There are a total of  items in the pencil case. On the first try, the probability of Jen's picking a pen is , because there are  pens in the case. On the second pick, there is now one less item in the case (because Jen does not replace what she picked the first time). So now, there are  items.  The probability of picking an eraser is .

To find the probability of Jen picking a pen and then an eraser:

Karen has two full decks of cards, this means there are cards total. In each deck there are two black ten cards and two black five cards. Since there are two decks, this means there are eight cards that are either a black ten or a black five. Therefore, the probability of choosing a black five or a black ten card is:

The probability that a randomly selected customer is in the moderate- or high-risk class is simply the sum of the number of clients in the moderate-risk class and the number of clients in the high-risk class divided by the  total number of clients:

This is also equivalent to just summing up the probability of being in the high-risk class and the probability of being in the moderate-risk class. Since the 3 classes are mutually exclusive, we do not have to worry about subtracting the probability of mutual elements.

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:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case, a "heads" outcome), and  is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

Thus the probability of three or more flips is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case rolling a four), and  is the probability of success (one in four).

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case, a "heads" outcome), and  is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

Therefore, the probability of six or fewer heads 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: