List out the the possibilities of combinations for rolling a total of 6 and the possibilities of combinations for rolling a total of 9. There are 5 possible outcomes for rolling a total of 6 and 4 possible outcomes for rolling a total of 9. There are a total of 36 possible outcomes, so the probability of rolling a 6 is  and the possibility of rolling a 9 is . 

The probability of an event not occurring is defined as 1 minus the probability of that event occurring. The probability of rolling three of a kind on every face is .

There are six different possible three-of-a-kinds, so the probability is therefore .

Therefore the probability of not rolling a three of the same number is .

This is a straightforward probability problem. Recall that to find the probability of related draws without replacement, we multiply the relative probabilities of each event. The first draw has a probability of , the second draw of , and the third draw of .

There are 40 total balls, and only 12 blue balls.

In a normally distributed curve, a standard deviation of 1 contains ~68% of all values within its range, and a standard deviation of 2 contains ~95% of all values within its range. Since the mean value is 500 for this situation, two standard deviations would occur at values 460 to 540 (i.e. ~95% of all values are within this range). This would mean that the value of 450 would fall outside this range, i.e. a 100%–95% = 5% probability. Since 5% expressed as a fraction is 1/20, Quantity B is far greater than the probability of landing at 450.

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: