Averages, Counting and Probability

🎲 Averages, Counting and Probability

Part 1 of 7 — Averages and Chance on the AR Subtest

Topic Overview

What These Questions Look Like

The Arithmetic Reasoning (AR) subtest is officially described as the "ability to solve basic arithmetic word problems." Averages show up as short stories about real numbers:

• Find the average: "A recruit's scores on four practice tests were 84, 90, 78 and 88. What was her average score?"
• Find the missing value: "What score does she need on the fifth test to raise her average to 86?"
• Combine two groups: "Twenty recruits average 80 on a test and thirty average 90. What is the average for all 50?"

Counting and probability questions tell a story too: how many different meal combos a menu allows, or the chance of pulling a certain name from a hat.

What is really being tested is whether you can turn the words into the right setup (add, divide, multiply) and carry it out cleanly by hand.

Basic probability and counting also appear on the Mathematics Knowledge (MK) subtest, whose official description lists probability. Everything in Parts 4 and 5 pays off twice.

The Conditions You Work Under

If you take a paper version of the test, the pace is much faster, so the shortcuts in this lesson matter even more.

AR is one of the four subtests in your AFQT score, the score that decides whether you can enlist, and it also feeds job line scores such as the Army GT score. Average questions follow a small set of patterns, which makes them some of the most winnable points on the subtest.

Seven Habits, Applied to Averages and Chance

Key Fact: The two most common ways to lose an average point are dividing by the wrong count and averaging two averages when the groups are different sizes. Parts 2 and 3 show you how to avoid both.

A Fully Worked Example

Problem. A squad leader recorded push-ups for 5 soldiers: 42, 38, 50, 45 and 35. What was the average number of push-ups?

Step 1 — Read the last sentence. The question asks for the average of 5 values.

Step 2 — Estimate. The values run from 35 to 50, so the average must be somewhere between them, probably in the low 40s.

Step 3 — Add in a column, then divide by the count.

$42 + 38 + 50 + 45 + 35 = 210$

$210 \div 5 = 42$

Step 4 — Check. 42 sits between 35 and 50, and it matches the estimate. ✓

The trap. Dividing 210 by 4 gives 52.5, which is bigger than the largest value (50). An average can never be outside the range of the data, so that choice is impossible.

Warm-up. 🎯 Read the last sentence first, then compute.

Warm-up: averages and totals. ✍️

1) What is the average of 14, 18, 22 and 26? (number only)

2) Five numbers have an average of 12. What is their sum? (number only)

Key Takeaways — Part 1

• Average, counting and probability questions on AR are word problems about scores, pay, loads and choices
• You have about 3.5 minutes per question on the computer test and no calculator, so write each value in a column on scratch paper
• An average must lie between the smallest and largest values; a probability must lie between 0 and 1
• Probability and counting also show up on the Mathematics Knowledge subtest

Up Next: Part 2 — finding the average (mean), the balance shortcut, and when the median is the better "typical" value.

🎲 Averages, Counting and Probability

Part 2 of 7 — Finding the Average (Mean)

Add, Then Divide by How Many

The average (also called the mean) of a list of numbers is

$\text{average} = \frac{\text{sum of the values}}{\text{number of values}}$

Example 1. A part-time worker logged 7, 9, 8, 6 and 10 hours on five shifts. What was the average shift length?

Step 1 — Add in a column. $7 + 9 + 8 + 6 + 10 = 40$ hours.

Step 2 — Count the values. There are 5 shifts.

Step 3 — Divide. $40 \div 5 = 8$ hours.

Reasonableness: the shifts run from 6 to 10 hours, and 8 is right in that range. ✓

Key Fact: An average always lies between the smallest and largest values. If your answer is outside that range, you divided by the wrong count or made an addition slip.

The Balance Shortcut

Adding big numbers by hand is where slips happen. The balance method keeps the numbers small:

1. Guess a round number near the middle of the data.
2. Write how far each value is above (+) or below (−) the guess.
3. Add those differences, divide by the count, and adjust the guess.

Example 2. Test scores of 88, 92, 95 and 85. Guess 90.

The differences add to $-2 + 2 + 5 - 5 = 0$. Nothing to adjust: the average is exactly 90.

Example 3. A convoy's five fuel stops took 48, 52, 55, 49 and 51 gallons. Guess 50.

Differences: $-2, +2, +5, -1, +1$, which add to $+5$.

Spread over 5 stops: $+5 \div 5 = +1$. The average is $50 + 1 = 51$ gallons.

Check the long way: $48 + 52 + 55 + 49 + 51 = 255$, and $255 \div 5 = 51$. ✓

Don't forget to divide the total difference. Adding all $+5$ to the guess (getting 55) is a common slip.

Running It Backward: Sum = Average × Count

If you know the average and the count, you know the total:

$\text{sum} = \text{average} \times \text{number of values}$

Example 4. Eight supply boxes average 15 pounds each. Their total weight is $15 \times 8 = 120$ pounds.

You will use this in almost every Part 3 problem, so it is worth memorizing now.

Median and Mode

Two other "typical value" words can appear:

• Median: the middle value after you put the list in order. With an even number of values, there are two middle values, so average them.
• Mode: the value that appears most often.

Example 5 — an odd count. Five apartments rent for $700, $650, $720, $680 and $1,750 a month.

Order first: $650, $680, $700, $720, $1,750. The median is $700.

The mean: $700 + 650 + 720 + 680 + 1{,}750 = 4{,}500$, and $4{,}500 \div 5 = 900$, so $900.

Four of the five rents are under $750, yet the mean is $900 because one very high rent pulls it up. When one value is far from the rest (an outlier), the median is the better "typical" value.

Example 6 — an even count. Find the median of 12, 15, 9 and 20.

Order first: 9, 12, 15, 20. The two middle values are 12 and 15:

$\frac{12 + 15}{2} = \frac{27}{2} = 13.5$

Key Fact: Always order the list before finding a median. Picking the middle of the list as it is printed is one of the most common wrong answers.

Find the typical value. 🎯

Work it out. ✍️

1) A recruiter made 23, 31, 27, 19 and 30 phone calls on five days. What was the average number of calls per day? (number only)

2) Nine packages have an average weight of 14 pounds. What is their total weight? (number only, in pounds)

Median, mode, or mean? 🔽

Key Takeaways — Part 2

• Average (mean) = sum ÷ count. Count the values carefully before you divide
• Balance shortcut: guess a round number, add the differences, divide by the count, adjust
• Sum = average × count works the formula backward
• Median: order the list first; with an even count, average the two middle values. Mode: most frequent value
• With one far-off value, the median is the better typical value
• An average always lies between the smallest and largest values

Up Next: Part 3 — the score you need on the last test, adding or removing a value, and weighted averages.

🎲 Averages, Counting and Probability

Part 3 of 7 — Missing Values and Weighted Averages

The Score You Need on the Last Test

This is one of the most common average questions on AR. The trick is to think in totals, not averages:

$\text{needed score} = (\text{target average} \times \text{total count}) - (\text{sum so far})$

Example 1. A recruit's first four practice tests average 82. What score does she need on the fifth test to raise her average to 85?

Step 1 — Target total. To average 85 over 5 tests she needs $85 \times 5 = 425$ points.

Step 2 — Points so far. Four tests averaging 82: $82 \times 4 = 328$ points.

Step 3 — Subtract. $425 - 328 = 97$.

Balance check: each of the 4 tests is 3 points short of 85, so she is $4 \times 3 = 12$ points short in all. The fifth test must make up the 12: $85 + 12 = 97$. ✓

Reasonableness: 97 is possible on a 100-point test, and it is above 85, as it must be to pull the average up.

Example 2. Scores so far are 78, 84 and 90. What score on the fourth test gives an average of 85?

Target total: $85 \times 4 = 340$. So far: $78 + 84 + 90 = 252$. Needed: $340 - 252 = 88$.

Key Fact: The needed score is not "the target plus a little." Every earlier test that fell short adds to what the last test must make up.

Adding or Removing One Value

Again, work with totals, and change the count.

Example 3 — removing a value. The average of 6 numbers is 20. One number, 35, is removed. What is the average of the other 5?

Total before: $20 \times 6 = 120$. Total after: $120 - 35 = 85$. New average: $85 \div 5 = 17$.

Reasonableness: 35 was above the average, so removing it should pull the average down. 17 is lower than 20. ✓

Example 4 — adding a value. A crew of 5 has an average age of 30. A 24-year-old joins. What is the new average age?

Total before: $30 \times 5 = 150$. Total after: $150 + 24 = 174$. New average: $174 \div 6 = 29$.

The trap: dividing by the old count (5 instead of 6 here, or 6 instead of 5 in Example 3).

Weighted Averages: Don't Average the Averages

When two groups of different sizes combine, the bigger group pulls harder.

Example 5. Twenty recruits average 80 on a test, and thirty recruits average 90. What is the average for all 50?

Step 1 — Total for each group. $20 \times 80 = 1{,}600$ and $30 \times 90 = 2{,}700$.

Step 2 — Combine. $1{,}600 + 2{,}700 = 4{,}300$ points for $20 + 30 = 50$ recruits.

Step 3 — Divide. $4{,}300 \div 50 = 86$.

The answer is 86, not 85. Averaging 80 and 90 gives 85, which would only be right if the groups were the same size. Here there are more recruits at 90, so the combined average lands closer to 90.

Shortcut: the averages differ by 10 points, and the 90-group is $\frac{30}{50} = \frac{3}{5}$ of everyone. Move $\frac{3}{5}$ of the way from 80 toward 90: $80 + 6 = 86$.

Key Fact: The combined average is always between the two group averages and closer to the bigger group. If the groups are different sizes, the simple midpoint is wrong.

Averages From a Frequency Table

Sometimes the data come as a table of values and how many times each occurred.

Example 6. Ten recruits did pull-ups:

Step 1 — Multiply each value by its count. $5 \times 2 = 10$, $6 \times 3 = 18$, $8 \times 4 = 32$, $10 \times 1 = 10$.

Step 2 — Add for the total. $10 + 18 + 32 + 10 = 70$ pull-ups.

Step 3 — Divide by the number of people, $2 + 3 + 4 + 1 = 10$: $70 \div 10 = 7$.

The traps: averaging just the four values in the left column ($29 \div 4 = 7.25$) ignores how many recruits did each; dividing 70 by 4 (the number of rows) gives 17.5, which is bigger than any value in the table.

Totals first. 🎯

Work it out. ✍️

1) A manager records how many shifts each employee worked this week:

What is the average number of shifts per employee? (number only)

2) A salesperson's sales averaged $300 a day for 5 days. How much must she sell on the sixth day to average $320 a day for all 6 days? (number only, in dollars)

Key Takeaways — Part 3

• Needed score = target average × total count − sum so far. Think in totals
• Adding or removing a value: change the total, then divide by the new count
• Weighted average: find each group's total, add, divide by everyone. Don't average the averages
• The combined average lands closer to the bigger group
• Frequency table: multiply each value by its count, add, divide by the number of people (not the number of rows)

Up Next: Part 4 — counting outcomes: combos, codes, arrangements and pairs.

🎲 Averages, Counting and Probability

Part 4 of 7 — Counting Outcomes

The Multiplication (Counting) Principle

If one choice can be made in $a$ ways and a second, separate choice in $b$ ways, the two choices together can be made in $a \times b$ ways. Add more choices by multiplying again.

Example 1 — outfits. A recruit has 5 shirts, 3 pairs of pants and 2 pairs of shoes. How many different outfits (one of each) are possible?

$5 \times 3 \times 2 = 30 \text{ outfits}$

Why multiply, not add? Each of the 5 shirts can go with each of the 3 pants: that is $5 \times 3 = 15$ shirt-and-pants pairs. Each of those 15 can go with either pair of shoes: $15 \times 2 = 30$. Adding ($5 + 3 + 2 = 10$) counts the items, not the outfits.

Example 2 — lock codes. A lock uses a 3-digit code, and each digit can be 0 through 9.

• Repeats allowed: each position has 10 choices: $10 \times 10 \times 10 = 1{,}000$ codes.
• No digit repeated: 10 choices for the first digit, then 9 left, then 8 left: $10 \times 9 \times 8 = 720$ codes.

Key Fact: Draw one blank for each position, write the number of choices in each blank, and multiply. If items cannot repeat, each blank gets one fewer choice than the one before.

Arranging a Group in a Line

How many ways can you line up a small group? Use the blanks again: every spot has one fewer person left to fill it.

Example 3. Four soldiers stand in a line for a photo. How many different orders are possible?

$4 \times 3 \times 2 \times 1 = 24$

This product has a name: 4 factorial, written $4!$. You only need the small ones:

Reasonableness: the number of orders grows fast. If a question asks how many ways 4 people can line up and you got 10 (from $4 + 3 + 2 + 1$), you added instead of multiplying.

Choosing 2 When Order Does Not Matter

Ask: if I swap the two people, is it a different outcome?

Example 4 — order matters. From 6 club members, choose a president and a vice president. Ann as president with Ben as vice president is different from Ben as president with Ann as vice president.

$6 \times 5 = 30 \text{ ways}$

Example 5 — order does not matter. Six people at a meeting each shake hands once with every other person. Ann shaking Ben's hand is the same handshake as Ben shaking Ann's.

$6 \times 5 = 30$ counts every handshake twice, so divide by 2:

$\frac{6 \times 5}{2} = 15 \text{ handshakes}$

In general, the number of unordered pairs from a group of $n$ is

$\frac{n \times (n - 1)}{2}$

List to check. For a small group, write the pairs in order so none are missed or repeated. From A, B, C and D:

AB, AC, AD, BC, BD, CD

That is 6 pairs, and the formula agrees: $\frac{4 \times 3}{2} = 6$. Start with every pair that includes A, then every new pair with B, and so on. Never list BA after AB.

Count the outcomes. 🎯

Work it out. ✍️

1) A parking tag has 2 letters followed by 1 digit. Each letter can be A, B, C, D or E (letters may repeat), and the digit can be 0 through 9. How many different tags are possible? (number only)

2) A coach picks 2 co-captains from 9 players. Both co-captains have the same role. How many different pairs of co-captains are possible? (number only)

Does order matter? 🔽 Each question uses a group of 5.

Key Takeaways — Part 4

• Counting principle: one blank per choice, write the number of options in each, multiply
• No repeats: each blank has one fewer choice ($10 \times 9 \times 8$)
• Lining up a group: $3! = 6$, $4! = 24$, $5! = 120$
• Pairs where order does not matter: $\frac{n(n - 1)}{2}$. If swapping makes a new outcome (different jobs), do not divide by 2
• List systematically (AB, AC, AD, BC, ...) to check small cases

Up Next: Part 5 — basic probability: favorable over total, "not," "and," and drawing without replacement.

🎲 Averages, Counting and Probability

Part 5 of 7 — Basic Probability

Favorable Over Total

When every outcome is equally likely,

$\text{probability} = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$

Write the fraction, then simplify it: answer choices are almost always in lowest terms.

Example 1. A standard die (faces 1 to 6) is rolled. What is the probability of rolling a number greater than 2?

Favorable: 3, 4, 5, 6, which is 4 outcomes. Total: 6 outcomes.

$\frac{4}{6} = \frac{2}{3}$

Example 2. A box holds 15 black pens, 9 blue pens and 6 red pens. One pen is picked without looking. What is the probability it is blue?

Total: $15 + 9 + 6 = 30$ pens. Favorable: 9 blue.

$\frac{9}{30} = \frac{3}{10}$

The denominator is the total of everything, not just the outcomes that are not favorable. Writing $\frac{9}{21}$ (blue over not-blue) is a common slip.

Key Fact: Every probability is between 0 and 1. A probability of 0 means the event cannot happen, and 1 means it is certain. An answer bigger than 1 is always wrong.

"Not": The Complement

The probability that something does not happen is 1 minus the probability that it does:

$P(\text{not A}) = 1 - P(\text{A})$

Example 3. Using the same box of 30 pens, what is the probability that the pen is not red?

$P(\text{red}) = \frac{6}{30} = \frac{1}{5} \qquad P(\text{not red}) = 1 - \frac{1}{5} = \frac{4}{5}$

Check by counting: $15 + 9 = 24$ pens are not red, and $\frac{24}{30} = \frac{4}{5}$. ✓

"And" for Separate Events: Multiply

Two events are independent when one has no effect on the other, like a coin toss and a die roll. The probability that both happen is the product:

$P(\text{A and B}) = P(\text{A}) \times P(\text{B})$

Example 4. A coin is tossed and a die is rolled. What is the probability of heads and a 6?

$\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Check by listing: the coin has 2 outcomes and the die has 6, so there are $2 \times 6 = 12$ equally likely pairs (H1, H2, ..., T6). Only H6 is favorable: $\frac{1}{12}$. ✓

Reasonableness: "both" is harder than either one alone, so the answer must be smaller than each fraction you multiplied. Adding ($\frac{1}{2} + \frac{1}{6} = \frac{2}{3}$) gives a bigger number, which is the sign of the wrong operation.

Two Draws Without Replacement

When you draw twice and do not put the first item back, the second draw has one fewer item in total, and possibly one fewer favorable item.

Example 5. A drawer has 4 black socks and 2 white socks. Two socks are pulled out without looking. What is the probability that both are black?

First draw: 4 black out of 6 socks: $\frac{4}{6}$.

Second draw: one black sock is gone, so 3 black out of 5 socks: $\frac{3}{5}$.

Multiply:

$\frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5}$

The trap: using $\frac{4}{6}$ twice gives $\frac{16}{36} = \frac{4}{9}$. That would be right only if the first sock went back into the drawer.

Expected Count

To predict how many times something will happen in a number of tries:

$\text{expected count} = \text{probability} \times \text{number of trials}$

Example 6. A die is rolled 60 times. About how many times should a 3 come up?

$\frac{1}{6} \times 60 = 10 \text{ times}$

Reasonableness: the expected count can never be more than the number of trials.

Find the probability. 🎯

Work it out. ✍️

1) A spinner has 8 equal sections, and 3 of them are blue. If it is spun 40 times, how many times should you expect it to land on blue? (number only)

2) A bag holds 12 tiles. The probability of drawing a red tile is $\frac{1}{4}$. How many red tiles are in the bag? (number only)

Probability rules. 🔽

Key Takeaways — Part 5

• Probability = favorable ÷ total, simplified. The total includes every outcome
• Every probability is between 0 and 1
• Not: $1 - P$
• And (independent events): multiply. The answer must be smaller than each piece
• Without replacement: the second denominator shrinks by 1, and the numerator may too
• Expected count = probability × number of trials

Up Next: Part 6 — test strategy and the common traps in average, counting and probability questions.

🎲 Averages, Counting and Probability

Part 6 of 7 — Test Strategy and Common Traps

The wrong answer choices on AR are not random. Each one is the result of a specific, common mistake. If you know the mistakes, you can see the traps coming.

The Seven Big Traps

Key Fact: Each of these traps produces a number that appears among the answer choices. Finding your answer in the list does not prove it is right. Your check does.

Timing, Estimation and Back-Solving

Timing

On the computer test you have 55 minutes for 15 questions, about 3.5 minutes each. A one-step average or probability question should take about a minute. That banked time pays for the multi-step ones: a needed score, a weighted average, or two draws without replacement. Remember that you cannot go back, so spend the extra seconds checking before you answer.

Estimate to eliminate choices

• An average must be between the smallest and largest values. Any choice outside that range is out.
• A combined average of two groups must be between the two group averages and closer to the bigger group.
• A probability must be between 0 and 1, and the chance of "both" must be smaller than the chance of either one.

Back-solve "needed score" questions

Example. Scores so far are 85, 79 and 88. What score on the fourth test gives an average of 86? Choices: 90, 92, 94, 96.

Try 92: $85 + 79 + 88 + 92 = 344$, and $344 \div 4 = 86$. ✓ The answer is 92.

Faster with the balance method: compared with 86, the scores so far are $-1, -7, +2$, a total of $-6$. The fourth test must be 6 above 86, which is 92.

Reasonableness checklist

Before you click, ask:

• Is my average between the smallest and largest values?
• Did I divide by the number of values (or people), not the number of rows or groups?
• Did I order the list before finding a median?
• Is my probability between 0 and 1, and is "both" smaller than each piece?
• Without replacement, did the second denominator shrink?
• Does swapping two people make a new outcome? If not, did I divide by 2?
• Did I answer the question in the last sentence (a probability, not a count)?

Spot the trap. 🎯

Key Takeaways — Part 6

• The wrong choices come from seven traps: wrong count, averaging averages, median without ordering, adding instead of multiplying, not shrinking the denominator, double counting pairs, a count instead of a probability
• Bank time on one-step items (about a minute) for multi-step items; you cannot go back, so check before you click
• Estimate with ranges: averages between low and high, probabilities between 0 and 1
• Back-solve needed-score questions by plugging in a choice
• Run the reasonableness checklist on every answer

Up Next: Part 7 — a mixed mastery set in true ASVAB style.

🎲 Averages, Counting and Probability

Part 7 of 7 — Mixed Mastery Set

This set mixes everything from Parts 2 through 5 the way the real AR subtest does: a few quick ones, several medium ones, and a few multi-step problems. Treat it like test day:

• No calculator. Use scratch paper for every sum and product.
• Read the last sentence first: average, total, missing value, count, or probability?
• Estimate before you compute (averages between low and high, probabilities between 0 and 1), and check before you answer, because on the real test you cannot go back.
• Aim for about 3.5 minutes per question or less.

Set A: warm-up to medium. 🎯

Set B: medium to hard. 🎯

One to work out. ✍️

In a league of 10 teams, each team plays every other team exactly once. How many games are played in all? (number only)

Key Takeaways — Averages, Counting and Probability

• Average = sum ÷ count; use the balance shortcut for big numbers; an average lies between the smallest and largest values
• Median: order first; even count, average the middle two. Better than the mean when one value is far off
• Needed score = target × count − sum so far; adding or removing a value changes the count too
• Weighted average: totals first; don't average the averages
• Counting: one blank per choice, multiply; no repeats means one fewer each time; unordered pairs $\frac{n(n - 1)}{2}$
• Probability = favorable ÷ total; not = $1 - P$; and = multiply; without replacement, shrink the second denominator; expected count = $P \times$ trials

What to Do Next

1. Take the exit quiz for this topic to lock in your score.
2. Review the flashcards for the average formulas and probability rules until they are automatic.
3. This is the last topic in the AR unit. Review any AR topic where your exit-quiz score was lowest, then move on to the Mathematics Knowledge unit, starting with Number Properties, Exponents and Roots. Probability and counting come back on that subtest, so this practice counts twice.