Discrete Probability

There are 4 aces in a standard deck of 52 cards. So, the probability of drawing an ace is 4/52 = 1/13.

The total number of balls in the bag is 10 + 15 + 5 = 30. The number of balls that are not green is 10 + 15 = 25. So, the probability of drawing a ball that is not green is 25/30.

There are four possible outcomes when a coin is tossed twice: HH, HT, TH, TT. Only one of these outcomes, HH, results in two heads. So, the probability of getting two heads is 1/4.

The probability of drawing a red ball on the first draw is 6/12 = 1/2. After the first draw, there are 5 red balls and 11 total balls remaining in the box. So, the probability of drawing a red ball on the second draw is 5/11. The probability of both events occurring is (1/2) * (5/11) = 1/5.

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

The probability of selecting a woman on the first draw is 6/10. After the first draw, there are 5 women and 9 total employees remaining. So, the probability of selecting a woman on the second draw is 5/9. After the second draw, there are 4 women and 8 total employees remaining. So, the probability of selecting a woman on the third draw is 4/8. The probability of all three events occurring is (6/10) * (5/9) * (4/8) = 1/20.

The probability of drawing a white ball on the first draw is 10/30. After the first draw, there are 9 white balls and 29 total balls remaining in the bag. So, the probability of drawing a white ball on the second draw is 9/29. After the second draw, there are 8 white balls and 28 total balls remaining in the bag. So, the probability of drawing a white ball on the third draw is 8/28. After the third draw, there are 7 white balls and 27 total balls remaining in the bag. So, the probability of drawing a white ball on the fourth draw is 7/27. The probability of exactly 2 of the balls being white is (10/30) * (9/29) * (8/28) * (19/27) = 2/7.

There are 36 possible outcomes when a fair six-sided die is rolled twice. The outcomes that result in a sum of 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 that none of the employees selected are women is (4/12) * (3/11) * (2/10) * (1/9) * (0/8) = 0. The probability that at least one of the employees selected is a woman is 1 - 0 = 1.

The probability of drawing a red ball on the first draw is 10/30. After the first draw, there are 9 red balls, 15 blue balls, and 4 green balls remaining in the bag. So, the probability of drawing a blue ball on the second draw is 15/29. After the second draw, there are 9 red balls, 14 blue balls, and 4 green balls remaining in the bag. So, the probability of drawing a green ball on the third draw is 4/28. The probability of all three events occurring is (10/30) * (15/29) * (4/28) = 3/14.

The probability of getting a head on a single toss is 1/2. The probability of getting a tail on a single toss is also 1/2. The probability of getting exactly 5 heads in 10 tosses is given by the binomial distribution: P(X = 5) = (10 choose 5) * (1/2)^5 * (1/2)^5 = 252 * 1/32 = 1/256.

The probability of drawing a red ball on the first draw is 12/36. After the first draw, there are 11 red balls, 18 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a red ball on the second draw is 11/35. After the second draw, there are 10 red balls, 18 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a blue ball on the third draw is 18/34. After the third draw, there are 10 red balls, 17 blue balls, and 6 green balls remaining in the box. So, the probability of drawing a blue ball on the fourth draw is 17/33. The probability of all four events occurring is (12/36) * (11/35) * (18/34) * (17/33) = 2/7.

The probability of selecting 3 women on the first, second, and third draws is (12/20) * (11/19) * (10/18) = 110/342. The probability of selecting 2 men on the fourth and fifth draws is (8/17) * (7/16) = 56/272. The probability of all five events occurring is (110/342) * (56/272) = 3/10.

There are 5 possible outcomes where there are more red balls than blue balls: (3 red, 1 blue), (3 red, 1 green), (2 red, 2 blue), (2 red, 1 green, 1 blue), and (1 red, 3 blue). The probability of each of these outcomes is given by the multinomial distribution: P(X = 3 red, 1 blue) = (4 choose 3) * (15 choose 3) * (10 choose 1) / (30 choose 4) = 1350 / 27405, P(X = 3 red, 1 green) = (4 choose 3) * (15 choose 3) * (5 choose 1) / (30 choose 4) = 675 / 27405, P(X = 2 red, 2 blue) = (4 choose 2) * (15 choose 2) * (10 choose 2) / (30 choose 4) = 2700 / 27405, P(X = 2 red, 1 green, 1 blue) = (4 choose 2) * (15 choose 2) * (5 choose 1) * (10 choose 1) / (30 choose 4) = 1350 / 27405, and P(X = 1 red, 3 blue) = (4 choose 1) * (15 choose 1) * (10 choose 3) / (30 choose 4) = 675 / 27405. The probability of more red balls than blue balls is the sum of these probabilities: 1350/27405 + 675/27405 + 2700/27405 + 1350/27405 + 675/27405 = 6750/27405 = 3/5.