Every time you say "there's a good chance it'll rain," check the odds of winning the lotto, or wonder if a coin is "due" to land heads, you are thinking about probability. And every time you hear "the average grade," "the middle income," or "the most common answer," you are using statistics. These two topics turn vague feelings about "likely" and "typical" into exact numbers — and the Civil Service Exam tests both. This chapter makes them clear, concrete, and quick.
Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain), often written as a fraction or percentage. Statistics (the descriptive kind tested here) summarizes a set of data with a few numbers: the mean, median, mode, and range.
On the Civil Service Exam, probability and basic statistics appear in Numerical Reasoning and Data Interpretation, typically 2 to 4 items. The core probability formula is short, and the four statistics measures are quick to compute — so this is a topic where a little memorization yields dependable marks.
The difficulty is beginner to intermediate. The traps are specific and learnable: knowing when to multiply ("and") versus add ("or"), adjusting for draws without replacement, using the complement to shortcut "at least one" problems, and not confusing the mean with the median. Handle those and the topic is friendly.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the idea probability rests on: a fair outcome means every possibility is equally likely. A fair coin has two equally likely sides; a fair die has six equally likely faces; a well-shuffled deck gives every card an equal chance. Probability counts the favorable equally-likely outcomes and divides by the total equally-likely outcomes. Keep "equally likely" in mind — it's the assumption that makes the simple formula valid.
Probability and statistics are how we reason about uncertainty and summarize information — everyday skills:
Understanding these topics makes you a sharper, less foolable reader of numbers — exactly the judgment the exam rewards.
We start with simple probability and build toward combined events, then cover the statistics measures.
Probability of an event = (Number of favorable outcomes) ÷ (Total number of possible outcomes).
A bag has 4 red balls and 6 blue balls. P(red) = 4 ÷ 10 = 2/5 (or 0.4, or 40%).
Every probability lands between 0 and 1. A probability of 0 means the event cannot happen; 1 means it is certain; 0.5 means it's as likely as not.
Analogy: Probability is like slicing a pizza where each equally-likely outcome is one slice. The chance of your event is just "how many slices are yours" out of all the slices. If 2 of 5 equal slices are yours, your probability is 2/5.
When you want two events to both happen, and they don't affect each other (independent), multiply their probabilities.
Rolling a die twice, P(6 then 6) = 1/6 × 1/6 = 1/36. Tossing two coins, P(heads and heads) = 1/2 × 1/2 = 1/4.
Why multiply? Each outcome of the first event branches into all outcomes of the second, so the total possibilities multiply — and only one combined branch is fully favorable.
When you want either of two events that can't both happen at once (mutually exclusive), add their probabilities.
Rolling a die once, P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3. Drawing one card, P(king or queen) = 4/52 + 4/52 = 8/52 = 2/13.
Memory tip: "And" → multiply, "or" → add. "And" narrows the target (both must happen), so the probability shrinks; "or" widens it (either counts), so the probability grows.
If you draw an item and put it back before the next draw, the situation resets — the same probabilities apply. If you don't put it back (without replacement), the total shrinks and the favorable count may too — recompute for the second draw.
From 4 red and 6 blue (10 total), draw 2 reds without replacement: P = (4/10) × (3/9) = 12/90 = 2/15. Note the second fraction: 3 reds left out of 9 total.
The probability an event does not happen is 1 minus the probability it does:
P(not A) = 1 − P(A).
This is a lifesaver for "at least one" problems, which are messy to count directly but easy through their opposite ("none").
A fair coin is flipped 3 times. P(at least one head)? Easier: P(no heads at all) = (1/2)³ = 1/8. So P(at least one head) = 1 − 1/8 = 7/8.
Two dice are rolled. P(at least one six) = 1 − P(no six) = 1 − (5/6)² = 1 − 25/36 = 11/36.
When a choice is made in stages, the total number of outcomes is the product of the choices at each stage.
With 3 shirts and 2 pairs of pants, the number of outfits = 3 × 2 = 6. A 4-digit PIN using digits 0–9 (repeats allowed) has 10 × 10 × 10 × 10 = 10,000 possibilities.
This principle also gives the denominator (total outcomes) in many probability problems.
These four numbers summarize a data set:
Data: 3, 7, 7, 9, 12. Mean = (3 + 7 + 7 + 9 + 12) ÷ 5 = 38 ÷ 5 = 7.6. Median = 7 (the middle of the sorted five). Mode = 7 (appears twice). Range = 12 − 3 = 9.
A single extreme value (an outlier) can drag the mean far from the "typical" value, but the median barely moves, because it only cares about the middle position, not the size of the extremes.
Incomes: 15k, 18k, 20k, 22k, and 500k. The mean is (15 + 18 + 20 + 22 + 500) ÷ 5 = 115k — which describes no one in the group. The median, 20k, is the honest "typical" value. This is why news reports use median income, not mean.
With the ideas in hand, let's picture, tabulate, and drill.
| Concept | Formula |
|---|---|
| Probability | favorable ÷ total |
| "and" (independent) | P(A) × P(B) |
| "or" (mutually exclusive) | P(A) + P(B) |
| Complement | P(not A) = 1 − P(A) |
| Without replacement | shrink the total (and favorable) each draw |
| Counting principle | multiply the choices at each stage |
| Mean | sum ÷ count |
| Median | middle of the sorted data |
| Mode | most frequent value |
| Range | maximum − minimum |
A note on "or" that overlaps: the simple "add" rule is for events that can't happen together. If they can overlap (e.g., "a heart or a king" in cards), subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). For "heart or king": 13/52 + 4/52 − 1/52 = 16/52 = 4/13 (the king of hearts was counted twice).
| When the problem says… | Do this… |
|---|---|
| "probability of drawing/rolling…" | favorable ÷ total |
| "both / and / then again" | multiply (independent) |
| "either / or" (can't both happen) | add |
| "without putting it back" | shrink the total each draw |
| "at least one" | 1 − P(none) (complement) |
| "how many ways / how many outfits/PINs" | multiply the stage choices |
| "average" | mean (sum ÷ count) |
| "middle / typical, resistant to extremes" | median (sort first) |
| "most common / most frequent" | mode |
Step 1 — Count the total equally-likely outcomes (the denominator). ↓ Step 2 — Count the favorable outcomes (the numerator). ↓ Step 3 — For combined events, decide "and" (multiply) or "or" (add), and adjust for replacement. ↓ Step 4 — Consider the complement if the phrasing is "at least one." ↓ Step 5 — Simplify and sanity-check: the answer must be between 0 and 1. For statistics, sort before taking the median.
Why Step 3 matters most: swapping "and" for "or" (multiply vs. add) is the defining probability error. Read the linking word carefully and decide before computing.
Example 1. A die is rolled. What is the probability of getting an even number? Solution: Favorable = {2, 4, 6} = 3; total = 6. P = 3/6 = 1/2. Difficulty: ★☆☆☆☆
Example 2. A bag has 5 red and 3 green balls. What is the probability of drawing a green ball? Solution: 3 ÷ 8 = 3/8. Difficulty: ★☆☆☆☆
Example 3 ("and"). Two coins are tossed. What is the probability that both land heads? Solution: 1/2 × 1/2 = 1/4. Difficulty: ★★☆☆☆
Example 4 ("or"). A card is drawn from a standard 52-card deck. What is the probability that it is a king or a queen? Solution: 4/52 + 4/52 = 8/52 = 2/13. Difficulty: ★★☆☆☆
Example 5 (statistics). Find the mean, median, mode, and range of 4, 8, 6, 8, 10. Thinking: Sort first: 4, 6, 8, 8, 10. Solution: Mean = 36 ÷ 5 = 7.2; median = 8 (middle); mode = 8; range = 10 − 4 = 6. Difficulty: ★★☆☆☆
Example 6 (without replacement). From 4 red and 6 blue balls, two are drawn without replacement. What is the probability both are red? Solution: (4/10) × (3/9) = 12/90 = 2/15. Difficulty: ★★★☆☆
Example 7 (complement). A fair coin is tossed 4 times. What is the probability of getting at least one tail? Thinking: Easier via the opposite: all heads. Solution: P(all heads) = (1/2)⁴ = 1/16, so P(at least one tail) = 1 − 1/16 = 15/16. Difficulty: ★★★☆☆
Example 8 (two dice sum). Two dice are rolled. What is the probability the sum is 7? Thinking: Total outcomes = 6 × 6 = 36. Favorable sums of 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6. Solution: 6/36 = 1/6. Difficulty: ★★★☆☆
Example 9 (counting principle). A cafeteria offers 3 rice meals, 4 viands, and 2 drinks. How many different meal combinations (one of each) are possible? Solution: 3 × 4 × 2 = 24. Difficulty: ★★★☆☆
Example 10 (at least one six). Two dice are rolled. What is the probability of getting at least one six? Solution: P(no six) = (5/6)² = 25/36, so P(at least one six) = 1 − 25/36 = 11/36. Difficulty: ★★★★☆
Example 11 (median, even count). Find the median of 12, 7, 9, 15, 20, 10. Thinking: Sort: 7, 9, 10, 12, 15, 20. Even count of 6 → average the two middle values (10 and 12). Solution: (10 + 12) ÷ 2 = 11. Difficulty: ★★★☆☆
Example 12 (mean vs. median with an outlier). Seven employees earn (in thousands) 14, 15, 16, 16, 18, 20, and 120. Which better describes a "typical" salary, the mean or the median, and what is it? Thinking: The 120 is an outlier that inflates the mean. Solution: Mean = 219 ÷ 7 ≈ 31.3k (higher than almost everyone). Median = 16k (the 4th of 7 sorted values) — the honest "typical" figure. The median is better here. Difficulty: ★★★★☆
Example 13 (dependent draw). From a deck of 52 cards, two are drawn without replacement. What is the probability both are kings? Solution: (4/52) × (3/51) = 12/2,652 = 1/221. Difficulty: ★★★★☆
Example 14 (combined "and" across draws). A box has 3 defective and 7 good bulbs. Two are drawn without replacement. What is the probability that both are good? Solution: (7/10) × (6/9) = 42/90 = 7/15. Difficulty: ★★★★☆
Probability is favorable ÷ total, always between 0 and 1. Combine events by the linking word: "and" → multiply (independent), "or" → add (mutually exclusive), and subtract the overlap when events can coincide. For draws without replacement, shrink the totals each time; for "at least one," use the complement (1 − P(none)). Count staged possibilities with the fundamental counting principle (multiply the choices). Statistics summarizes data with the mean (average), median (sorted middle), mode (most frequent), and range (max − min); the median is the honest "typical" value when outliers distort the mean. Read the linking word, mind replacement, use the complement, and sort before the median — and this becomes a dependable scoring area.
| Item | Key point |
|---|---|
| Probability | favorable ÷ total (0 to 1) |
| "and" | multiply |
| "or" (exclusive) | add |
| "or" (overlapping) | add, then subtract the overlap |
| without replacement | shrink totals each draw |
| "at least one" | 1 − P(none) |
| counting | multiply stage choices |
| mean / median / mode / range | avg / sorted middle / most frequent / max − min |
| outliers | move the mean, not the median |
When do I multiply and when do I add probabilities? Multiply for "and" (both events happen, independent). Add for "or" (either event, mutually exclusive). If the two "or" events can happen together, add and then subtract the overlap.
What changes when a draw is "without replacement"? The item isn't returned, so the total (and possibly the favorable count) drops by one for the next draw. Recompute the fraction for each draw.
Why is the complement rule useful? "At least one" is hard to count directly but easy as 1 − P(none). It turns a messy problem into a quick subtraction.
What's the difference between mean and median? The mean is the arithmetic average (sum ÷ count); the median is the middle value once the data is sorted. They differ when the data has outliers — the median stays near the "typical" value.
Can a data set have more than one mode? Yes. If two values tie for most frequent, the set is "bimodal"; if none repeats, there is no mode. The mode is simply the most frequent value(s).
Tick them all and probability and statistics become quick, confident points — and you'll read real-world odds and averages far more critically.
Put it to the test with 1,545 practice questions on this topic.