Probability

There are 5 numbers that are not multiples of 5 on a standard six-sided die (1, 2, 3, 4, and 6), and 2 of them are even (2 and 4). Therefore, the probability of rolling a 6, given that the die has already landed on a number that is not a multiple of 5, is 2/5.

There are 5 numbers that are not multiples of 6 on a standard six-sided die (1, 2, 3, 4, and 5), and 2 of them are even (2 and 4). Therefore, the probability of rolling a 6, given that the die has already landed on a number that is not a multiple of 6, is 2/5.

There are a total of 6 + 4 + 2 = 12 balls in the bag. The number of ways to choose 2 red balls out of 6 is C(6, 2) = 15. The number of ways to choose 1 blue ball out of 4 is C(4, 1) = 4. Therefore, the total number of ways to choose 2 red balls and 1 blue ball is 15 x 4 = 60. The probability of getting exactly 2 red balls and 1 blue ball is 60 / 12^3 = 1/15.

The probability of getting at least 1 red ball is equal to 1 minus the probability of getting no red balls. The probability of getting no red balls is given by (8/24)^4 = 1/256. Therefore, the probability of getting at least 1 red ball is 1 - 1/256 = 4/5.

We can first choose the president of the company, which can be done in 1 way. Then, we can choose the other 2 members of the committee from the remaining 11 employees, which can be done in C(11, 2) = 55 ways. Therefore, the total number of different committees is 1 x 55 = 220.

There are a total of 10 + 8 + 6 = 24 balls in the bag. The number of ways to choose 2 red balls out of 10 is C(10, 2) = 45. The number of ways to choose 1 blue ball out of 8 is C(8, 1) = 8. The number of ways to choose 1 green ball out of 6 is C(6, 1) = 6. Therefore, the total number of ways to choose 2 red balls, 1 blue ball, and 1 green ball is 45 x 8 x 6 = 2160. The probability of getting exactly 2 red balls, 1 blue ball, and 1 green ball is 2160 / 24^4 = 1/20.

To solve this probability problem, we can use conditional probability. First, we calculate the probability of selecting 2 red balls from the 6 red balls: P(2 red) = C(6, 2) / C(12, 2) = 15 / 66. Then, given that we have selected 2 red balls, we calculate the probability of selecting 1 blue ball from the remaining 10 balls: P(1 blue | 2 red) = C(4, 1) / C(10, 1) = 4 / 10. Multiplying these probabilities, we get P(2 red and 1 blue) = P(2 red) * P(1 blue | 2 red) = (15 / 66) * (4 / 10) = 1/10.

This is a classic probability problem involving binomial distribution. The formula for binomial distribution is P(x) = (n! / (x!(n-x)!)) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure. In this case, n = 10, x = 5, p = 1/2, and q = 1/2. Plugging these values into the formula, we get P(5) = (10! / (5!(10-5)!)) * (1/2)^5 * (1/2)^(10-5) = 252 / 1024 = 1/4.

The probability of getting exactly 5 heads in 10 tosses is (P(X = 5) = \binom{10}{5} (0.5)^5 (0.5)^5 = 0.246).

The probability of getting exactly 3 sixes in 6 rolls is (P(X = 3) = \binom{6}{3} (1/6)^3 (5/6)^3 = 0.143).

This is a combination problem with restrictions. First, you need to select 1 red ball from 6 red balls, which can be done in 6 ways. Then, you need to select 1 blue ball from 4 blue balls, which can be done in 4 ways. Finally, you need to select 1 green ball from 2 green balls, which can be done in 2 ways. So, the total number of ways is 6 * 4 * 2 = 144.

This is a combination problem with restrictions. First, you need to select 2 red balls from 10 red balls, which can be done in C(10, 2) = 45 ways. Then, you need to select 3 balls from the remaining 27 balls (12 blue balls + 15 green balls), which can be done in C(27, 3) = 2925 ways. So, the total number of ways is 45 * 2925 = 3003.

This is a combination problem with restrictions. First, you need to select 1 red ball from 8 red balls, which can be done in 8 ways. Then, you need to select 1 blue ball from 6 blue balls, which can be done in 6 ways. Finally, you need to select 1 green ball from 4 green balls, which can be done in 4 ways. So, the total number of ways is 8 * 6 * 4 = 420.

This is a combination problem with restrictions. First, you need to select 1 red ball from 10 red balls, which can be done in 10 ways. Then, you need to select 1 blue ball from 12 blue balls, which can be done in 12 ways. Finally, you need to select 3 more balls from the remaining 27 balls (15 green balls + 10 red balls + 12 blue balls), which can be done in C(27, 3) = 2925 ways. So, the total number of ways is 10 * 12 * 2925 = 4320.

This is a combination problem with restrictions. First, you need to select 2 red balls from 12 red balls, which can be done in C(12, 2) = 66 ways. Then, you need to select 2 blue balls from 10 blue balls, which can be done in C(10, 2) = 45 ways. Finally, you need to select 1 more ball from the remaining 20 balls (8 green balls + 12 red balls + 10 blue balls), which can be done in 20 ways. So, the total number of ways is 66 * 45 * 20 = 2880.