This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For  trials, each of which has a probability  of a 'successful' outcome, the probability of exactly  successes is given by the function:

Quantity A:

The probability of getting heads is 

Quantity B:

The probability of getting tails is also 

The two quantities are equal.

This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For  trials, each of which has a probability  of a 'successful' outcome, the probability of exactly  successes is given by the function:

Heads and tails have an equal probability of occurence of , so flipping heads twice or tails twice has an equal probability; do not let that specification complicate the problem.

Now, when there is language such as 'at least' or 'no more than', a cumulative probability can be assumed. I.e. we sum up the probabilities of each event that satisfies the condition. For example, if we're asked for the probability of rolling a three on a die at least two times out of four, we'd sum up the probability of rolling  a three twice, a three thrice, and a three four times. If we're asked for the probability rolling a three no more than once out of four times, we'd sum up the probabilities of rolling a three once, and of rolling a three zero times. 

Now, for this problem, before actually calculating probabilities, let's right them out

Quantity A:

Added to

Quantity B:

Added to

Notice how the probabilities come out the same for A and B ( for the curious), although the terms are slightly reorganized.

A visual examination shows that they're the same, and it saves the time of actual calculation.

The two quantities are equal.

This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For  trials, each of which has a probability  of a 'successful' outcome, the probability of exactly  successes is given by the function:

Note that if Harold flips a coin and gets head exactly three times, he'll have to get four tails, and vice-versa. Calculate the probability of one and you get the probability of the other; there is no need to calculate both. The probability of getting heads is one half. For the three heads:

This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For  trials, each of which has a probability  of a 'successful' outcome, the probability of exactly  successes is given by the function:

Now, when there is language such as 'at least' or 'no more than', a cumulative probability can be assumed. I.e. we sum up the probabilities of each event that satisfies the condition. For example, if we're asked for the probability of rolling a three on a die at least two times out of four, we'd sum up the probability of rolling  a three twice, a three thrice, and a three four times. If we're asked for the probability rolling a three no more than once out of four times, we'd sum up the probabilities of rolling a three once, and of rolling a three zero times. 

Now, to calculate the probability of flipping a head at least once out of ten times, it would be time consuming to calculate the probability of once, twice, thrice, etc.

Realize, rather, that flipping the coin heads at least once is the complement to flipping it zero times. That is to say it is one minus the probability of flipping it zero times.

The probability of flipping heads is one half.

So the complement, the probability of one or more heads, is

This problem is dealing with the repeated trials for an event that only has two potential outcomes. This scenario describes a binomial distribution.

For  trials, each of which has a probability  of a 'successful' outcome, the probability of exactly  successes is given by the function:

Now, when there is language such as 'at least' or 'no more than', a cumulative probability can be assumed. I.e. we sum up the probabilities of each event that satisfies the condition. For example, if we're asked for the probability of rolling a three on a die at least two times out of four, we'd sum up the probability of rolling  a three twice, a three thrice, and a three four times. If we're asked for the probability rolling a three no more than once out of four times, we'd sum up the probabilities of rolling a three once, and of rolling a three zero times. 

The probability of rolling a sixteen or greater is . This can be seen by dividing the number of events that satisfy this by the number of total possible events: 

Now, to get this sixteen or greater at least twice out of three rolls:

The probability of Abel winning is

Looks like the odds favor the orc.

Consider first all of the possible ways the men may be arranged, which is

Now, consider all of the ways that Aaron and Gary could be seated beside each other; it may be easier to visualize by drawing it out:

1. A G _ _
2. G A _ _
3. _ A G _
4. _ G A _
5. _ _ A G
6. _ _ G A

As seen, there are six possibilities.

Finally, for each of these cases, Craige and Boone could be seated in one of two ways.

So the probability that Aaron and Gary will be seated beside each other is:

To solve this problem, employ the fundamentals of combinatorics. For the first digit of the number, you might think that there are 10 options (the digits 0-9), but remember that no seven-digit integer would start with 0 (e.g. you don't represent one-hundred thousand as 0,100,000). So there are 9 choices available for the first digit.

For the second digit, there are also 9 choices: 0 is now a possibility, but you cannot repeat the original digit.

For each remaining spot, you lose one option because you cannot match the unique digits you've already chosen.  For the third digit you have two of the ten options unavailable because you're chosen two already; for the fourth digit you cannot match the previous three, and so on.  So the calculation becomes:

9 * 9 * 8 * 7 * 6 * 5 * 4 = 544,320. 

This problem is dealing with binomial probabilities. For a probability of success, , and  trials, the probability of  successes is:

For this problem there are three trials, and we're interested in at least two successes. Note that this probability equation gives the probability of getting EXACTLY k succeeses, so we must consider getting three successes at well.

The probability of rolling at least a six on a twenty-sided die is 

Therefore, the probability of rolling this at least twice out of three trials is:

Answer: –1
Explanation: Recall that the median of an even-numbered set of numbers is the arithmetic mean of the pair of middle terms. Thus (4 + x)/2 = median of the first set and (5 + y)/2 = median of the second set. Since both medians are equal, we can set the equations equal to eachother. (4 + x)/2 = (5 + y)/2. Multiply both sides by 2 and we get 4 + x = 5 + y. We also know that 4 < x < 8 and 5 < y < 9, since the sets are arranged in ascending order. This narrows our options for x and y down significantly. Plugging in various values will eventually get you to x = 7 and y = 6, since 7 + 4 = 11 and 5 + 6 = 11, and thus the median in both cases would be 5.5. Thus, y – x = –1.

16, 19, 16, 7, 2, 20, 9, 5

Order the numbers from smallest to largest.

2,5,7,9,16,16,19,20

The median is the number in the middle.

In this case, there is a 9 and 16 in the middle.

When that happens, take the average of the two numbers.