2026 Applied Information Technology Engineer Examination, Subject A — Part 3: Probability, Expected Value, and Statistics

Last revised: 2026-10-20 This study material targets the 2026 examination and is scoped to IPA’s official “応用情報技術者試験（レベル3）シラバス Ver.7.2” (Applied Information Technology Engineer Examination, Level 3, Syllabus Ver.7.2). Motitory Labs is not affiliated with, approved by, or sponsored by IPA (Information-technology Promotion Agency, Japan), and this article is not official IPA material. Studying with this article does not guarantee that you will pass. Exam guidelines and the syllabus may be revised, so always check the latest information on IPA’s official website before your exam. All examples and practice questions in this article are Motitory originals, not official IPA past exam questions.

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Learning index

1. Learning objectives
2. Terminology map (Korean↔Japanese↔English)
3. Counting: permutations and combinations
4. Basic probability rules, conditional probability, and Bayes’ theorem
5. Random variables, expected value, and variance
6. Bernoulli and binomial distributions; discrete vs continuous
7. Descriptive statistics: central tendency, spread, and charts
8. Correlation and key inference terms
9. Worked examples
10. Four original practice questions
11. Formula sheet
12. Common traps
13. Final self-check
14. 1/3/7-day spaced review suggestion
15. IPA syllabus mapping
16. Sources

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1. Learning objectives

After this installment you should be able to:

• Decide whether order matters and choose correctly between permutations nPr and combinations nCr.
• Compute probabilities using complements, the addition rule (including overlap), and the multiplication rule.
• Explain the difference between independence and mutual exclusivity.
• Use conditional probability and Bayes’ theorem for problems such as “the probability an item is actually defective given a positive inspection.”
• Compute the expected value, variance, and standard deviation of a discrete random variable, and understand that the expected value need not be a possible outcome.
• Apply the core binomial formulas and explain that, for a continuous distribution, a probability density is not the probability of a single point.
• Interpret the mean, median, and mode; variance and standard deviation; histograms and box plots.
• Interpret the range and sign of a correlation coefficient and explain why correlation is not causation.
• Correctly interpret inference terms such as p-value, significance level, and Type I/Type II errors.

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2. Terminology map (Korean↔Japanese↔English)

This table helps you recognize the expressions used in the Japanese exam paper.

Korean	Japanese	English
경우의 수	場合の数	number of cases (counting)
계승(팩토리얼)	階乗	factorial
순열	順列	permutation
조합	組合せ	combination
확률	確率	probability
여사건	余事象	complementary event
덧셈정리	加法定理	addition rule
곱셈정리	乗法定理	multiplication rule
동시확률	同時確率	joint probability
배반(상호배반)	排反	mutually exclusive
독립	独立	independent
조건부확률	条件付き確率	conditional probability
베이즈 정리	ベイズの定理	Bayes’ theorem
확률변수	確率変数	random variable
확률분포	確率分布	probability distribution
베르누이 분포	ベルヌーイ分布	Bernoulli distribution
이항분포	二項分布	binomial distribution
확률밀도	確率密度	probability density
정규분포	正規分布	normal distribution
기댓값	期待値	expected value
분산	分散	variance
표준편차	標準偏差	standard deviation
평균값	平均値	mean
중앙값	中央値（メジアン）	median
최빈값	最頻値（モード）	mode
도수분포표	度数分布表	frequency distribution table
히스토그램	ヒストグラム	histogram
상자수염그림	箱ひげ図	box plot (box-and-whisker plot)
사분위수	四分位数	quartile
이상값	外れ値	outlier
상관계수	相関係数	correlation coefficient
의사상관(허위상관)	擬似相関	spurious correlation
모집단 / 표본	母集団 / 標本	population / sample
점추정 / 구간추정	点推定 / 区間推定	point estimation / interval estimation
귀무가설	帰無仮説	null hypothesis
대립가설	対立仮説	alternative hypothesis
유의수준	有意水準	significance level
p값	p値（有意確率）	p-value
제1종 오류	第1種の誤り	Type I error
제2종 오류	第2種の誤り	Type II error
검정력	検出力（検定力）	power

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3. Counting: permutations and combinations

3.1 Basic principles

• Product rule: if step 1 can be done in m ways and, for each of those, step 2 can be done in n ways, there are m × n ways in total.
• Sum rule: if case A (m ways) and case B (n ways) cannot happen at the same time (no overlap), “A or B” can happen in m + n ways.

3.2 Factorial

n! = n × (n−1) × … × 2 × 1, with 0! = 1 by definition. Example: 4! = 24. The number of ways to arrange n items in a row is n!.

3.3 Permutations and combinations

Symbols: n = total number of items, r = number chosen (0 ≤ r ≤ n)

Type	Order	Formula	Example (n=5, r=2)
Permutation nPr	matters	nPr = n! / (n−r)!	5P2 = 5 × 4 = 20
Combination nCr	does not matter	nCr = n! / (r!(n−r)!) = nPr / r!	5C2 = 20 / 2 = 10
Permutation with repetition	matters, repeats allowed	n^r	5^2 = 25

How to decide: if swapping the order of the chosen items gives a “different case,” use permutations; if it gives “the same case,” use combinations.

• When roles are distinct (e.g., chair and vice-chair), order matters → permutation
• When roles are identical (e.g., two committee members), order does not matter → combination

Remembering nCr = nC(n−r) also speeds up calculation (e.g., 8C6 = 8C2 = 28).

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4. Basic probability rules, conditional probability, and Bayes’ theorem

Symbols: P(A) = probability of event A, Aᶜ = complement of A, A∩B = both A and B occur, A∪B = A or B occurs.

4.1 Basics

• When all outcomes are equally likely: P(A) = (number of outcomes in A) / (total number of outcomes)
• 0 ≤ P(A) ≤ 1
• Complement: P(Aᶜ) = 1 − P(A). “At least one” problems are often fastest via the complement.

4.2 Addition rule (watch the overlap)

P(A∪B) = P(A) + P(B) − P(A∩B)

Only when A and B are mutually exclusive (cannot occur together, P(A∩B)=0) does P(A∪B) = P(A) + P(B).

4.3 Multiplication rule and independence

• Multiplication rule: P(A∩B) = P(A) × P(B|A)
• Independence: whether one event occurs does not affect the probability of the other → P(A∩B) = P(A) × P(B)

Independent ≠ mutually exclusive: if two events with P(A)>0 and P(B)>0 are mutually exclusive, then P(A∩B)=0 ≠ P(A)P(B), so they are not independent. “If A happens, B definitely does not” is a strong dependence.

4.4 Conditional probability

P(B|A) = P(A∩B) / P(A) (provided P(A) > 0)

This is “the probability of B given that A occurred.” The key point is that the denominator is not the whole space but the world narrowed down to A. In general, P(B|A) and P(A|B) are different.

4.5 Bayes’ theorem

P(A|B) = P(B|A) × P(A) / P(B) where, by total probability: P(B) = P(B|A)P(A) + P(B|Aᶜ)P(Aᶜ)

Use it to reason backward from an observed result (B, e.g., a positive inspection) to the probability of a cause (A, e.g., actually defective). If the prior probability P(A) is small, P(A|B) can be surprisingly small even for an accurate test (see Example 2).

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5. Random variables, expected value, and variance

5.1 Discrete random variables

When a random variable X takes values x₁, x₂, …, xₙ with probabilities P(X=xᵢ), this correspondence is its probability distribution. For a discrete variable:

• each P(X=xᵢ) ≥ 0
• Σ P(X=xᵢ) = 1 (all probabilities sum to 1)

5.2 Expected value, variance, and standard deviation

Measure	Formula	Meaning
Expected value	E[X] = Σ x · P(X=x)	probability-weighted average
Variance	V[X] = E[(X−E[X])²] = E[X²] − (E[X])²	spread around the expected value
Standard deviation	σ = √V[X]	spread in the original units

Here E[X²] = Σ x² · P(X=x). Note that (E[X])² and E[X²] are different.

Linear transformation properties (a, b are constants):

• E[aX + b] = aE[X] + b
• V[aX + b] = a²V[X] (adding a constant b does not change the variance)

The expected value need not be a possible outcome. For example, a fair die has expected value 3.5, yet no face shows 3.5. The expected value is “the average value over many repetitions.”

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6. Bernoulli and binomial distributions; discrete vs continuous

6.1 Bernoulli distribution

A single trial with success (1) with probability p and failure (0) with probability 1−p. E[X] = p, V[X] = p(1−p)

6.2 Binomial distribution B(n, p)

The number of successes X in n independent Bernoulli trials.

P(X = k) = nCk · p^k · (1−p)^(n−k) (k = 0, 1, …, n) E[X] = np, V[X] = np(1−p)

Example: the probability of exactly 2 heads in 3 tosses of a fair coin = 3C2 × (1/2)² × (1/2)¹ = 3/8. Here E[X] = 1.5 and V[X] = 0.75.

6.3 Discrete vs continuous (overview)

	Discrete	Continuous
Values	countable values (0, 1, 2, …)	every real number in an interval
Description	probability mass P(X=x)	probability density function f(x)
Computing probability	add the probabilities of the values	area over an interval (integral) P(a≤X≤b)=∫ₐᵇ f(x)dx
Probability of a single point	can be greater than 0	P(X=a) = 0

A probability density f(x) is not the probability of a single point. A density value can even exceed 1; it is the total area that equals 1. For the normal distribution, a typical continuous distribution, it is handy to remember the approximation that about 68% of values fall within mean ±1σ and about 95% within ±2σ.

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7. Descriptive statistics: central tendency, spread, and charts

7.1 Measures of central tendency

Symbols: data x₁, …, xₙ, number of data points n

• Mean: x̄ = (x₁ + … + xₙ) / n — strongly affected by extreme values (outliers)
• Median: the middle value after sorting (if n is even, the average of the two middle values) — robust to outliers
• Mode: the most frequent value

In a distribution with a long right tail (outliers on the large side), usually mean > median.

7.2 Spread

• Range = maximum − minimum
• Variance = mean of squared deviations: s² = Σ(xᵢ − x̄)² / n
• Standard deviation = √variance
• When estimating a population variance from a sample, the unbiased variance divides by n−1 instead of n. Check which one the question uses.

7.3 Charts

• Frequency table and histogram: divide the data into classes (bins) and draw frequencies as bars. Shows the shape of the distribution (peaks, skew). Bars are typically drawn without gaps.
• Box plot: shows the minimum, first quartile Q1, median (Q2), third quartile Q3, and maximum.
  • Interquartile range IQR = Q3 − Q1 (the width of the middle 50% of the data)
  • A common convention marks values below Q1 − 1.5×IQR or above Q3 + 1.5×IQR as outliers. Quartile calculation methods and outlier criteria can vary by source, so follow the definition given in the question.

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8. Correlation and key inference terms

8.1 Correlation coefficient

The correlation coefficient r shows the direction and strength of the linear relationship between two variables.

• −1 ≤ r ≤ 1
• r > 0: positive correlation (when one is large, the other tends to be large); r < 0: negative correlation
• The closer |r| is to 1, the stronger the linear relationship
• r ≈ 0 does not mean “no relationship at all” (the relationship may be curved)

Correlation ≠ causation: a high correlation does not establish that one variable causes the other. A correlation produced because a third variable affects both is called a spurious correlation.

8.2 Inference glossary

Term	Brief explanation
Population / sample	the entire group of interest / the part actually observed
Point estimation	estimating a parameter with a single value (e.g., sample mean for population mean)
Interval estimation	giving an interval expected to contain the parameter, with a confidence level
Null hypothesis H₀	the reference hypothesis, such as “no difference / no effect”
Alternative hypothesis H₁	the hypothesis you want to support by rejecting the null
Significance level α	the pre-set limit on the risk of rejecting a true null hypothesis (e.g., 5%)
p-value	assuming the null hypothesis is true, the probability of a result at least as extreme as the one observed
Decision	if p-value ≤ α, reject the null; if p-value > α, do not reject it
Type I error	rejecting a null hypothesis that is true. The significance level α is an upper bound on this error probability; the actual probability equals α only when the test is calibrated to have exact size α.
Type II error	failing to reject a null hypothesis that is false (probability β)
Power	the probability of correctly rejecting a false null = 1 − β

Caution: the p-value is not “the probability that the null hypothesis is true.” Also, “not rejecting” does not mean “the null hypothesis has been proven true.”

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9. Worked examples

All examples below are Motitory originals, not IPA past exam questions.

Example 1 (counting and probability): choosing 2 of 5 people

There are 5 people: A, B, C, D, and E. (1) How many ways are there to choose one chair and one vice-chair? (2) How many ways are there to choose two committee members? (3) If two committee members are chosen at random, what is the probability that A is included?

Solution (1) Roles are distinct, so use a permutation: 5P2 = 5 × 4 = 20 ways (2) Roles are identical, so use a combination: 5C2 = 20 / 2! = 10 ways (3) Cases including A: fix A, then choose 1 of the other 4 → 4C1 = 4 ways. Probability = 4/10 = 2/5 Check via the complement: cases without A = 4C2 = 6 → 1 − 6/10 = 2/5 ✔

Example 2 (conditional probability and Bayes’ theorem): defect inspection

At a factory, 2% of products are defective. An inspection machine flags a defective item as “defective” with probability 90%, and also wrongly flags a good item as “defective” with probability 5%. What is the probability that an item flagged “defective” is actually defective?

Solution Let D = actually defective and + = flagged “defective.”

• P(D) = 0.02, P(Dᶜ) = 0.98
• P(+|D) = 0.90, P(+|Dᶜ) = 0.05

Total probability: P(+) = 0.90 × 0.02 + 0.05 × 0.98 = 0.018 + 0.049 = 0.067 Bayes’ theorem: P(D|+) = 0.018 / 0.067 = 18/67 ≈ 0.269 (about 26.9%)

Even though the machine seems fairly accurate, when the defect rate (prior probability) is low, most “defective” flags are false alarms on good items. With 10,000 items: 180 of the 200 defective items and 490 of the 9,800 good items are flagged → 180/670 ≈ 26.9%.

Example 3 (expected value and variance): a discrete random variable

The distribution of a random variable X is:

x	0	1	2
P(X=x)	0.2	0.5	0.3

Solution

• Check the probabilities sum to 1: 0.2 + 0.5 + 0.3 = 1 ✔
• E[X] = 0×0.2 + 1×0.5 + 2×0.3 = 0 + 0.5 + 0.6 = 1.1
• E[X²] = 0²×0.2 + 1²×0.5 + 2²×0.3 = 0 + 0.5 + 1.2 = 1.7
• V[X] = 1.7 − 1.1² = 1.7 − 1.21 = 0.49
• σ = √0.49 = 0.7

X only takes the values 0, 1, or 2, yet its expected value is 1.1. The expected value need not be a possible outcome.

Example 4 (descriptive statistics): central tendency and an outlier

Data: 2, 4, 4, 5, 10

Solution

• Mean = (2+4+4+5+10)/5 = 25/5 = 5
• Median = the 3rd of the 5 sorted values = 4
• Mode = 4
• Deviations: −3, −1, −1, 0, 5 → sum of squared deviations = 9+1+1+0+25 = 36
• Variance dividing by n = 36/5 = 7.2 (standard deviation ≈ 2.68)
• Unbiased variance dividing by n−1 = 36/4 = 9 (standard deviation = 3)

The large value 10 pulls the mean up, so mean (5) > median (4). When outliers are present, the median is a more stable measure of central tendency.

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10. Four original practice questions

The following four questions are Motitory originals, not official IPA past exam questions. Try them yourself first, then check the explanations.

Question 1 (counting and order)

From 8 members, one leader, one sub-leader, and one recorder are chosen. No person may take more than one role. How many ways are there to make the selection?

• A. 24
• B. 56
• C. 336
• D. 512

Answer and explanation

Answer: C

The roles are distinct, so order matters, and the same person cannot be repeated → permutation 8P3 = 8 × 7 × 6 = 336.

• A. 24 = 3! counts only the ways to assign roles to 3 people already chosen. It omits the step of choosing 3 of the 8.
• B. 56 = 8C3 is a combination that ignores the distinct roles (order). It should be 56 × 3! = 336.
• C. Correct. 8P3 = 336.
• D. 512 = 8³ is a permutation with repetition, allowing one person to hold several roles. That violates the condition.

Question 2 (overlap and conditional probability)

Of 200 employees, 120 use service X, 80 use service Y, and 50 use both services. An employee chosen at random is a user of service X. What is the probability that this employee also uses service Y?

• A. 1/4
• B. 5/12
• C. 5/8
• D. 3/4

Answer and explanation

Answer: B

The condition is “uses X,” so the denominator is the 120 X users. P(Y|X) = P(X∩Y)/P(X) = 50/120 = 5/12.

• A. 1/4 = 50/200 is the joint probability P(X∩Y) of using both. It does not narrow the denominator by the condition.
• B. Correct. 50/120 = 5/12.
• C. 5/8 = 50/80 is the conditional probability in the opposite direction, P(X|Y).
• D. 3/4 = 150/200 is the probability of using at least one, P(X∪Y) = (120+80−50)/200. It subtracts the overlap correctly, but answers a different question.

Question 3 (expected value and variance)

In a game, the points X you earn are 0 with probability 0.6 and 5 with probability 0.4. Which pair gives the correct expected value E[X] and variance V[X]?

• A. E[X] = 2, V[X] = 4
• B. E[X] = 2, V[X] = 10
• C. E[X] = 2.5, V[X] = 6.25
• D. E[X] = 2, V[X] = 6

Answer and explanation

Answer: D

E[X] = 0×0.6 + 5×0.4 = 2 E[X²] = 0²×0.6 + 5²×0.4 = 10 V[X] = E[X²] − (E[X])² = 10 − 4 = 6

• A. V[X] = 4 mistakes (E[X])² = 2² for the variance.
• B. V[X] = 10 is E[X²] without subtracting (E[X])².
• C. This ignores the probabilities and treats 0 and 5 as equally likely (0.5 each). That gives mean 2.5 and variance 6.25, which do not match the given probabilities.
• D. Correct. E[X] = 2, V[X] = 6.

Question 4 (interpreting inference)

A hypothesis test at the 5% significance level gives a p-value of 0.03. Which interpretation of this result is most appropriate?

• A. Assuming the null hypothesis is true, the probability of a result at least as extreme as the one observed is 0.03, so the null hypothesis is rejected at the 5% significance level.
• B. The probability that the null hypothesis is true is 3%.
• C. The alternative hypothesis has been proven true with 97% probability.
• D. Because the p-value is smaller than the significance level, the null hypothesis is not rejected.

Answer and explanation

Answer: A

Since p-value 0.03 ≤ significance level 0.05, the null hypothesis is rejected. The p-value is “the probability of a result at least as extreme as the one observed,” computed on the premise that the null hypothesis is true.

• A. Correct. It states both the definition of the p-value and the decision rule correctly.
• B. The p-value is not the probability that the null hypothesis is true. It is a probability computed while assuming the null hypothesis is true.
• C. 1 − p-value is not the probability that the alternative hypothesis is true, and a test does not “prove” a hypothesis.
• D. The decision is reversed. If p-value ≤ significance level, the null hypothesis is rejected.

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11. Formula sheet

Item	Formula
Factorial	n! = n(n−1)…1, 0! = 1
Permutation	nPr = n!/(n−r)!
Combination	nCr = n!/(r!(n−r)!) = nPr/r!, nCr = nC(n−r)
Permutation with repetition	n^r
Complement	P(Aᶜ) = 1 − P(A)
Addition rule	P(A∪B) = P(A) + P(B) − P(A∩B)
Multiplication rule	P(A∩B) = P(A)P(B|A)
Independence	P(A∩B) = P(A)P(B)
Conditional probability	P(B|A) = P(A∩B)/P(A)
Bayes’ theorem	P(A|B) = P(B|A)P(A) / [P(B|A)P(A) + P(B|Aᶜ)P(Aᶜ)]
Expected value	E[X] = Σ x·P(X=x)
Variance	V[X] = E[X²] − (E[X])²
Standard deviation	σ = √V[X]
Linear transformation	E[aX+b] = aE[X]+b, V[aX+b] = a²V[X]
Bernoulli	E = p, V = p(1−p)
Binomial	P(X=k) = nCk p^k (1−p)^(n−k), E = np, V = np(1−p)
Mean	x̄ = Σxᵢ / n
Interquartile range	IQR = Q3 − Q1
Correlation coefficient	−1 ≤ r ≤ 1
Test decision	p-value ≤ α → reject the null hypothesis

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12. Common traps

1. Using combinations when roles are distinct, or permutations when roles are identical.
2. Forgetting to subtract the overlap P(A∩B) in the addition rule.
3. Treating mutual exclusivity and independence as the same concept.
4. Confusing P(B|A) with P(A|B) (choosing the wrong denominator for a conditional probability).
5. Ignoring the prior probability (base rate) in Bayes problems.
6. Stopping at E[X²] for the variance, or confusing it with (E[X])².
7. Thinking the expected value must be a possible outcome.
8. Reading the density f(a) as P(X=a) for a continuous distribution.
9. Judging what is “typical” from the mean alone when the data contain outliers.
10. Concluding causation because the correlation coefficient is large.
11. Interpreting the p-value as “the probability the null hypothesis is true,” or interpreting “not rejected” as “the null hypothesis is proven.”
12. Swapping the definitions of Type I and Type II errors.

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13. Final self-check

• I can choose between permutations and combinations by asking “does order matter?”
• I can state the P(A∪B) formula and the condition under which it simplifies to P(A)+P(B).
• I can explain why two mutually exclusive events (both with positive probability) are not independent.
• I can redo Example 2 using a table or by assuming 10,000 items.
• I can compute E[X], E[X²], V[X], and σ in order.
• I can write the binomial formulas for P(X=k), E, and V.
• I can explain why a probability density is not the probability of a single point.
• I can read Q1, the median, Q3, and the IQR from a box plot.
• I can state the definition of the p-value and Type I/Type II errors precisely.

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14. 1/3/7-day spaced review suggestion

This suggestion collects no learner data at all. Record your progress only in your own notes or calendar.

• After 1 day: cover the formula sheet and write out the permutation, combination, addition rule, Bayes’ theorem, and variance formulas on paper. Redo Questions 1 and 2.
• After 3 days: redo Examples 2 and 3 without looking at the solutions, and do Question 3. Check which items in the common traps list apply to you.
• After 7 days: redo all four questions and explain why each wrong option is wrong. Confirm you can tick every self-check item.

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15. IPA syllabus mapping

This installment corresponds to the following items in IPA’s “応用情報技術者試験（レベル3）シラバス Ver.7.2”:

• 大分類1：基礎理論／中分類1：基礎理論／2. 応用数学／（1）確率と統計／①確率
• 大分類1：基礎理論／中分類1：基礎理論／2. 応用数学／（1）確率と統計／②統計

This installment prioritizes the core calculations and interpretation that matter for Subject A. It does not claim to exhaustively cover every term listed in these broad syllabus subsections (e.g., Markov processes, the Poisson, exponential, and chi-square distributions, regression analysis, analysis of variance, principal component analysis, maximum likelihood estimation, t-tests).

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16. Sources

• IPA, “応用情報技術者試験（レベル3）シラバス Ver.7.2” (used for scope; checked 2026-10-20): https://www.ipa.go.jp/shiken/syllabus/omgdg50000005kq5-att/syllabus_ap_ver7_2.pdf

The source above was used only to confirm the scope of study. The explanations, examples, and practice questions in this article were independently authored by Motitory Labs.