Remember that the probability of an event is given by dividing the number of desired outcomes by the number of total outcomes. 

The probability of Khalil spinning one of the three primary colors (Red, Blue, or Yellow) is . 

The probability of Khalil rolling an odd number (1, 3, or 5) is . 

We want to know the probability of Khalil spinning a primary color AND rolling an odd number. To do this compound probability, we multiply the two probabilities together. 

The probability of an event is given by dividing the number of desired outcomes by the number of total outcomes. 

In this case, we want to find all the possibilities of Raul spinning anything but an A. That means, he can spin a B, C, or D. 

Therefore, there are three desired outcomes out of a total four possible outcomes. 

This means that the probability of Raul not spinning an A is . 

If Adam and Sam have already pulled out one card a piece and not put them back in the deck, that means that there are  cards left. 

We also know that one of the Aces has already been pulled. Since there are four Aces in a deck, that means that there are  left.

We know that the probability of an event is the number of desired outcomes divided by the total number of possible outcomes. Therefore, the probability of Kristi pulling an Ace is . 

A standard deck of cards has  cards in it, and it is divided equally into four suits: Hearts, Spades, Clubs, and Diamonds. In this problem, all the Hearts are taken away before Kristen draws her card. That means that our deck has  total cards in it. 

Remember that the probability of an event occurring is expressed as the number of desired outcomes divided by the total number of possible outcomes. In our case, we want to know the probability that Kristen draws a Queen. In our modified deck, there are three Queens (one of Diamonds, Spades, and Clubs), so there are three outcomes that would give us our desired result.  

Therefore, the probability that she draws a Queen is . 

Here, we have a compound probability problem. Remember that the probability of a single event is given by dividing the number of desired outcomes by the number of total outcomes. 

With that in mind, the probability of Yousef's spinner landing on Blue is  since there are section possible outcomes and one desired outcome. 

For the first die, the probability of him rolling a six is . The same goes for the second die since the two rolls are independent of one another. 

The probability of multiple events happening together can be expressed like this:

Therefore, the probability of Yousef spinning Blue and rolling two sixes is:

There are twenty even whole numbers from 1 to 40 - 2, 4, 6, up to 40. Also, there are eleven odd primes - 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 (there is one even prime, 2, but it has already been counted). Therefore, the box will contain 31 blue balls out of 40, and the probability that the draw of one ball will result in the ball being blue is . This same probability holds in both the first and second draws.

Since the draws are independent events, the probabilities can be multiplied:

,

the correct choice.

5 of the 26 balls are marked with vowels, so the probability that the draw of one ball will result in the ball being marked with a vowel is ; this probability is the same for each of the three draws. Since the draws are independent events, the probabilities can be multiplied:

.

There are 10 even numbers and 10 odd numbers in the set of whole numbers from 1 to 20. Therefore, there are 20 even-numbered balls and 10 odd-numbered balls - a total of 30 balls in the box.

The probability that the draw of one ball will result in an even-numbered ball is

This probability is the same for each of the two draws. Since the draws are independent events, the probabilities can be multiplied:

,

the correct response.

There are 10 even numbers and 10 odd numbers in the set of whole numbers from 1 to 20. Therefore, there are 20 even-numbered balls and 10 odd-numbered balls - a total of 30 balls in the box.

Assume that the balls are drawn one after the other. The probability that the first ball will be odd is .

Since the ball is not replaced, there are now 29 balls total, 9 of which are marked with odd numbers. The probability that the second ball will be odd is .

Multiply these probabilities to get the probability that both balls will be marked with odd numbers:

.

We can assume that the three balls are drawn one at a time.

When the first ball is drawn, there are 26 balls in the box, 21 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .

Since there is no replacement, when the second ball is drawn, there are 25 balls in the box, 20 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .  

Similarly, when the third ball is drawn, there are 24 balls in the box, 19 of which are marked with consonants. The probability of drawing a ball marked with a consonant is therefore .

Multiply the three probabilities together. The probability that all three balls will be marked with consonants is 

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