Probability

To find the probability of selecting a blue ball, we divide the number of blue balls by the total number of balls: 10 blue balls / (15 red balls + 10 blue balls + 5 green balls) = 10/30 = 1/3.

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.

First, select 1 red ball from 6 red balls, which can be done in 6 ways. Then, select 1 blue ball from 4 blue balls, which can be done in 4 ways. Now, we have 2 balls left to select, and we can choose either 2 green balls or 1 green ball and 1 red ball or 1 green ball and 1 blue ball. The number of ways to select 2 green balls from 2 green balls is 2C2 = 1. The number of ways to select 1 green ball and 1 red ball from 2 green balls and 5 red balls is 2C1 * 5C1 = 10. The number of ways to select 1 green ball and 1 blue ball from 2 green balls and 3 blue balls is 2C1 * 3C1 = 6. Therefore, the total number of ways to select 4 balls, including at least 1 red ball and at least 1 blue ball, is 6 * 4 * (1 + 10 + 6) = 240.

Since there must be an equal number of red, blue, and green balls, we need to select 2 balls from each color. The number of ways to select 2 red balls from 8 red balls is 8C2 = 28. The number of ways to select 2 blue balls from 6 blue balls is 6C2 = 15. The number of ways to select 2 green balls from 4 green balls is 4C2 = 6. Therefore, the total number of ways to select 6 balls, 2 of each color, is 28 * 15 * 6 = 2520.

The probability of an event occurring is the likelihood of it happening. In this case, the probability of the event occurring is 30%, which is equal to 0.3.

For a fair coin, the probability of obtaining a head or a tail is equal, which is 0.5.

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.

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.

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 an event is the ratio of the number of favorable outcomes to the total number of possible outcomes. In this case, the favorable outcomes are the number of green balls, which is 24. The total number of possible outcomes is the total number of balls, which is $12 + 18 + 24 = 54$. Therefore, the probability of selecting a green ball is $rac{24}{54} = rac{1}{2}$.

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 (\frac{1}{2}).

A fair die has six equally likely outcomes: 1, 2, 3, 4, 5, and 6. Therefore, the probability of obtaining a 6 is 1 / 6.