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Averages, Counting and Probability

Averages, counting outcomes and basic probability in word problems.

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🎲 Averages, Counting and Probability

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

Topic Overview

PartTopic
1Averages and chance on the AR subtest
2Finding the average (mean)
3Missing values and weighted averages
4Counting outcomes
5Basic probability
6Test strategy and common traps
7Mixed mastery set

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

FactWhat it means for this topic
15 scored questions in 55 minutes on the computer test (CAT-ASVAB)About 3.5 minutes per question: enough time to write out every sum
No calculator, at any version of the testAdding a list of scores and dividing is done on scratch paper
Two sheets of scratch paper and two pencils (ask for more if you need them)Write each value in a column so nothing gets skipped
4 answer choices per questionThe wrong choices come from real slips, such as dividing by the wrong count
You cannot go back once you answerCheck your average against the smallest and largest values before you click

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.

🎲 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

average=sum of the valuesnumber of values\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=407 + 9 + 8 + 6 + 10 = 40 hours.

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

Step 3 — Divide. 40÷5=840 \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.

🎲 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:

needed score=(target average×total count)−(sum so far)\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 85×5=42585 \times 5 = 425 points.

Step 2 — Points so far. Four tests averaging 82: 82×4=32882 \times 4 = 328 points.

Step 3 — Subtract. 425−328=97425 - 328 = 97.

Balance check: each of the 4 tests is 3 points short of 85, so she is 4×3=124 \times 3 = 12 points short in all. The fifth test must make up the 12: 85+12=9785 + 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: 85×4=34085 \times 4 = 340. So far: 78+84+90=25278 + 84 + 90 = 252. Needed: 340−252=88340 - 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.

🎲 Averages, Counting and Probability

Part 4 of 7 — Counting Outcomes

The Multiplication (Counting) Principle

If one choice can be made in aa ways and a second, separate choice in bb ways, the two choices together can be made in a×ba \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?

5×3×2=30 outfits5 \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 5×3=155 \times 3 = 15 shirt-and-pants pairs. Each of those 15 can go with either pair of shoes: 15×2=3015 \times 2 = 30. Adding (5+3+2=105 + 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: 10×10×10=1,00010 \times 10 \times 10 = 1{,}000 codes.
  • No digit repeated: 10 choices for the first digit, then 9 left, then 8 left: 10×9×8=72010 \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.

🎲 Averages, Counting and Probability

Part 5 of 7 — Basic Probability

Favorable Over Total

When every outcome is equally likely,

probability=number of favorable outcomestotal number of outcomes\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.

46=23\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=3015 + 9 + 6 = 30 pens. Favorable: 9 blue.

930=310\frac{9}{30} = \frac{3}{10}

The denominator is the total of everything, not just the outcomes that are not favorable. Writing 921\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.

🎲 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

TrapWhat it looks likeHow to dodge it
Dividing by the wrong countFive scores added, then divided by 4; a frequency table divided by the number of rowsCount the values (or people) before you divide, and check that the average is between the low and high values
Averaging the averages20 people at 80 and 30 people at 90 answered as 85Find each group's total, add, divide by everyone; the answer leans toward the bigger group
Median without orderingTaking the middle number of the list as printedRewrite the list from smallest to largest first
Adding instead of multiplying"Heads and a 6" answered as 12+16\frac{1}{2} + \frac{1}{6}"And" for separate events means multiply; "both" must be smaller than each piece
Not shrinking the denominatorTwo draws without replacement, both out of the full totalAfter the first draw, the total drops by 1 (and the favorable count may too)
Double counting pairs6 people shaking hands counted as 6×5=306 \times 5 = 30If swapping the two gives the same outcome, divide by 2
A count instead of a probabilityAnswering 8 (the number of outcomes) when asked for a probabilityRead the last sentence: a probability is a fraction between 0 and 1

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.

🎲 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.
Explain using:

📌 Related Topics in Arithmetic Reasoning (AR)

❓ Frequently Asked Questions

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Averages, counting outcomes and basic probability in word problems.
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Averages, Counting and Probability is part of the ASVAB course on Study Mondo, specifically in the Arithmetic Reasoning (AR) section. You can explore the full course for more related topics and practice resources.