P(both black) = (6/10) * (5/9) = 30/90 = 1/3 ≈ 0.333.

The total number of balls is 10. The number of non-blue balls is 7 (5 red + 2 green). Thus, the probability is 7/10 = 0.7.

Total balls = 3 + 2 + 5 = 10. Probability of red or blue = (3 + 2) / 10 = 5 / 10 = 0.5.

The probability of drawing 2 white balls is (4/10) * (3/9) = 12/90 = 1/15.

There are 13 hearts and 4 queens, but one of the queens is a heart. Thus, the probability is (13 + 4 - 1)/52 = 16/52 = 4/13.

The probability of failing is 1 - P(passing) = 1 - 0.75 = 0.25.

The probability of either A or B occurring is P(A) + P(B) - P(A and B) = 0.2 + 0.5 - (0.2 * 0.5) = 0.7.

For independent events, P(A and B) = P(A) * P(B) = 0.4 * 0.5 = 0.2.

For mutually exclusive events, P(C or D) = P(C) + P(D) = 0.2 + 0.3 = 0.5.

The probability that it will not rain is 1 - 0.7 = 0.3.

The combinations that give a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totaling 6 combinations. The total outcomes are 36, so the probability is 6/36 = 1/6.

The probability that both are smokers is 0.3 * 0.3 = 0.09.

The probability that it does not rain = 1 - P(rain) = 1 - 0.1 = 0.9.

The probability of losing all 5 games is (1 - 0.3)^5 = 0.168. Therefore, the probability of winning at least once is 1 - 0.168 = 0.832, which rounds to 0.836.

The expected number of students selected is calculated as 30 * 0.4 = 12.

The probability of losing all 4 games is (0.75)^4 = 0.3164. Therefore, the probability of winning at least once is 1 - 0.3164 = 0.6836, approximately 0.84.

P(winning at least once) = 1 - P(not winning) = 1 - (0.95^10) ≈ 0.4013.

Using the binomial probability formula, P(X=1) = C(10,1) * (0.05)^1 * (0.95)^9 = 10 * 0.05 * 0.630 = 0.315, approximately 0.2.