You multiply each number of students in each class by how many times that class is entered into the drawing, then add up the totals. There will be a total of 1800 entrants and 800 will be seniors.

There are four possible numbers that can be chosen from S, and there are four that can be chosen from T. This means that there are 4 * 4, or 16, pairs of numbers that can be drawn from S and T. We need to find the sum of these pairs and determine if each sum is prime. Let's find the sum of the 16 pairs. Remember that a number is prime if it divisible only by itself and 1.

0 + 2 = 2, which is prime

0 + 3 = 3, prime

0 + 4 = 4, not prime

0 + 7 = 7, prime

1 + 2 = 3, prime

1 + 3 = 4, not

1 + 4 = 5, prime

1 + 7 = 8, not

5 + 2 = 7, prime

5 + 3 = 8, not

5 + 4 = 9, not

5 + 7 = 12, not

9 + 2=11, prime

9 + 3 = 12, not

9 + 4 = 13, prime

9 + 7 = 16, not

Of the sixteen pairs, 8 have a sum that equals a prime number. Thus, because each of these pairs has an equal chance of being drawn randomly, the probability that the sum will be prime is 8 out of 16, or 8/16 = 1/2.

The answer is 1/2.

We can find the individual probabilities of these three events occuring first. 

P(rolling a 6) = 1/6

P(head) = 1/2

P(spade) = 13/52 = 1/4

Now, to find the probability of rolling a 6 AND flipping a head AND drawing a spade, we must multiply the individual probabilities. So the answer is 1/6 * 1/2 * 1/4 = 1/48.

There are a total of 7 + 5 + 3 + 6 = 21 marbles. The probability of picking a yellow marble is 7/21 = 1/3. Then we put it back and choose a blue marble with probability 3/21 = 1/7. We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now 6/20 = 3/10, because now we are choosing from 20 marbles instead of 21. So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is 1/3 * 1/7 * 3/10 = 1/70. 

There are 4 aces in the 52 card deck, so the probability of drawing an ace is 4/52. Then we put this ace back in the deck and draw again. The probability of drawing an ace is again 4/52. Similarly, on the third draw, the probability of getting an ace is 4/52. So the probability of drawing an ace on the 1st draw AND the 2nd draw AND the 3rd draw is 4/52 * 4/52 * 4/52.

The probability of an event is the ratio of the number of desired outcomes to the total number of possible outcomes. In this problem, the total number of outcomes is equal to the total number of marbles in the bag. There are four blue, eight red, and six orange marbles, so the total number of marbles is the sum of four, eight, and six, or eighteen.

We are asked to find the probability of choosing a marble that is either red or orange. This means we have to consider the number of marbles that are either red or orange. Because there are eight red and six orange, there are fourteen marbles that are either red or orange. 

The probability is thus fourteen out of eighteen, because there are fourteen red or orange marbles, out of a total of eighteen marbles. We will need to simplify the fraction 14/18.

Probability = 14/18 = 7/9

The answer is 7/9.

The possible dice combinations that sum to 4 are .

The number of all possible dice combinations is . (6 numbers on each of the two dice.)

So the probability that the numbers sum to 4  =

There are 5 black marbles worth $1000 dollars out of the total of 200 marbles.  or 2.5%

There are 13 diamonds in a standard deck of 52 cards. So, you have a 13/52 (1/4) chance of getting a diamond; then a 12/51 (4/17) chance of pulling the next diamond; last, there is a 11/50 chance of getting the third diamond. When you combine probabilities, you multiply the individual probabilities together

Probability = # Gum Drops / # Total Number of Candies