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 

.

The odd composite numbers  - that is, the odd numbers that have more than two factors - are all represented by red balls. The odd composite whole numbers in the set  comprise the set  - a set of eight elements. The ball marked "1" is also red, since 1 is odd and neither composite nor prime. Therefore, 9 balls of the 40 are red.

We can assume that the two balls are drawn one at a time. The probability of the first ball being red is . Since there are now 39 balls, 8 of which are red, the probability that the second ball will be red is . The probabilities are multiplied:

.

This is the probability that both balls will be red.

Add the numbers of tiles corresponding to each of the letters from "A" to "M":

The probability that one of these tiles will be drawn on the first draw is . On the second draw, there will be 52 such tiles out of 99, making the probability of the same event . The probability of both draws resulting in one of these letters is the product of these individual probabilities:

Out of the 100 tiles, there are 9 tiles marked with "A", 12 with "E", 9 with "I", 8 with "O", and 4 with "U", as well as 2 blank tiles; this is a total of

tiles not marked with a consonant.

The probability that the first draw will result in a vowel or a blank is .

After this draw, there will be 99 tiles, 43 of which are vowels or blanks; the probability that the second draw will be one of these is .

Use the multiplication principle to find the probability that both tiles will be vowels or blanks:

A six-sided die has three sides with odd numbers and three sides with even numbers. Thus, the probability of the die landing on an odd number is 

which simplifies to 

.

The probability of the die landing on an even number is also 

which we can simplify to 

. 

Because the probabilities are independent of each other, we can find the probability of the two events happening in a row by multiplying the fractions. 

Therefore, the probability of the first roll landing on an odd number, and the second roll landing on an even number is  .

There are four different types of cake. In this type of problem we want to guarantee we have one of each, so we need to assumbe we have very bad luck. We start with the red velvet since that is the type with the most cakes. If we open those 6 we are not guaranteed to have different ones. Then say we opened all five chocolate cakes, then all four carrot cakes. We still have only three types of cakes but opened 15 boxes. When we open the next box (16) we will be guaranteed to have one of each.

Remember that the "probability" of an event happening is the number of outcomes which result in that event divided by the total number of possible outcomes. That is,

.

Hence, the probability of an event happening is described as a fraction. To solve this problem, we must first recognize that three separate events are being described: drawing an ace, drawing a king, and drawing a jack. Because each of these events do not overlap (you cannot, for example, draw a card that is both a jack and a king at the same time), calculating the probability  that you draw any one of them is a simple matter of adding together their respective probabilities , , and .

There are  cards in a deck of playing cards,  of which are aces,  of which are kings, and  of which are jacks. Hence, 

, 

, and

,

and therefore

This means that the probability of drawing either an ace, a king, or a jack from a standard deck of playing cards is .

Mrs. Smith assigns homework to her class once a week and records each student's results. For last week's assignment, the histogram depicts these results. If the homework set was worth twenty points, what is the range of grades for last week's assignment?

Note: For each error, one point is subtracted from the total points possible. 

To solve this problem, first identify what the highest and lowest grade on the assignment was. The highest grade will be the one with the least amount of errors made and the lowest grade will be the one with the most amount of errors made.

The red bar in the histogram says that students made between zero and two errors. In other words, either: 0, 1, 2 errors.

The yellow bar in the histogram says that the students made between two and four errors. Since the red bar includes two errors the yellow bar only includes 3 or 4 errors.

The green bar in the histogram says that students made 5 or 6 errors.

The teal bar in the histogram says that students made 7 or 8 errors.

The blue bar in the histogram says that students made 9 or 10 errors.

The purple bar in the histogram says that students made 11 or 12 errors.

This means that at least one student made zero errors on the assignment and therefore did not loose any points. Therefore, the highest grade on the assignment was twenty.

The histogram shows that at least one student made twelve errors on the test. Therefore, the lowest grade on the assignment would be:

From here, calculate the range.