) = E(X^2) - [E(X)]^2. First, we find E(X^2): E(X^2) = ∫[0, 1] 3x^2 * x dx = ∫[0, 1] 3x^3 dx = [3x^4/4]_[0, 1] = 3/4. Then, we find [E(X)]^2: [E(X)]^2 = (3.5)^2 = 12.25. Therefore, Var(X) = 3/4 - 12.25 = 0.25.

We can use the standard normal distribution (Z) to calculate this probability: P(X < 80) = P((X - 95)/10 < (80 - 95)/10) = P(Z < -1.5) = 0.0228.