Probability

The probability of not rolling a 4 on one die is 5/6. Since the dice are independent, the probability of neither being a 4 is (5/6) * (5/6) = 25/36.

For 5 flips, the total outcomes are 2^5 = 32. Outcomes with more heads than tails: 3 heads (10), 4 heads (5), 5 heads (1). Total = 16. Probability = 16/32 = 1/2.

Total balls = 12. P(first blue) = 7/12. P(second blue | first blue) = 6/11. P(both blue) = (7/12) * (6/11) = 42/132 = 7/22.

Total ways to pick 2 cards = 52C2 = 1326. Ways to pick 2 from same suit = 4 * 13C2 = 4 * 78 = 312. Probability = 312/1326 = 4/17.

Total outcomes for 4 tosses are 2^4 = 16. Sums less than 15: (3,3,3,3)=12, (3,3,3,5)=14, (3,3,5,3)=14, (3,5,3,3)=14, (5,3,3,3)=14. There are 5 outcomes with sum < 15 and 11 outcomes with sum >= 15. The odds against are 11:5.

Total outcomes = 36. Sums that are prime: 2 (1,1), 3 (1,2; 2,1), 5 (1,4; 4,1; 2,3; 3,2), 7 (1,6; 6,1; 2,5; 5,2; 3,4; 4,3), 11 (5,6; 6,5). Total prime outcomes = 1 + 2 + 4 + 6 + 2 = 15. Probability = 15/36 = 5/12.

Doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 such outcomes.

We need to choose 5 red balls from 10 (10C5) and 4 white balls from 8 (8C4). The number of ways is 10C5 * 8C4.

Total arrangements of ASHOKA = 6! / 2! = 360. Treating the two A's as a single unit, we have 5 units (AA, S, H, O, K), which can be arranged in 5! = 120 ways. Probability = 120 / 360 = 1/3.

We need to select 6 players from 10, but 2 are already included. We only need to choose 4 more players from the remaining 8. The number of ways is 8C4 = (8*7*6*5)/(4*3*2*1) = 70.

Total outcomes for 3 dice = 6^3 = 216. Outcomes with no 2 = 5^3 = 125. Outcomes with at least one 2 = 216 - 125 = 91.

Count all ways to choose 3 balls from 9, then subtract selections containing no white ball. This gives C(9,3) - C(6,3) = 84 - 20 = 64.

We need to select 1 red (from 4), 1 white (from 3), and 1 blue (from 2). The number of ways is 4C1 * 3C1 * 2C1 = 4 * 3 * 2 = 24.

Total ways to draw 3 balls from 6 (2 white, 4 red) is 6C3 = 20. Ways to draw 3 balls with no red (all white) is 2C3 = 0. Since there are only 2 white balls, it is impossible to draw 3 white balls. Thus, all combinations of 3 balls must contain at least one red ball. Total ways = 20.