CHAPTER 16

Statistics on the GMAT

Tracy has 3 bags of marbles, each bag containing at least 1 blue marble, at least 1 red marble, and no marbles of another color. If Tracy selects 1 marble at random from each bag, what is the probability that all 3 marbles that she selects will be red?

Above is a typical Data Sufficiency question dealing with probability, one of the categories of statistics that we will discuss in this chapter. In this chapter, we’ll look at how to apply the Kaplan Method to this question, discuss the probability rules being tested, and go over the basic principles and strategies that you want to keep in mind on every Quantitative question involving statistics. But before you move on, take a minute to think about what you see in this question and answer some questions about how you think it works:

PREVIEWING STATISTICS ON THE GMAT

What Does Probability Measure?

Probability measures the likelihood of a certain event occurring. Probability can be expressed as a fraction or decimal between 0 and 1 or as a percentage between 0% and 100%. The lower the probability, the less likely that event is to occur. On the GMAT, the testmakers often incorporate this concept into high-difficulty questions that ask about the probability of multiple events happening or not happening. In addition to learning the rules, operations, and formulas specific to probability, remember that the same rules you already know for proportions and fractions continue to apply.

HOW Can Ratios Help You Calculate Probability on This Question?

In a situation like the one described in this question stem, the probability of a given outcome of random selection is equal to the ratio of the items from which the selection is being made. In other words, if a bag contains only two red marbles and two blue marbles, the probability of selecting a red marble from this bag is the same as the ratio of red marbles to total marbles in the bag, which is 2:4, or image.

What Kaplan Strategy Can You Use to Solve This Question?

Since different scenarios are possible for how many marbles of each color are in each bag, Picking Numbers will help you test out the possible distributions of marbles among the bags and determine whether the information is sufficient to solve for a single answer to the question stem.

Remember that whenever you pick numbers for a Data Sufficiency question, the numbers you pick must be permissible—they must obey any rules or restrictions given in the stem and the statement you’re testing. For example, if a statement tells you that there cannot be more than 6 marbles in one bag, you cannot pick the numbers red = 4 and blue = 5 for that bag.

What GMAT Core Competencies Are Most Essential to Success on This Question?

Critical Thinking will help you determine what information is necessary to calculate the desired probability: are ratios sufficient, or do you need actual numbers of the marbles? Do you need information on each bag individually, or is information about all three bags combined enough? Even though Data Sufficiency questions don’t ask you to calculate an exact value for the answer to the question stem, you do need to know how you would calculate that answer if you needed to.

Pattern Recognition is crucial here, as it is on all statistics questions, since the GMAT often asks questions that hinge on the same concepts and formulas that you can memorize. Here, you need to remember the probability formula, as well as how to calculate the joint probability of independent events (Tracy’s selection from each of the three bags).

Finally, Attention to the Right Detail is necessary whenever you work with probabilities, which are essentially fractions or ratios. You need to be clear whether you are dealing with part-to-part or part-to-whole ratios and keep track of which items are in the numerator and denominator.

Here are the main topics we’ll cover in this chapter:

Handling GMAT Statistics

Now let’s apply the Kaplan Method to the statistics question we saw earlier:

Tracy has 3 bags of marbles, each bag containing at least 1 blue marble, at least 1 red marble, and no marbles of another color. If Tracy selects 1 marble at random from each bag, what is the probability that all 3 marbles that she selects will be red?

Step 1: Analyze the Question Stem

This is a Value question. You are asked for the probability that when one marble is chosen at random from each of the three bags, each marble chosen is red. You know that each bag contains at least one blue marble, at least one red marble, and no other marbles of any other color. Now let’s look at the statements.

Step 2: Evaluate the Statements Using 12TEN

Statement (1) tells you that a total of five red marbles and five blue marbles are distributed among the three bags. Because there are different ways to distribute the marbles among the bags, there are different possible probabilities and, therefore, different possible answers to the question.

You can use Picking Numbers to test out these different scenarios; if the different permissible scenarios result in different answers to the question stem, then the statement is insufficient.

For example, let’s call the bags Bag 1, Bag 2, and Bag 3. If Bag 1 contains three blue marbles and three red marbles, while Bags 2 and 3 each contain one blue marble and one red marble, then the probability that all three marbles chosen are red can be calculated by multiplying together the individual probabilities of selecting a red marble from each of the three bags:

image.

If, however, Bag 1 contains three blue marbles and one red marble, Bag 2 contains one blue marble and three red marbles, and Bag 3 contains one blue marble and one red marble, then the probability that all three marbles chosen are red is image × image × image = image.

Because different answers to the question are possible when you pick different permissible numbers, Statement (1) is insufficient. Eliminate choices (A) and (D).

Now look at Statement (2). You are told the ratio of red to blue marbles in each bag. So you can say that the probability of picking a red marble from one bag is image, the probability of picking a red marble from another bag is image, and the probability of picking a marble from the remaining bag is image. The probability of picking a red marble from each of the three bags is image × image × image. You do not have to calculate the value of this expression. Just knowing that you can calculate this probability is enough. Statement (2) is sufficient, and (B) is correct.

Now let’s look at each of the areas of statistics that show up on the GMAT Quantitative section, starting with median, mode, range, and standard deviation.

MEDIAN, MODE, RANGE, AND STANDARD DEVIATION

Median

If an odd group of numbers is arranged in numerical order, the median is the middle value.

Example:  What is the median of 4, 5, 100, 1, and 6?

First, arrange the numbers in numerical order: 1, 4, 5, 6, and 100. The middle number (the median) is 5.

If a set has an even number of terms, then the median is the average (arithmetic mean) of the two middle terms after the terms are arranged in numerical order.

Example:  What is the median value of 2, 9, 8, 17, 11, and 37?

Arrange the values in numerical order: 2, 8, 9, 11, 17, 37. The two middle numbers are 9 and 11. The median is the average of 9 and 11, that is, 10.

The median can be quite different from the average. For instance, in the first set of numbers above, {1, 4, 5, 6, 100}, the median is 5, but the average is image.

Mode

The mode is the number that appears most frequently in a set. For example, in the set {1, 2, 2, 2, 3, 4, 4, 5, 6}, the mode is 2.

It is possible for a set to have more than one mode. For example, in the set {35, 42, 35, 57, 57, 19}, the two modes are 35 and 57.

Range

The range is the positive difference between the largest term in the set and the smallest term. For example, in the set {2, 4, 10, 20, 26}, the range is 24.

Standard Deviation

Standard deviation measures the dispersion of a set of numbers around the mean. If you’ve worked with the standard deviation formula in high school or college, you might remember that the formula looks like this:

image

It can be intimidating, so let’s paraphrase it in plain English. Calculating the standard deviation involves the following steps:

  1. Find the average (arithmetic mean) of the set.

  2. Subtract the average of the set from each term in that set.

  3. Square each result.

  4. Take the average of those squares.

  5. Calculate the positive square root of that average.

Fortunately, the GMAT will only very rarely require you to apply this formula, and only on the most difficult questions. When standard deviation is tested, all you will generally need to understand is the basic concept: standard deviation represents how close or far the terms in a list are from the average.

Thus, {1, 2, 3} and {101, 102, 103} have the same standard deviation, since they both have one term on the average and two terms exactly one unit away from the average. A quickly sketched number line can confirm this:

image

The set {1, 3, 5} will have a smaller standard deviation than {0, 3, 6}. Both sets have an average of 3, but the first has terms 2, 0, and 2 units from the average while the second has terms 3, 0, and 3 units from the average. Again, you can confirm this with a quick sketch:

image

You could also calculate the standard deviation for the sets {1, 3, 5} and {0, 3, 6} to confirm:

image

Since 6 is larger than image, you can see that the standard deviation of the second, more widely spaced set of numbers is larger than that of the first set. However, you will save much time on Test Day by using your understanding of how standard deviation works to estimate rather than calculating the answers to questions such as these.

In other words, the GMAT testmakers are more interested in whether you understand the concept of standard deviation and can apply Critical Thinking to situations that involve it than in whether you can perform complex mathematical calculations.

In-Format Question: Median, Mode, Range, and Standard Deviation on the GMAT

Now let’s use the Kaplan Method on a Data Sufficiency question dealing with median, mode, range, and standard deviation:

List L consists of 12, 8, 10, 14, and 6. If two numbers are removed from list L, what is the standard deviation of the new list?

Step 1: Analyze the Question Stem

This is a Value question. You know that two numbers are removed from the list 12, 8, 10, 14, 6. You want to determine the standard deviation of the new list. There is no other information given, so let’s look at the statements.

Step 2: Evaluate the Statements Using 12TEN

Statement (1): You are told that the range of the new list is 8. The only way to have a range of 8 is for both 6 and 14 to appear in the new list, because any other pair of numbers in the original list will be closer to one another than 8. So the new list must contain 6 and 14.

The new list will also contain one of the numbers 8, 10, and 12. If, for example, the third number in the list is 10, then the list will be 6, 10, and 14. If the third number in the list is 12, then the numbers in the list will be 6, 12, and 14. The standard deviation of the list 6, 10, and 14 will be different from the standard deviation of the list 6, 12, and 14 because the two sets have different means and the numbers in the two lists are dispersed differently around the mean.

More than one standard deviation is possible, and therefore more than one answer to the question is possible. Statement (1) is insufficient. Eliminate choices (A) and (D).

Statement (2) says that the median of the new set is 8. Because there will be an odd number (3) of members of the new list, 8 will have to appear in the list in order for 8 to be the median. Additionally, since 8 must be in the middle of the new list and there is only one possible value less than 8, you know that 6 must also be in the list. The new list could be 6, 8, 10, which has a median of 8, and the new list could also be 6, 8, 14, which also has a median of 8.

The standard deviation of the numbers in the list 6, 8, 10 is less than the standard deviation of the list 6, 8, 14. More than one standard deviation is possible, and therefore more than one answer to the question is possible. So Statement (2) is insufficient. You can eliminate choice (B).

Now that you’ve determined that Statement (1) and Statement (2) are both insufficient on their own, look at the statements together: you know from Statement (1) that 6 and 14 must be in the list, and you know from Statement (2) that 6 and 8 must be in the list. The statements taken together tell you that the list must consist of 6, 8, and 14. Because there is only one possible list, there can only be one standard deviation. The statements taken together are sufficient. (C) is correct.

TAKEAWAYS: MEDIAN, MODE, RANGE, AND STANDARD DEVIATION

 

Practice Set: Median, Mode, Range, and Standard Deviation on the GMAT

Answers and explanations at end of chapter

  1. Set S contains the elements {−4, x, 0, 17, 1}. What is the median of set S?

    • (1)  x > 1

    • (2) x < 2

  2. In the list 3, 4, 5, 5, 5, 5, 7, 11, 21, what fraction of the data is less than the mode?

    • image

    • image

    • image

    • image

    • image

  3. The range of the set {60, 120, −30, x, 0, 60} is 210. If a is the median, b is the mode, and c is the average (arithmetic mean), is a = b?

    • (1)  abc > 0

    • (2) b = 60

  4. If x ≤ 4 and y ≥ 10, what is the range of the numbers in the set {5, 8, x, y}?

    • (1)  3x + 5y = 72

    • (2)  −7x + 7y = 56

SEQUENCES OF INTEGERS

Average and Median of Consecutive Integers

The average, or mean, of a set of consecutive integers will equal its median.

Example: What is the average of 1, 2, 3, 4, and 5?

image

For this set, 3 is both the mean and the median. This is always the case for sets of consecutive or evenly spaced integers.

There is another way to calculate the average of this set. The average of a series of consecutive or evenly spaced integers is also the same as the average of its first and last terms. For the last example, that works out to image.

Sum of Numbers

If you know the average of a group of numbers and how many numbers are in the group, you can find the sum of the numbers. You can think of it as if all the numbers in the group are equal to the average value.

Sum of values = Average value × Number of values

Example:  What is the sum of all the integers between 1 and 66 inclusive?

Since this is a series of consecutive integers, you can find its average by averaging the first and last terms: image.

To find the number of terms in a set of consecutive integers, take the difference between the largest and the smallest numbers and add 1.

Sum of values = Average value × Number of values

Sum = image

Granted, you still have to calculate 67 × 33, but that is much easier than adding all 66 numbers!

You may also have to find a sum in what might seem to be a normal average problem.

Example:  The average daily temperature for the first week in January was 31 degrees. If the average temperature for the first six days was 30 degrees, what was the temperature on the seventh day?

The sum for all 7 days = 31 × 7 = 217 degrees.

The sum of the first 6 days = 30 × 6 = 180 degrees.

The temperature on the seventh day = 217 − 180 = 37 degrees.

In-Format Question: Sequences of Integers on the GMAT

Now let’s use the Kaplan Method on a Problem Solving question dealing with sequences of integers:

In a new housing development, trees are to be planted in a straight line along a 166-foot stretch of sidewalk. Each tree will be planted in a square plot with sides measuring one foot. If there must be 14 feet between each plot, what is the maximum number of trees that can be planted?

Step 1: Analyze the Question

Though this is not a geometry question, a quick sketch of the situation will help illustrate how to solve it.

image

So you know that the unit of one tree and one space is 1 foot + 14 feet = 15 feet.

Step 2: State the Task

To find how many trees can be planted, determine the feet required for a tree and the space between trees. Then divide the total length of the sidewalk by the unit of one tree and the space between trees.

Step 3: Approach Strategically

Each tree takes up 1 foot, and each space takes up 14 feet. Together they take up 15 feet. Now find how many times 15 goes into the total number of feet on one side of the sidewalk:

166 ÷ 15 = 11, with a remainder of 1 foot.

You can plant one last tree in the remaining foot, bringing the total number of trees to 12. This means along the sidewalk, you can plant 12 trees with 11 spaces between them, as long as you start and end with a tree. (E) is correct.

Step 4: Confirm Your Answer

Make sure your answer makes sense in the context of the question. Did you take into account the remainder of the division? Will an entire tree fit in the remaining space? You can use these questions to confirm your work.

TAKEAWAYS: SEQUENCES OF INTEGERS

 

Practice Set: Sequences of Integers on the GMAT

Answers and explanations at end of chapter

  1. What is the sum of the integers 45 through 175 inclusive?

    • 12,295

    • 13,000

    • 14,300

    • 14,410

    • 28,820

  2. What is the sum of the multiples of 4 between 13 and 125 inclusive?

    • 1,890

    • 1,960

    • 2,200

    • 3,780

    • 4,400

  3. If t is the sum of three consecutive positive integers, is t a multiple of 24?

    • (1)  The smallest of the 3 integers is even.

    • (2) t is a multiple of 3.

  4. If the sum of five consecutive positive integers is a, then what is the sum of the next five consecutive integers in terms of a?

    • a + 1

    • a + 5

    • a + 25

    • 2a

    • 5a

COMBINATIONS AND PERMUTATIONS

Some GMAT questions ask you to count the number of possible ways to select a small subgroup from a larger group. If the selection is unordered, then it’s a combinations question. But if the selection is ordered, it is a permutations question.

For example, if a question asks you to count the possible number of different slates of officers who could be elected to positions in class government, then order matters—President Bob and Vice President Thelma is a different slate than President Thelma and Vice President Bob. You would use the permutations formula in this scenario. But if you had to count the number of pairs of flavors of jelly beans, you are solving a combinations question; cherry and lemon is the same pair as lemon and cherry, so order does not matter (i.e., you wouldn’t count those as two different pairs).

The very first thing you need to do is use Critical Thinking to figure out whether a given question calls for an ordered or an unordered selection—otherwise, you won’t know which formula to use.

Combinations

The combinations formula is used when solving for the number of k unordered selections one can make from a group of n items, where k ≤ n. This is usually referred to as nCk, which is often said as “n choose k.” Not that the GMAT is an oral exam, of course, but it’s helpful to say it that way to yourself because that’s what making combinations is—an act of choosing small groups from a larger group.

Here’s the formula, in which n!, or n-factorial, is the product of n and every positive integer smaller than n (for example, 5! = 5 × 4 × 3 × 2 × 1), where n is a positive integer:

image

Example:  A company is selecting 4 members of its board of directors to sit on an ethics subcommittee. If the board has 9 members, any of whom may serve on the subcommittee, how many different selections of members could the company make?

Since the order in which you select the members doesn’t change the composition of the committee in any way, this is a combinations question. The size of the group from which you choose is n, and the size of the selected group is k. So n = 9 and k = 4.

image

image

image

Save yourself some work. There’s rarely a need to multiply out factorials, since many of the factors can quickly be canceled.

image

9 × 2 × 7 = 126

Sometimes the problems will be more complicated and will require multiple iterations of the formula.

Example:  County X holds an annual math competition, to which each county high school sends a team of 4 students. If school A has 6 boys and 7 girls whose math grades qualify them to be on their school’s team, and competition rules stipulate that the team must consist of 2 boys and 2 girls, how many different teams might school A send to the competition?

The order of selection doesn’t matter here, so you can use the combinations formula. But be careful: if you lump all the students together in a group of 13 and calculate 13C4, you’d wind up including some all-boy teams and all-girl teams. The question explicitly says you can only select 2 boys and 2 girls. So you aren’t really choosing 4 students from 13 but rather choosing 2 boys from 6 and 2 girls from 7.

image

image

image

3 × 5 × 7 × 3 = 315 possible teams consisting of 2 boys and 2 girls.

Permutations

If the order of selection matters, you can use the permutation formulas:

Number of permutations (arrangements) of n items = n!

Number of permutations of k items selected from n = nPk = image

Example:  How many ways are there to arrange the letters in the word ASCENT?

Clearly order matters here, since ASCENT is different from TNECSA. There are 6 letters in the word, so you must calculate the permutations of 6 items:

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Another way to solve is to draw a quick sketch of the problem with blanks for the arranged items. Then write in the number of possibilities for each blank in order. Finally, multiply the numbers together. Many high-difficulty GMAT permutation questions resist formulaic treatment but are easier to complete with the “draw blanks” approach.

Here’s how you would use this technique to solve the ASCENT problem:

ASCENT has 6 letters, so you need 6 blanks:

image

There are 6 letters you might place in the first blank (A, S, C, E, N, or T):

image

No matter which letter you placed there, there will be 5 possibilities for the next blank:

image

There will be 4 for the next, 3 thereafter, 2 after that, and just 1 letter left for the last:

image

Notice that you wind up reproducing the arrangements formula, n!. It’s quite possible to solve most permutation problems without figuring out the right formula ahead of time, although knowing the formulas can save you time on Test Day.

Example:  There are 6 children at a family reunion, 3 boys and 3 girls. They will be lined up single-file for a photo, alternating genders. How many arrangements of the children are possible for this photo?

You may not be sure how to approach this with a formula, so draw a picture. You know you’ll have 6 “blanks,” but you don’t know whether to begin with a boy or with a girl. It could be either one. So try both:

bgbgbg or gbgbgb

image

Any of the three boys could go in the first spot, and any of the three girls in the second:

image

The next spot can be filled with either of the remaining two boys; the one after by either of the two remaining girls. Then the last boy and the last girl take their places:

image

That’s the boy-first possibility. The same numbers of boys and girls apply to the girl-first possibility, and so you get:

image

9 × 4 × 1 + 9 × 4 × 1

36 + 36 = 72 possible arrangements of 3 boys and 3 girls, alternating genders.

Hybrids of Combinations and Permutations

Some questions involve elements of both ordered and unordered selection.

Example:  How many ways are there to arrange the letters in the word ASSETS?

Earlier you saw that the rearrangement of ASCENT was a permutation. But what about ASSETS? The order of the E, the A, the T, and the S’s matter … but the order of the three S’s themselves does not.

Think about it this way: put a “tag” on the S’s … AS1S2ETS3. If you just calculated 6! again, you’d be counting AS1S2ETS3 and AS3S1ETS2 as different words, even though with the tags gone, you can clearly see that they aren’t (ASSETS is the same as ASSETS). So you’ll need to eliminate all the redundant arrangements from the 6! total.

Since there are 3 S’s in the word ASSETS, there are 3! ways to rearrange those S’s without changing the word. You need to count every group of 3! within the 6! total as only 1 arrangement.

So instead of 6! arrangements, as there were for ASCENT, the word ASSETS has image.

image

The same logic would apply to the arrangements of ASSESS: image.

And if two letters repeat, you need two corrections to eliminate the redundant arrangements. For instance, the number of arrangements of the letters in the word REASSESS is image.

Some tricky GMAT problems boil down to this “rearranging letters” problem.

Example:  A restaurant is hanging 7 large tiles on its wall in a single row. How many arrangements of tiles are possible if there are 3 white tiles and 4 blue tiles?

This problem essentially asks for the arrangements of WWWBBBB. Remember that although there are 7 total tiles to arrange, all the white tiles are indistinguishable from one another, as are the blue tiles. Therefore, you will need to divide out the number of redundant arrangements from the 7! total arrangements:

image = 7 × 5 = 35

Did you notice that image is the same as 7C3? Which is also the same as image?

Whether you think “rearrange WWWBBBB,” “choose 4 of the 7 tiles to be blue (the rest will be white),” or “choose 3 of the 7 tiles to be white (the rest will be blue),” the result is the same. As you’ll soon see, many probability questions involve just this kind of calculation.

In-Format Question: Combinations and Permutations on the GMAT

Now let’s use the Kaplan Method on a Problem Solving question dealing with combinations and permutations:

Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?

Step 1: Analyze the Question

You have to arrange six children in six chairs, but two of the children can’t sit together. You’re asked to calculate the number of different arrangements of children.

Step 2: State the Task

You need to calculate the total number of possible arrangements of the children. Then, you’ll subtract the number of ways Betsy can sit next to Emily.

Step 3: Approach Strategically

The possible number of arrangements of six elements is 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

Now you’ll have to calculate the number of arrangements that would violate the question stem by putting Betsy next to Emily. If you number the seats from left to right, there are 5 ways they could sit together if Betsy is on the left and Emily is on the right:

Seats 1 & 2

Seats 2 & 3

Seats 3 & 4

Seats 4 & 5

Seats 5 & 6

And there are 5 more ways if Emily is on the left and Betsy is on the right, for a total of 10. Now, for any one of those 10 ways, the four remaining children can be seated in 4! ways: 4! = 4 × 3 × 2 × 1 = 24. So you need to subtract 24 × 10 = 240 ways that have Betsy and Emily sitting together from your original total of 720: 720 − 240 = 480. The correct answer is (B).

Step 4: Confirm Your Answer

You made several calculations for this problem, so review your work to confirm your calculations are correct.

TAKEAWAYS: COMBINATIONS AND PERMUTATIONS

 

Practice Set: Combinations and Permutations on the GMAT

Answers and explanations at end of chapter

  1. Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?

    • 9

    • 12

    • 14

    • 15

    • 60

  2. A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. If there is only one way to present each letter, how many different codes are possible?

    • 42

    • 210

    • 420

    • 840

    • 5,040

  3. Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent?

    • (1)  The display contains an equal number of gold and silver coins.

    • (2)  If only the silver coins were displayed, 5,040 different arrangements of the silver coins would be possible.

  4. Of the 5 distinguishable wires that lead into an apartment, 2 are for cable television service and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?

    • 6

    • 7

    • 8

    • 9

    • 10

PROBABILITY

Probability is the likelihood that a desired outcome will occur.

image

Example:  If you have 12 shirts in a drawer and 9 of them are white, the probability of picking a white shirt at random is image = image. The probability can also be expressed as 0.75 or 75%.

To find the probability that one or another of mutually exclusive events occurs, add the probabilities of the events.

Example:  Of the 12 shirts in a drawer, 4 are white, 5 are blue, and 3 are green. If you choose one shirt at random, what is the probability that the shirt you choose is either white or green?

If you choose only one shirt, that shirt cannot be more than one color. This is what the phrase “mutually exclusive events” means; the shirt you choose will either be white, or blue, or green. In this case, the desired outcomes are white and green. The probability of picking a white shirt at random is image. The probability of picking a green shirt at random is image. The probability that the chosen shirt is either white or green is therefore image + image = image.

Many hard probability questions involve finding the probability of a certain outcome after multiple repetitions of the same experiment or different experiments (a coin being tossed several times, etc.).

Broadly speaking, there are five approaches you can take to probability questions.

Approach 1

Multiply the probabilities of individual events.

This works best when you know the probability for each event and you need to find the probability of all the events occurring (e.g., the probability that the first flip of a coin lands heads up and the second flip lands tails up). Make sure to pay attention to what effect, if any, the outcome of the first event has on the second, the second on the third, and so on.

Example:  If 3 students are chosen at random from a class with 6 girls and 4 boys, what is the probability that all 3 students chosen will be girls?

The probability that the first student chosen will be a girl is image, or image.

After that girl is chosen, there are 9 students remaining, 5 of whom are girls. So the probability of choosing a girl for the second student is image.

Finally, for the third pick, there are 8 total students remaining, 4 of whom are girls. The probability of choosing a girl for the third student is therefore image = image.

The probability that all three students chosen will be girls is image × image × image = image = image.

Approach 2

Subtract the probability of the undesired outcomes from the total.

This approach works best when you cannot readily calculate the probability of the desired outcomes but you can readily do so for the undesired ones. In probability, the total of all possible outcomes is always 1.

Example:  If a fair coin is flipped three times, what is the probability of getting at least one tail?

Note that “fair” means that every outcome is equally possible. A fair coin will land heads up 50 percent of the time.

What’s desired is one tail, two tails, or three tails in three flips. That’s a lot to keep track of. But what’s undesired is very clear—three heads in a row.

Total − Undesired

1 − HHH

1 − image× image× image

1 − image

image

The “Total − Undesired” approach works well for many kinds of GMAT questions, not just for probability. For example, it’s how you’d calculate the shaded area of a figure such as image.

Approach 3

Solve for the probability of one possible desired outcome, then multiply by all the permutations of that outcome.

This works best when you need to know the probability that an event will occur a certain number of times, but the order of those occurrences doesn’t matter.

Example:  If a fair coin is flipped 5 times, what is the probability of getting exactly 3 heads?

It’s clear what you desire—3 heads and 2 tails—but you don’t have to get it in any particular order. HTHTH would be fine, as would HHHTT, TTHHH, and so forth. So Approach #1 wouldn’t work well for this problem. But you can use Approach #1 to figure out the probability of one of these outcomes:

image image × image × image × image × image = image

Now you can multiply this result by the number of ways you could get this outcome—in other words, by the number of ways you could rearrange the letters in HHHTT. You saw how to do this earlier, in our discussion of “hybrid combinations and permutations” problems a few pages ago.

Use the combinations formula to solve for the number of arrangements of HHHTT (which is the same as calculating the total number of arrangements, 5!, and dividing out the indistinguishable outcomes):

image

There are 10 different ways to get 3 heads and 2 tails. Each one of those outcomes has a probability of image.

The probability of getting exactly 3 heads in 5 coin flips is image.

Approach 4

You could work with the numerator and denominator of the probability formula separately, then lump the results together into one calculation. You calculate the total number of possible outcomes and the total number of desired outcomes, then put them together in one big fraction (instead of multiplying lots of little fractions together).

Like Approach 3, this one also works best when what you want is very specific but the order in which it happens is not.

Example:  A bag holds 4 red marbles, 5 blue marbles, and 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles, and 1 green marble?

image

Start by thinking about the possible outcomes. You are reaching into a bag of 11 marbles and pulling out 5. (The fact that they are pulled out one by one doesn’t change anything; in the end, you still have 5 marbles.) Quite literally, this is “11 choose 5,” 11C5.

image

Save yourself some work: don’t multiply out factors that may later cancel. You can leave the expression as 11 × 2 × 3 × 7 for now.

Now, what is desired? You want 2 of the 4 red marbles, literally “4 choose 2,” or 4C2. You also want 2 of the 5 blue marbles (5C2) and 1 of the 2 green marbles (2C1).

# desired selections = 4C2 and 5C2 and 2C1

image

image

6 × 10 × 2

Now put the fraction together:

image

Approach 5

Don’t do any math at all. Just count the various outcomes.

This approach works best when the numbers involved are small. In this circumstance, there’s really no need to waste time thinking of the right arithmetic calculation.

Example:  If a fair coin is flipped two times, what is the probability of getting exactly one head?

Two coin flips aren’t very much. List out the possible results:

HH, HT, TH, or TT

Of these 4 possibilities, 2 have exactly one head—HT and TH. So the probability of getting exactly 1 head in 2 flips is image, or image.

Take your time thinking probability questions through—often the hardest part is figuring out which approach you want to take.

In-Format Question: Probability on the GMAT

Now let’s use the Kaplan Method on a Problem Solving question dealing with probability:

Robert tossed a fair coin 3 times. What is the probability that the coin landed heads up exactly twice?

Step 1: Analyze the Question

As with any probability question, think carefully about your approach before you start any calculations. You know that each coin flip has a image chance of being heads and a image chance of being tails.

Step 2: State the Task

You need to calculate the probability of exactly two heads in three flips. There are many approaches you could take. But even if you didn’t see a math-based one, the low number of flips means that you could just count possibilities.

Step 3: Approach Strategically

You can write out all the ways of flipping three coins and count how many have exactly two heads. As long as you are systematic about it, so you don’t miss anything, you can get the answer well within two minutes. HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. That’s 8 possible outcomes, 3 of which have exactly two heads: image, or 0.375, is the answer.

Of course, there are math-based solutions as well. One is to calculate the probability of one desired outcome, then multiply by the number of desired outcomes. HHT has a probability of image× image× image = image, or 0.125. The number of ways of arranging 2 H’s and 1 T is calculated in the same manner as the rearranged letter problems. In this case, that’s

image

So the probability of getting exactly two heads in three flips is 0.125 × 3 = 0.375. The answer is (C).

Step 4: Confirm Your Answer

Whichever method you chose, there weren’t many calculations to make. Double-check your arithmetic so that you avoid careless errors and quickly reread the question stem, confirming that you answered the question that was asked.

TAKEAWAYS: PROBABILITY

 

Practice Set: Probability on the GMAT

Answers and explanations at end of chapter

  1. A bowl has only 5 apples and 5 bananas. If one piece of fruit is selected from the bowl at random, and a second piece is selected from the bowl without replacing the first, what is the probability that both pieces of fruit chosen are apples?

    • image

    • image

    • image

    • image

    • image

  2. A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

    • image

    • image

    • image

    • image

    • image

  3. A student is entered in a college housing lottery for 2 consecutive years. What is the probability that the student receives housing through the lottery for at least 1 of these years?

    • (1)  Eighty percent of the students in the lottery do not receive housing through the lottery in any given year.

    • (2) Each year, 1 of 5 students receives housing through the lottery.

  4. Each person in Room A is a student, and image of the students in Room A are seniors. Each person in Room B is a student, and image of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

    • image

    • image

    • image

    • image

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Answer Key

Practice Set: Median, Mode, Range, and Standard Deviation on the GMAT

  1. A

  2. A

  3. E

  4. B

Practice Set: Sequences of Integers on the GMAT

  1. D

  2. B

  3. A

  4. C

Practice Set: Combinations and Permutations on the GMAT

  1. E

  2. C

  3. C

  4. D

Practice Set: Probability on the GMAT

  1. C

  2. D

  3. D

  4. C

Answers and Explanations

Practice Set: Median, Mode, Range, and Standard Deviation on the GMAT

1.    (A)

Set S contains the elements {−4, x, 0, 17, 1}. What is the median of set S?

Step 1: Analyze the Question Stem

In this Value question, we are asked to determine the median of Set S. The median of a set with an odd number of terms is the middle term when the terms are arranged in ascending or descending order. Sufficiency means being able to determine one and only one value for the median.

Step 2: Evaluate the Statements Using 12TEN

Statement (1) states that x > 1. This results in either the set {−4, 0, 1, x, 17} or the set {−4, 0, 1, 17, x}. In either case, the median is 1, so this statement is sufficient. Eliminate (B), (C), and (E).

Statement (2) states that x < 2. If x < 0, the set could be {x, −4, 0, 1, 17} or {−4, x, 0, 1, 17}; either way, the median is 0. However, if 0 < x < 1, the set is {−4, 0, x, 1, 17} and the median is x. If 1 < x < 2, the set is {−4, 0, 1, x, 17} and the median is 1. This statement permits multiple possibilities for the median, so it is insufficient. Eliminate (D).

Statement (1) alone is sufficient, so the correct answer is (A).

2.    (A)

In the list 3, 4, 5, 5, 5, 5, 7, 11, 21, what fraction of the data is less than the mode?

Step 1: Analyze the Question

This relatively straightforward question can be answered by using the definitions of mode and proportions.

Step 2: State the Task

To answer this question, we need to determine the mode, the total number of items in the list, and the number of items in the set less than the mode.

Step 3: Approach Strategically

The mode of any list of numbers is the number that appears most frequently in the list. In this case, the mode is 5.

The total number of items in the list is 9.

Finally, the number of items in the list less than the mode, or 5, is 2.

So our final fraction is image, and (A) is correct.

Step 4: Confirm Your Answer

While this is a straightforward question, be careful to review your calculations to confirm that you did not make any simple counting errors.

3.    (E)

The range of the set {60, 120, −30, x, 0, 60} is 210. If a is the median, b is the mode, and c is the average (arithmetic mean), is a = b?

Step 1: Analyze the Question Stem

In this Yes/No question, the range of the set {60, 120, −30, x, 0, 60} is 210. Let’s write the given numbers of the list in increasing order. We have {−30, 0, 60, 60, 120}. Because no pair of numbers from −30, 0, 60, 60, and 120 differs by 210, we must have one of these possibilities:

If x < −30, then 120 − x = 210, in which case x = −90.

If x > 120, then x − (−30) = 210, in which case x = 180.

These are the only possible values of x. Our task is now much clearer. We must determine if x is equal to either of these values. Let’s look at the statements.

Step 2: Evaluate the Statements Using 12TEN

Statement (1): We are told that abc > 0. We can test our two possible values for x to determine whether this is sufficient.

If x = −90, then the members of the list written in increasing order are −90, −30, 0, 60, 60, 120. The median a is image = 30, so the median is positive.

The mode b is 60, which is positive. The average c of the numbers is also positive. Statement (1) is true because abc is positive, and it is not true that a = b. In this case, the answer to the question is “no.”

If x = 180, then the members of the list written in increasing order are −30, 0, 60, 60, 120, and 180. Here, the median a is 60, so the median is positive.

The mode b is 60, which is positive. The average c of the numbers is also positive. So here, Statement (1) is true because abc is positive, and it is true that a = b. In this case, the answer to the question is “yes.”

Because different answers to the question are possible, Statement (1) is insufficient. We can eliminate choices (A) and (D).

Statement (2): From Statement (1), we have seen that if x = −90, then the members of the list written in increasing order are −90, −30, 0, 60, 60, and 120. Statement (2) is true because the mode b is 60. The median a is 30, which is not equal to the mode b. In this case, the answer to the question is “no.”

We have also seen that if x = 180, then the members of the list written in increasing order are −30, 0, 60, 60, 120, 180. Statement (2) is true because the mode b is 60. The median a is 60, which is equal to the mode b, which is 60. In this case, the answer to the question is “yes.”

Because different answers to the question are possible, Statement (2) is insufficient. We can eliminate choice (B).

Each statement permitted both x = 180 and x = −90. So when combined, both values are still possible. The statements together are insufficient, so choice (E) is correct.

4.    (B)

If x ≤ 4 and y ≥ 10, what is the range of the numbers in the set {5, 8, x, y}?

Step 1: Analyze the Question Stem

This is a Value question. We’re asked for the range of numbers in a set that includes two variables, and we’re given inequalities expressing the range of possible values for each of the variables. The inequalities allow us to arrange the numbers in the set from smallest to largest: x, 5, 8, y. Remembering that the definition of the range of a set of numbers is the difference between the largest and smallest value, there are two ways we’ll be able to answer the question with a single value: get the individual values of x and y, or find the value of the expression y − x.

Step 2: Evaluate the Statements Using 12TEN

Picking Numbers illustrates that Statement (1) is insufficient. If, for example, x = 0, we would get 3(0) + 5y = 72. This is just the same as saying that 5y = 72. Dividing each side by 5 leaves us with y = 14.4, so y − x would equal 14.4. The exact calculation is unimportant. For a Value question, we just need to confirm that another permissible pair of numbers will yield a different result to know that the statement is insufficient. Let’s try x = 4. This gives us 3(4) + 5y = 72. Do the multiplication and subtract 12 from each side to get 5y = 60. Divide each side by 5, and we have y = 12. In this case, y − x would equal 8. Since two solutions are possible, Statement (1) is insufficient. Eliminate answer choices (A) and (D).

Simplifying Statement (2) shows that it is sufficient. Start by rewriting the equation to read 7y − 7x = 56. Now factor out the 7 to get 7(y − x) = 56. Divide each side by 7, and you’re left with y − x = 8. Since the expression y − x has the same value as the range of numbers in the set, Statement (2) is sufficient. The correct answer is (B).

Practice Set: Sequences of Integers on the GMAT

5.    (D)

What is the sum of the integers 45 through 175 inclusive?

Step 1: Analyze the Question

We are asked to sum up a very large list of numbers. When the GMAT presents us with seemingly onerous calculations, it generally means there is a much easier way if we apply Critical Thinking.

Step 2: State the Task

We need to find the sum of all the integers from 45 to 175, inclusive.

Step 3: Approach Strategically

To find the sum of a series of consecutive integers, multiply the average of the largest and smallest terms (which is equal to the average of all the terms in the set) by the total number of terms. The largest and smallest terms in this sequence are 175 and 45, and their average is image = image = 110. There are 175 − 45 + 1 = 131 terms in the set, so the sum of all the integers in the set is 110 × 131 = 14,410. The correct answer is (D).

Step 4: Confirm Your Answer

Double-check your calculations to ensure that you haven’t made any avoidable mistakes.

6.    (B)

What is the sum of the multiples of 4 between 13 and 125 inclusive?

Step 1: Analyze the Question

When working with consecutive sets, look for ways to save time and work by recognizing patterns and first and last terms in the set.

To find the sum of terms in a set of consecutive numbers, we can apply the average formula.

image

Sum of Terms = (Average)(Number of Terms)

Step 2: State the Task

We need to calculate the average of the terms in the set and the number of items in the set. To do so, we need to determine the smallest and largest terms in the sequence.

Step 3: Approach Strategically

Find the largest and smallest multiple of 4 between 13 and 125. In this range, the smallest multiple of 4 is 16, and the largest is 124.

The average of a consecutive set equals the average of the smallest and largest terms. The average of these two numbers is

image

So the average of all terms in this set is 70.

Now find the number of multiples of 4 between 16 and 124 inclusive. We could simply list all of the terms in the sequence, but that would be time-consuming. A great shortcut to determine this value is to find the difference between the smallest and largest terms and add 1. Because the numbers in this set are multiples of 4, we must divide the difference by 4 before adding 1.

The number of terms in this set is

image

There are 28 multiples of 4 between 16 and 124 inclusive.

So the average of the terms in the set is 70, and the number of terms in the set is 28. The sum of the terms in the set is simply the product of these two values:

28 × 70 = 1,960

(B) is correct.

Step 4: Confirm Your Answer

Pay careful attention to the wording of this question stem. The set is inclusive, so confirm that you added 1 when you calculated the number of terms.

7.    (A)

If t is the sum of three consecutive positive integers, is t a multiple of 24?

Step 1: Analyze the Question Stem

This is a Yes/No question. If t is the sum of three consecutive positive integers, we can call the smallest integer x and write the other 2 as (x + 1) and (x + 2), respectively. Therefore, the sum of the three consecutive integers is x + (x + 1) + (x + 2) or, combining like terms, 3x + 3. Finally, we can factor and get 3(x + 1).

So the question really asks, “Is 3(x + 1) a multiple of 24?”

Step 2: Evaluate the Statements Using 12TEN

Statement (1): We are told that x is an even number. If x is even, then x + 1 must be odd since an even plus an odd is always odd. Also, an odd times an odd is always odd; therefore, 3(x + 1) is odd. Since an odd number is never divisible by an even number, 3(x + 1) cannot be a multiple of 24. Hence, the answer is “always no,” and Statement (1) is sufficient. Eliminate answers (B), (C), and (E).

Statement (2): We are told that t is a multiple of 3. Since we already know that t = 3(x + 1), where x is an integer, this statement gives us no new information. Statement (2) is insufficient. The correct answer is (A).

8.    (C)

If the sum of five consecutive positive integers is a, then what is the sum of the next five consecutive integers in terms of a?

Step 1: Analyze the Question

We’re given the sum of five consecutive integers, and we have to find the sum of the next five integers and express it in terms of a, the sum of the first five integers.

Step 2: State the Task

We only have to add five numbers, so it’s easier just to list them and add them rather than use our rule for adding consecutive numbers. We’ll calculate the two sums and express the second in terms of a.

Step 3: Approach Strategically

This question is easily answered by Picking Numbers. To make it easy, start your sequence of integers with 1: 1 + 2 + 3 + 4 + 5 = 15. So a = 15. The next five consecutive integers are 6, 7, 8, 9, and 10. They sum to 40. 40 = a + 25. Answer choice (C) matches.

If you solved with algebra, you would call your first variable x. The sum of the five consecutive integers starting with x is x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10. The sum of the next five consecutive integers is (x + 5) + (x + 6) + (x + 7) + (x + 8) + (x + 9) = 5x + 35. We’re told the sum of the first five consecutive integers is a. So 5x + 10 = a. We can add 25 to each side of the equation to get our second sum in terms of a: 5x + 35 = a + 25.

Step 4: Confirm Your Answer

No matter which approach you take, there isn’t much calculation, so you can afford a quick check of your arithmetic. Also, go back over the question stem to make sure you didn’t misread anything.

You might even use Critical Thinking to note that each number in the second set is 5 higher than the corresponding number in the first set, so the sum of the second set is 5 × 5 = 25 more than the sum of the first set.

Practice Set: Combinations and Permutations on the GMAT

9.    (E)

Amanda goes to the toy store to buy 1 ball and 3 different board games. If the toy store is stocked with 3 types of balls and 6 types of board games, how many different selections of the 4 items can Amanda make?

Step 1: Analyze the Question

Amanda is buying 1 ball and 3 different board games from a selection of 3 balls and 6 types of board games. Since we are solving for the number of different subgroups that can be chosen from a group, this appears to be a combinations or permutations question.

Step 2: State the Task

We need to find the number of different selections of 4 items that Amanda can make.

Step 3: Approach Strategically

Amanda is choosing 3 board games from a total of 6, so we can use the combination formula (since a different arrangement of the same board games is not considered a different selection) to find the number of combinations of board games:

nCk = image

6C3 = image

6C3 = image

6C3 = 20

There are 20 ways for Amanda to choose 3 board games from a total of 6. For each of those 20 ways, Amanda can choose 1 of 3 balls, so there are 20 × 3 = 60 different ways for Amanda to choose 1 ball and 3 games from 3 balls and 6 games. Choice (E) is correct.

Step 4: Confirm Your Answer

With combinations and permutations questions, check to ensure that you used the correct formula and that you didn’t make a mistake when canceling out terms from the numerator and denominator.

10.   (C)

A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. If there is only one way to present each letter, how many different codes are possible?

Step 1: Analyze the Question

We have to make a seven-letter code, but some of our letters are repeated. We have three As, two Bs, one C, and one D. We have to calculate the possible number of different codes.

Step 2: State the Task

We’ll calculate the number of permutations, remembering to take the repeated letters into account.

Step 3: Approach Strategically

To calculate the number of permutations where some of the elements are indistinguishable, we’ll divide the total number of permutations by the factorial of the number of indistinguishable elements. So we have image. Answer choice (C) is correct.

Step 4: Confirm Your Answer

Review your calculations to confirm your answer is correct.

11.   (C)

Each of the coins in a collection is distinct and is either silver or gold. In how many different ways could all of the coins be displayed in a row, if no 2 coins of the same color could be adjacent?

Step 1: Analyze the Question Stem

This is a Value question. We know that all the coins are distinguishable from one another and that every coin is gold or silver. No gold coins can be adjacent, and no silver coins can be adjacent. The question asks whether we can determine how many arrangements are possible. If we are given information that will lead us to a single number of possible arrangements, then we have sufficiency. If the given information leads to more than one possible number of arrangements, then we have insufficiency. Now let’s look at the statements.

Step 2: Evaluate the Statements Using 12TEN

Statement (1): We are told that the display contains an equal number of gold and silver coins. Because there are many more possible displays if there are 10 gold coins and 10 silver coins than if there are 3 gold coins and 3 silver coins, there is more than one possible answer to the question. Statement (1) is insufficient. Eliminate (A) and (D).

Statement (2): We are told that 5,040 different arrangements of only the silver coins are possible. If we let n be the number of silver coins, then n! = 5,040. This will lead to a unique value for n. (Actually, n = 7, because 7! = 5,040. However, you do not have to determine the value of n.)

From this, we could determine the number of possible arrangements of n silver coins and n gold. But what if there were n + 1 or n − 1 gold coins? We would then have a different number of possible arrangements while still obeying the restrictions in the question stem. We can eliminate (B).

Combine statements: We know that equal numbers of gold and silver coins are to be displayed, and we can determine the number of silver coins to be displayed. Therefore, we also know the number of gold coins. Since we know the numbers of each kind of coin, we could calculate their arrangements. Happily, since this isn’t Problem Solving, we don’t actually have to do the calculation. The statements together are sufficient, so (C) is the answer.

12.   (D)

Of the 5 distinguishable wires that lead into an apartment, 2 are for cable television service and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?

Step 1: Analyze the Question

For any question that asks you to make subsets or groups of a larger pool of items, pay careful attention to the wording of the question stem. We need to determine whether the order of the items in each subgroup matters. In this question, we are asked to make distinct combinations of three wires from a set of five wires. Because we want distinct combinations, the order of the wires will not matter, and we can apply the combinations formula to solve the question. Notice the phrase “at least” in the question stem. When questions use that phrase, solving for the total and then subtracting the undesired outcomes is usually a very efficient approach. The total is the number of ways one could select any three wires from five. The undesired outcome is selecting only telephone wires.

Step 2: State the Task

Apply the combinations formula to determine how many subsets of three wires can be created from the pool of five wires. Then determine how many ways there are to select only 3 telephone wires and subtract.

Step 3: Approach Strategically

Use the combinations formula, image, to find the total number of ways to choose three wires out of five. Substituting 5 for n and 3 for k, we find that the total number of ways to choose three wires out of five is

image

We now know that there are 10 subsets of three wires. Now turn your attention to what is undesired—selecting only telephone wires. Because there are only three telephone wires, there is only one possible way to choose three of the five wires such that all are telephone wires. Thus, there are 10 − 1 = 9 ways to choose three wires such that at least one of the wires would be for cable. (D) is correct.

Step 4: Confirm Your Answer

Pay careful attention to exactly what the question stem is asking. Notice that (E) is a trap for the unwary test taker who simply calculates the total number of possible outcomes.

Practice Set: Probability on the GMAT

13.   (C)

A bowl has only 5 apples and 5 bananas. If one piece of fruit is selected from the bowl at random, and a second piece is selected from the bowl without replacing the first, what is the probability that both pieces of fruit chosen are apples?

Step 1: Analyze the Question

Two pieces of fruit are randomly chosen, one at a time without replacement, from a bowl containing 5 apples and 5 bananas.

Step 2: State the Task

We must find the probability of selecting two apples.

Step 3: Approach Strategically

There are 5 apples and 5 bananas in the bowl, so the probability of choosing the first apple is image = image. After removing the apple, there are now 4 apples and 5 bananas left, so the probability of selecting the second apple is image. To find the probability of both events happening, multiply the probabilities: image. Choice (C) is correct.

Step 4: Confirm Your Answer

Reread the question stem to ensure that you understood the scenario correctly and double-check that you didn’t make any avoidable mistakes in calculating the probabilities.

14.   (D)

A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

Step 1: Analyze the Question

In this question, we must analyze the various scenarios in which a coin lands heads up at least twice in five tosses. However, keep in mind that with many probability questions, it is sometimes easier to calculate the probability of the situation not occurring and subtract this probability from 1. As you think through the various scenarios that can occur, determine whether there are fewer scenarios that will not occur. If so, working out the undesired probabilities will save you time on Test Day.

Step 2: State the Task

Determine the probability of each undesired outcome, then subtract the sum of these values from 1 to calculate the probability the question stem asks for.

Step 3: Approach Strategically

To determine the probability of at least two heads in five tosses, we must calculate the sum of the following probabilities: exactly two heads, exactly three heads, exactly four heads, and exactly five heads.

On the other hand, the undesired outcome can occur in only two possible ways: exactly one head is tossed, or no heads are tossed. This will be much simpler to calculate.

No heads: since the probability of heads or tails is image for a fair coin, the probability of no heads tossed, or all five tosses landing tails up, is image× image× image× image× image = image.

Exactly one head: if we let H represent heads and T represent tails, then in a sequence of five tosses, here are all the ways in which exactly one head can be tossed: HTTTT, THTTT, TTHTT, TTTHT, TTTTH.

Thus, in tossing a coin five times, the single toss of heads could occur on the first, second, third, fourth, or fifth toss. The probability of the sequence HTTTT occurring when a fair coin is tossed five times is

image× image× image× image× image = image

Now the probability of each of the sequences HTTTT, THTTT, TTHTT, TTTHT, and TTTTH is image, so the probability of tossing exactly one head is 5 image, which is image.

The probability of no heads is image, and the probability of exactly one head is image, so the probability of tossing fewer than two heads, which is the probability of tossing no heads or one head, is

image+ image = image = image

Therefore, the probability of tossing at least two heads is 1− image = image. (D) is correct.

Step 4: Confirm Your Answer

While this approach is a much more efficient strategy, be careful to remember to subtract the calculated value from 1 at the end of your work. The undesired outcome is often a wrong answer trap as seen in answer choice (B).

15.   (D)

A student is entered in a college housing lottery for 2 consecutive years. What is the probability that the student receives housing through the lottery for at least 1 of these years?

Step 1: Analyze the Question Stem

This Value question asks for the probability that a student receives housing in at least one of two consecutive years. In order to calculate that probability, we would need to know the probability of getting housing in each year. (Another possibility that could be useful is the chance of not getting housing, since 1 − Undesired Probability = Desired Probability.)

Step 2: Evaluate the Statements Using 12TEN

Statement (1): This gives us the probability of not getting into housing. An 80% chance of not getting housing each year means a 20% chance of getting housing each year. The exact answer isn’t something that we need to worry about, since this isn’t Problem Solving. We know an answer can be calculated, so Statement (1) is sufficient. Eliminate choices (B), (C), and (E).

Statement (2): This isn’t given in percentage form, but any proportion will serve. This gives us the likelihood of getting housing in any given year. We now know that Statement (2), which gives us the same information that Statement (1) does, must be sufficient. (D) is correct.

16.   (C)

Each person in Room A is a student, and image of the students in Room A are seniors. Each person in Room B is a student, and image of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

Step 1: Analyze the Question

This is a complex question, but it can be broken down into simple steps. As with any probability question, we must first consider all of the scenarios in which the desired outcome can occur. In this question, there are two different ways in which exactly one of two students chosen is a senior. Either (i) a senior is chosen from Room A and a non-senior is chosen from Room B or (ii) a non-senior is chosen from Room A and a senior is chosen from Room B.

Step 2: State the Task

We will determine the probabilities of the two scenarios above and add them together.

Step 3: Approach Strategically

Let’s start with scenario (i) and find the probability that a senior is chosen from Room A and a non-senior is chosen from Room B.

The probability that the student chosen from Room A is a senior is image.

The probability that the student chosen from Room B is not a senior is image.

So the probability that the student chosen from Room A is a senior and the student chosen from Room B is not a senior is image.

We won’t simplify this yet, because we can expect that the probability we will find when working with scenario (ii) will also have a denominator of 42.

Now let’s calculate the probability of scenario (ii). Let’s find the probability that a non-senior is chosen from Room A and a senior is chosen from Room B.

The probability that the student chosen from Room A is not a senior is image.

The probability that the student chosen from Room B is a senior is image.

So the probability that the student chosen from Room A is a not a senior and the student chosen from Room B is a senior is image.

Now we sum the total desired outcomes. The probability that exactly one of the students chosen is a senior is image + image = image = image. (C) is correct.

Step 4: Confirm Your Answer

Be careful to review your calculations and confirm that these two scenarios are indeed the only two ways in which the desired outcome can occur.

GMAT BY THE NUMBERS: STATISTICS

Now that you’ve learned how to approach statistics questions on the GMAT, let’s add one more dimension to your understanding of how they work.

Take a moment to try the following question. The next page features performance data from thousands of people who have studied with Kaplan over the decades. Through analyzing this data, we will show you how to approach questions like this one most effectively and how to avoid similarly tempting wrong answer choice types on Test Day.

A company plans to award prizes to its top 3 salespeople, with the largest prize going to the top salesperson, the next-largest prize to the next salesperson, and a smaller prize to the third-ranking salesperson. If the company has 12 salespeople on staff, how many different arrangements of winners are possible?

  • 1,728

  • 1,440

  • 1,320

  • 220

  • 6

Explanation

Questions that ask test takers to count possibilities are often the most frustrating. This is because most people try to shoehorn the problem into a pre-existing arithmetic formula without first analyzing the problem to see whether the formula applies. Understand the question before you jump into solving.

This problem asks you to calculate the number of ways you can distribute first-, second-, and third-place prizes among 12 people. Many test takers try to apply the combinations formula here. But doing so is not appropriate because that formula counts the number of unordered selections one can make from a group. You’re asked to order your selections into first-, second-, and third-place, so you need a different approach: simply count the number of possibilities for each prize.

Any of 12 people might win first place, leaving 11 possible choices for second. This leaves 10 possible winners for third. Hence, these prizes could be awarded 12 × 11 × 10 = 1,320 possible ways. Choice (C) is correct.

QUESTION STATISTICS
3% of test takers choose (A)
4% of test takers choose (B)
58% of test takers choose (C)
33% of test takers choose (D)
2% of test takers choose (E)
Sample size = 5,473

Unsurprisingly, the only other commonly selected answer on this problem, choice (D), is the result of erroneously applying the combinations formula to this problem. When solving a statistics question, understand the scenario before you start any arithmetic. You’ll dramatically improve your odds of avoiding the common traps and selecting the right answer.

More GMAT by the Numbers …

To see more questions with answer choice statistics, be sure to review the full-length CATs in your online resources.