Probability

Total balls = 7. Choosing 2 balls: 7C2 = 21. Choosing 2 balls that are not blue (from 2 red + 3 green = 5 balls): 5C2 = 10. Probability = 10/21.

Total ways to pick 2 cards = 52C2 = (52*51)/2 = 1326. Ways to pick 2 spades = 13C2 = (13*12)/2 = 78. Probability = 78 / 1326 = 1/17.

Total balls = 9. Total ways to draw 3 balls = 9C3 = 84. Ways to draw 3 balls with no black balls (only white or red) = 6C3 = 20. Ways with at least one black ball = 84 - 20 = 64.

This is a stars and bars problem: selecting 10 items from 4 categories with replacement. Formula: (n + k - 1)C(k - 1) where n=10, k=4. (10 + 4 - 1)C(4 - 1) = 13C3 = (13 * 12 * 11) / (3 * 2 * 1) = 286.

Total outcomes = 6^4 = 1296. Outcomes with no 2 = 5^4 = 625. Outcomes with at least one 2 = 1296 - 625 = 671.

Total balls = 10 (3 red, 7 blue). We need a sample of 6 with at most 2 red balls. Cases: 0 red (7C6 = 7), 1 red (3C1 * 7C5 = 3 * 21 = 63), 2 red (3C2 * 7C4 = 3 * 35 = 105). Total = 7 + 63 + 105 = 175. We must exclude samples with 6 balls of the same color. Only 6 blue balls is possible (7C6 = 7). 175 - 7 = 168.

The probability of getting 4, 5, or 6 on the first toss is 3/6 = 1/2. The probability of getting 1, 2, 3, or 4 on the second toss is 4/6 = 2/3. The joint probability is (1/2) * (2/3) = 2/6 = 1/3.

The number of ways to get a sum of 10 with 4 dice is the coefficient of x^10 in (x + x^2 + x^3 + x^4 + x^5 + x^6)^4. This is equivalent to the coefficient of x^6 in (1 + x + x^2 + x^3 + x^4 + x^5)^4, which is 80.

Total outcomes for 3 dice = 6^3 = 216. Outcomes with no prime numbers (only 1, 4, 6) = 3^3 = 27. Outcomes with at least one prime = 216 - 27 = 189.

The number of ways to select one or more balls from 10 white, 9 black, and 7 red balls is (10+1)(9+1)(7+1) - 1. This is 11 * 10 * 8 - 1 = 880 - 1 = 879.

This is a multinomial coefficient problem. We need the number of ways to choose six numbers from {-1, 0, 1} such that their sum is 0. This is the coefficient of x^0 in (x^-1 + 1 + x)^6.

There are 13 diamond cards in a standard deck. The number of ways to choose 3 cards from 13 is given by the combination formula 13C3 = (13 * 12 * 11) / (3 * 2 * 1) = 13 * 2 * 11 = 286.

This is a game of drawing without replacement. The probability of A winning on the first draw is 1/6. The probability of B winning on the second draw is (5/6)*(1/5) = 1/6. Since the probabilities are equal for each turn, and the game must end, the chances are 1/2 and 1/2.