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:

For Quantity A:

The probability of rolling a four or greater on a six sided die is  since three out of six values on the die satisfy the condition. The probability of getting this roll two times out of two rolls of the die is

The probability is 

For Quantity B:

The probability of rolling a six is . Rolling a six once or twice would satisfy the condition of rolling a six at least once out of two times, so the total probability is the sum of these two.

The probability is 

Quantity B is greater.

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:

For Quantity A:

The probability of rolling a three or greater on a six sided die is  since four out of six values on the die satisfy the condition: . The probability of getting this roll two times out of two rolls of the die is

The probability is 

For Quantity B:

The probability of rolling a six is . Rolling a six once or twice would satisfy the condition of rolling a six at least once out of two times, so the total probability is the sum of these two.

The probability is 

Quantity A is greater.

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:

For Quantity A:

The probability of drawing a face card is , since there are four suits, and each suit has the same sequence of cards: 

Drawing a face card once, twice, or three times out of three, satisfies the condtion of drawing at least once face card, so the total probability will be the sum of these three probabilities.

The probability is

However, there is a quicker way to find this value. The probability of getting at least once face card is the complement of getting no face cards:

This is a faster of method of getting the same result

For Quantity B:

The probability of drawing a spade is  in a standard fifty-two card deck.

To compare the quantities, cross multiply denominators:

A: 

B: 

Quantity A is greater.

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:

For Quantity A:

The probability of drawing a face card is , since there are four suits, and each suit has the same sequence of cards: 

Drawing a face card twice or three times out of three satisfies the condtion of drawing at least once face card, so the total probability will be the sum of these two probabilities.

The probability is

For Quantity B:

The probability of drawing a spade is  in a standard fifty-two card deck.

To compare the quantities, cross multiply denominators:

A: 

B: 

Quantity B is greater.

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:

Regardless of the actual number on the die, one, two, or three the chance of getting a particular roll is .

For Quantity A:

Rolling a one three times or four times still satisfies the condition of rolling a one at least thrice, and so the total probability will be the sum of the probabilities of these two rolls

The probability is

For Quantity B:

Similarly, rolling a two three or four times out of four satisfies the condition of rolling it at least thrice:

The probability is

Quantity A is greater.

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 

Quantity A is greater.

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