Probability is a fraction between 0 and 1. The denominator is the total number of possible outcomes and the numerator is the total number of desired outcomes. For two dice, the total number of ways to roll the dice is 36 (6 on the first die can go with any of the 6 on the second die)

The following outcomes result in a sum of less than 5:  {1, 1}  {1, 2}  {1, 3}  {2, 1}  {2, 2}  {3, 1}, for 6 possible outcomes

The following outcomes result in a sum of at least 10:  {4, 6}  {5, 6}  {5, 5}  {6, 6}  {6, 5}  {6, 4} for 6 possible outcomes

Since we are looking for a sum of less than five OR a sum of at least ten, we sum up the total number of desired outcomes

So the probability of getting less than five or at least ten will be 12/36, or 1/3

There are 6 pink marbles out of 14 total, which reduces to 3/7.

There are 8 possible sequences that can occur given this scenario, listed as {HHH, HHT, HTT, HTH, THT, THH, TTH, TTT} 

Only one element in this set is the desired outcome of 3 heads. So we get 1/8 is our probability.

Another way to look at this problem is to consider that the probability of flipping heads once is 1/2, so the probability of flipping heads 3 times consecutively is 1/2 * 1/2 * 1/2 = 1/8

To get a number less than 4 on a die, 3/6 is the probability. To get a heads is a 1/2 chance.

So, since both events are independent, multiply the two probabilities together to get your answer.

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).