Each flip of the coin is an independent event, and does not affect the other flips; all that matters is that the coin is fair. Because of that fact, the probability that the tenth flip comes up heads is .

This can more easily be solved by finding the probability that neither will be green, then subtracting it from 1.

The probability that the first ball will be not be green is five out of ten, or .

This will leave nine balls, four not green, making the probability that the second ball will then not be green .

Multiply these probabilities:

This is the probability that neither ball is green; subtract this from 1: 

The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are 

red marbles out of

marbles total, the probability of choosing a red marble is

The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are 

white marbles out of

marbles total, the probability of choosing a white marble is

The equivalent question is this: What is the probability that all three dice will show 1, 2, 3, or 4? 

The probability that one fair die will show a 1, 2, 3, or 4 is , so the probability that all three will show 1, 2, 3, or 4 is 

.

If four more threes are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and eight threes. This makes sixteen ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of 

.

This can more easily be solved by finding the probability that neither draw will result in a blue ball, then subtracting it from 1.

The probability that the first ball will not be blue will be seven (balls that are not blue) out of ten (balls total), or . Since the ball is replaced, this is also the probability that the second will not be blue. Therefore, the probability that neither will be blue is . 

The probability that at least one draw will result in a blue ball is therefore .

The best way to find this probability is to find the probability that both will show a 1, 2, 3, or 4, and subtract that probability from 1. 

The probability that one fair die will show a 1, 2, 3, or 4 is , so the probability that both will show 1, 2, 3, or 4 is 

.

The probability that at least one die will show a 5 or a 6 is therefore 

.

If four more tens are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and four threes. This makes twelve ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of 

.

There are 19 winning cards: the 13 clubs, the other 3 queens, and the other 3 kings (the queen and king of clubs are already counted). There are subsequently 33 losing cards, so the odds against a winning card are the number of losing cards, 33, to the number of winning cards, 19. This cannot be reduced, so the correct choice is 33 to 19.