Let's first look at the two probabilities separately.

There are 36 possible combinations of the two dice: (1, 1), (1, 2), (1, 3),...,(1, 6); (2, 1), (2, 2), (2, 3),...; ....; (6, 1), ..., (6, 6).  Getting a 4 on the red die can happen 6 different ways: (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), and (4, 6). So P(4 on red die)

.

Now, getting an odd sum on the dice can happen 18 different ways: (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), and (6, 5).  So P(odd sum) =

.

Putting them together, P(4 on red AND odd sum) =

.

The probability of drawing a diamond first is

The probabilty of drawing a diamond second when the first card is not replaced is .

To determine the probability of 2 INDEPENDENT events, we multiply them.

The prime numbers that can be rolled are 2, 3, 5, 7, and 11. The rolls that result in one of these numbers, out of a possible  rolls, are:

Therefore, there are 15 rolls out of 36 that result in a prime number, making the probability of a prime result 

There are 36 possible rolls. The only perfect squares that can be rolled are 4 and 9; the only perfect cube that can be rolled is 8. So the roll must be one of the following:

This adds up to 12 rolls out of 36, for a probability of 

The probability that all three coins will come up heads is the product of the individual probabilities:

Let  be the number of red marbles. Then there are  blue marbles and  yellow ones. Since there are 40 green marbles, and 100 marbles total, we can write this equation, simplifying and solving for :

Therefore, there are  yellow marbles, and the probability of drawing a yellow marble is 

The probability of drawing a club from a standard deck of cards is , since one-fourth of the cards - thirteen out of fifty-two - are clubs.

If the two red kings are removed, thirteen out of the remaining fifty cards are clubs, and the probability is .

If the thirteen hearts are removed, thirteen out of the remaining thirty-nine cards are clubs, and the probability is .

If the ace of spades is removed, thirteen out of the remaining fifty-one cards are clubs, and the probability is .

If the joker is added, thirteen out of the fifty-three cards are clubs, and the probability is .

If the four fours are removed, this leaves twelve clubs out of forty-eight cards, so the probability is . This is the corect choice.

Call the fair coins, which come up heads or tails with probability  , Coins 1, 2, and 3; call the loaded coin, which comes up heads with probability  and tails with probability , Coin 4. We are looking for the probability of one of the following four outcomes:

The probability of heads on Coins 1, 2, and 3 and tails on Coin 4: 

The probability of tails on Coin 1 and heads on Coins 2, 3, and 4:

The probability of tails on Coin 2 and heads on Coins 1, 3, and 4:

The probability of tails on Coin 3 and heads on Coins 1, 2, and 4:

Add these:

For each coin, the probability of a head is 0.6, and the probability of a tail is 0.4. 

The probability of a head coming up on the first coin and a tail coming up on the second is ; the same holds for the reverse case. Therefore, the probability of one head and one tail is 

Let  be the probability that the die will come up 1. Then  is also the common probability for each of the rolls of 2, 3, 4, or 5, and  is the probability of the die coming up 6. To determine the value of , add these six probabilities and set the sum to 1, then solve:

So 1, 2, 3, 4, and 5 each will come up with probability  and 6 will cme up with probability .

An even number will come up with probability . Two even rolls will happen with probability .