For this problem, we will multiply together the odds of each draw (assuming he draws one card at a time...the odds won't change if he draws three at once, but it's easier to visualize this way) resulting in a card that works for Jeff's goal of having one player from each team. The first draw cannot fail, as he needs one player from each team and the first card he draws must be from one of the teams. After this draw, he has 14 cards remaining, and 10 of these are players on the two teams that can still offer a player.

So the odds of a successful second draw are .

The last draw is the trickiest, as there would now be 13 cards remaining, with only 5 being players from the team that he still needs represented. When we multiply all of these odds together, we get

which is 27.5%.

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is .

There are thirteen spades in the deck, so the probability of drawing a spade is .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is 

Thus the probability of drawing a face card or a spade is:

Set A: 

Set B: 

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

Set A: 

Set B: 

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

The interesection is:

So, we can find the probability of drawing at least one consonant:

Set A: 

Set B: 

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, .

Therefore, the probability of drawing a matching set is:

First calculate the number of students:

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

First calculate the number of students:

The percentage of seniors that do not attend honors classes is:

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

Each toss and roll of the dice are independent events.  The probability of rolling either a head or tail on the first and second trial is one-half.

The probability of rolling any number on the dice is  since there is one number on each of the six faces of the dice.

Multiply all the probabilities.

Adding the joker and the four aces to a 53-card deck will yield a deck of 58 cards, which contains 8 aces and 50 other cards. The odds against drawing an ace at random are therefore  - that is, 25 to 4.

Before the addition of the hearts, the deck had 4 jacks and 48 other cards. The thirteen hearts from the other deck include 1 jack and 12 other cards, so the altered deck contains 5 jacks and 60 other cards. That makes the odds against drawing a jack at random 

 - that is, 12 to 1, just as in the unaltered deck.