There are 26 red cards in a standard deck of 52 cards. Therefore, the probability of drawing a red card is 26/52 = 1/2.

There are 8 possible outcomes when a coin is tossed three times: HHH, HHT, HTH, THH, THT, TTH, HTT, TTT. There are 3 outcomes with exactly two heads: HHT, HTH, THH. Therefore, the probability of getting exactly two heads is 3/8.

There are a total of 10 + 15 + 20 = 45 balls in the bag. The probability of selecting a red ball is 10/45. Therefore, the probability of selecting a ball that is not red is 1 - 10/45 = 4/5.

Let C be the event that a person likes chocolate and V be the event that a person likes vanilla. Then, P(C or V) = P(C) + P(V) - P(C and V) = 60/100 + 40/100 - 20/100 = 80/100 = 0.8.

The probability of the company winning at least one contract is 1 - the probability of the company winning no contracts. The probability of the company winning no contracts is (0.9)^10 = 0.3487. Therefore, the probability of the company winning at least one contract is 1 - 0.3487 = 0.95.

This is a binomial distribution problem. The probability of exactly 5 defective items is given by the formula P(X = 5) = (100 choose 5) * (0.05)^5 * (0.95)^95 = 0.201.

This is a hypergeometric distribution problem. The probability of exactly 2 defective light bulbs is given by the formula P(X = 2) = (10 choose 2) * (10/100)^2 * (90/100)^8 = 0.308.

The sample mean follows a normal distribution with a mean of 100 and a standard deviation of 15/sqrt(100) = 1.5. Therefore, the probability that the sample mean will be between 95 and 105 is P(95 < X < 105) = P((95 - 100)/1.5 < Z < (105 - 100)/1.5) = P(-3.33 < Z < 3.33) = 0.841.

The product lifetime follows a normal distribution with a mean of 120 hours and a standard deviation of 10 hours. Therefore, the probability that a randomly selected product will last for less than 100 hours is P(X < 100) = P((100 - 120)/10 < Z) = P(Z < -2) = 0.023.

Let C be the event that a person prefers coffee and T be the event that a person prefers tea. Then, P(C or T) = P(C) + P(T) - P(C and T) = 0.6 + 0.3 - 0.1 = 0.8.

The employee's vacation days follow a normal distribution with a mean of 12 days and a standard deviation of 2 days. Therefore, the probability that an employee will take less than 10 days of vacation in a given year is P(X < 10) = P((10 - 12)/2 < Z) = P(Z < -1) = 0.023.

This is a binomial distribution problem. The probability of between 5 and 10 defective items is given by the formula P(5 ≤ X ≤ 10) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.383.

Let D be the event that a person prefers dogs and C be the event that a person prefers cats. Then, P(D or C) = P(D) + P(C) - P(D and C) = 0.4 + 0.3 - 0.2 = 0.9.

The employee's vacation days follow a normal distribution with a mean of 12 days and a standard deviation of 2 days. Therefore, the probability that an employee will take more than 15 days of vacation in a given year is P(X > 15) = P((15 - 12)/2 < Z) = P(Z > 1.5) = 0.067.