Probability

There are 6 possible outcomes when rolling a die, and only one of them is a 6. Therefore, the probability of rolling a 6 is 1/6.

There are 3 even numbers on a standard six-sided die (2, 4, and 6), and only one of them is a 6. Therefore, the probability of rolling a 6, given that the die has already landed on an even number, is 1/3.

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 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 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 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.

The probability of getting exactly 3 heads when flipping a fair coin 5 times can be calculated using the binomial distribution. The formula for the binomial distribution is P(x) = (nCx) * 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 = 5, x = 3, p = 0.5, and q = 0.5. Plugging these values into the formula, we get P(3) = (5C3) * 0.5^3 * 0.5^(5-3) = 0.3125.

The probability of getting at least 2 heads when flipping a fair coin 4 times can be calculated by subtracting the probability of getting 0 or 1 head from 1. The probability of getting 0 heads is (0.5)^4 = 0.0625, and the probability of getting 1 head is 4 * (0.5)^4 * (0.5)^3 = 0.25. Therefore, the probability of getting at least 2 heads is 1 - 0.0625 - 0.25 = 0.875.

When flipping a coin, there are two possible outcomes: head or tail. Since both outcomes are equally likely, the probability of getting a head is $1/2 = 0.5$.

When flipping two coins, there are four possible outcomes: HH, HT, TH, TT. Since each outcome is equally likely, the probability of getting two heads is $1/4 = 0.25$.

There are four possible outcomes when flipping a coin twice: HH, HT, TH, TT. Only one of these outcomes (TT) does not have at least one head. Therefore, the probability of getting at least one head is 3/4.

There are 36 possible outcomes when rolling two six-sided dice. The outcomes that sum to 10 or less are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (6, 1). Therefore, the probability of getting a sum of 10 or less is 21/36, which simplifies to 5/6.