Probability

There is 1 queen of spades and 1 king of diamonds in a deck of 52 cards. The probability of drawing either is (1 + 1) / 52 = 2 / 52 = 1 / 26.

Total outcomes when rolling two dice is 36. Sums >= 10 are (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 6 such outcomes. Probability = 6/36 = 1/6.

Total balls = 5+4+3 = 12. The number of non-red balls is 5 (black) + 4 (white) = 9. The probability is 9/12, which simplifies to 3/4.

There is only one '2' of spades in a standard 52-card deck.

Probability of drawing a diamond = 13/52 = 1/4. Since cards are replaced, each draw is independent. Probability of 4 diamonds = (1/4)^4 = 1/256.

A standard six-sided die has faces numbered from 1 to 6. The only outcome on the die that is strictly greater than 5 is 6. Therefore, the event of getting a number greater than 5 contains only the outcome 6.

Total ways to pick 3 cards is 52C3 = 22100. Ways to pick 3 cards of different suits is 4C3 * 13^3 = 4 * 2197 = 8788. Probability = 8788 / 22100 = 169 / 425.

Binomial distribution: P(X >= 3) = 1 - [P(0) + P(1) + P(2)]. P(k) = 10Ck * (0.5)^10. P(0) = 1/1024, P(1) = 10/1024, P(2) = 45/1024. Sum = 56/1024. P(X >= 3) = 1 - 56/1024 = 968/1024 = 121/128.

The probability of drawing two kings without replacement is (4/52) × (3/51). This simplifies to 1/221, because there are 4 kings among 52 cards and then 3 kings among the remaining 51 cards.

Total outcomes for 6 dice is 6^6. Favorable outcomes where all show different faces is the number of permutations of 6 distinct items, which is 6!. Probability = 6! / 6^6.

Probability of white (p) = 4/10 = 0.4. Probability of red (q) = 6/10 = 0.6. We want exactly one white in three draws (with replacement). This is a binomial distribution: 3C1 * (0.4)^1 * (0.6)^2 = 3 * 0.4 * 0.36 = 1.2 * 0.36 = 0.432. 0.432 = 432/1000 = 54/125.

This is a probability problem involving state transitions. The silver coin moves between purses. The calculation leads to 5/9.