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Digital SAT Prep · Quantitative Aptitude

Probability, Statistics & Permutation-Combination

Basic probability, measures of central tendency, dispersion and counting via permutation and combination.

Six concepts covering three exam clusters that share one habit: count the sample space before you touch a formula. Probability is favourable over total; permutations and combinations are the same product with or without order; mean, median, mode and standard deviation are summaries of a list you have already ordered. Get the count right and the arithmetic collapses.

  • Digital SAT Prep
  • Hard level
  • 6 concepts
  • 42 practice questions

1Classical probability is favourable over total

When every outcome is equally likely, P(E) is the number of favourable outcomes divided by the number of possible outcomes. The answer always lands in [0, 1]: 0 for impossible, 1 for certain, and every exam option outside that range is already wrong.

The complement is the rest of the sample space: P(\bar E) = 1 - P(E). It is not a second formula so much as a reminder that "not E" is often shorter to count than "E" — which is why the next concept exists.

Figure. One bar is the whole deck. The short segment is the favourable set and the long segment is everything else; classical probability is the short length divided by the full length.

How it works

  1. Name the sample spaceCount every equally likely outcome — 52 cards, 6 faces, 2^n coin sequences.
  2. Count the favourable onesList or classify only the outcomes the event allows; do not invent extra cases.
  3. Divide, then reduceWrite P = \text{favourable}/\text{total} and cancel to simplest form — that is the option the key lists.

Probability with a Deck of Cards

One card is drawn from a well-shuffled standard deck of 52 cards. Find the probability that it is a king.

  • Total outcomes52
  • Favourable outcomes (kings)4
  • P(king) = 4/521/13
  • P(not a king) = 1 − 1/1312/13

Pro tip. Reduce the fraction fully; examiners usually list the simplest form as the correct option. Leaving 4/52 unreduced looks like a different answer and loses the mark even though the value is right.

One card is drawn from a deck of 52. P(ace or king) is
  1. 2/13
  2. 1/13
  3. 8/13

Aces and kings are disjoint, so favourable = 4 + 4 = 8 and P = 8/52 = 2/13. Answering 1/13 counts only one rank; answering 8/13 forgets to divide by 52 and treats the count as if it were already a probability.

2Addition for unions, multiplication for independent intersections

The general addition rule is P(A \cup B) = P(A) + P(B) - P(A \cap B). The intersection is subtracted once because it was counted in both P(A) and P(B). When A and B are mutually exclusive, that intersection is empty and the rule collapses to P(A) + P(B).

Independence is a different claim: A does not change the odds of B, which is written P(A \cap B) = P(A) \times P(B). Mutually exclusive events with positive probability are never independent — if one happened the other cannot — so do not multiply and add the same pair.

Figure. Die sample space from the worked ledger, drawn in sixths so the heights are exact face counts: even faces P(A)=3/6, multiples of three P(B)=2/6, shared face 6 at P(A\cap B)=1/6. Adding the first two bars double-counts that face; subtracting the intersection bar once yields the union bar at 4/6=2/3. Bars are counts on a common scale — not a Venn shape.

How it works

  1. Decide exclusive or overlappingIf both cannot happen together, add. If they can, you need P(A \cap B) before you add.
  2. Apply additionP(A \cup B) = P(A) + P(B) - P(A \cap B), and drop the last term only when the intersection is empty.
  3. Multiply only if independentFor "both", write P(A) \times P(B) only when the problem states independence or sampling with replacement / separate trials.

A die: even or multiple of three

A fair die is rolled once. Let A be "even" and B be "a multiple of 3". Find P(A \cup B).

  • P(A) = |{2,4,6}|/61/2
  • P(B) = |{3,6}|/61/3
  • P(A ∩ B) = |{6}|/61/6
  • P(A ∪ B) = 1/2 + 1/3 − 1/62/3

Pro tip. Adding 1/2 + 1/3 without subtracting 1/6 double-counts the face 6 and answers 5/6. The subtraction is not optional decoration — it is the face that sits in both lists.

Two fair coins are tossed. P(both heads), treating the tosses as independent, is
  1. 1/4
  2. 1/2
  3. 1

Independence multiplies: (1/2) \times (1/2) = 1/4. Answering 1/2 is the probability of heads on a single toss; answering 1 confuses "possible" with "certain".

3For "at least one", compute one minus none

An "at least one" event is a union of many favourable cases, and listing them all is how the clock runs out. The complement is a single case: none. So P(\text{at least one}) = 1 - P(\text{none}).

This is the same complement rule as classical probability, aimed at the wording that most often tempts a long sum. If you can describe "none" in one product — no six on either roll, no head in three tosses — you are already most of the way to the answer.

Figure. The top bar is the sample space. The middle bar is the complement event "no six", and the bottom bar is what remains — at least one six. The method is subtraction of bar lengths, not a case-split over the eleven favourable pairs.

How it works

  1. Name the "none" eventRewrite "at least one six" as the complement of "no sixes at all".
  2. Compute P(none)Usually a product of independent failure probabilities, or a single count over the sample space.
  3. Subtract from 1P(\text{at least one}) = 1 - P(\text{none}); reduce the fraction.

At least one six in two rolls

A fair die is rolled twice. Find the probability of getting at least one six.

  • Total outcomes6 × 6 = 36
  • P(no six on one roll)5/6
  • P(no six on either roll) = (5/6) × (5/6)25/36
  • P(at least one six) = 1 − 25/3611/36

Pro tip. Counting the favourable cases directly — six on the first only, six on the second only, six on both — also gives 5 + 5 + 1 = 11 out of 36, but only because the sample space is tiny. The complement route stays one product long when the direct count has grown into a case-split.

A fair coin is tossed three times. P(at least one head) is
  1. 7/8
  2. 1/8
  3. 1/2

P(no heads) = (1/2)^3 = 1/8, so P(at least one head) = 1 - 1/8 = 7/8. Answering 1/8 stops at the complement; answering 1/2 is a single-toss probability pasted onto a three-toss question.

4Order decides permutation or combination

Ask one question before writing a formula: does order matter? Arrangements, rankings and passwords are permutations, ^nP_r = n!/(n-r)!. Selections, committees and hands of cards are combinations, ^nC_r = n!/(r!(n-r)!).

The two are linked by ^nP_r = {}^nC_r \times r!: every unordered selection of r items can be lined up in r! orders. The identity ^nC_r = {}^nC_{n-r} is the same link used the other way — choosing whom to leave out is choosing whom to take, and the smaller of r and n-r is the one you compute.

Figure. Same six people, same three seats to fill. The tall bar counts ordered arrangements (^6P_3=120); the short bar counts unordered committees (^6C_3=20). The factor between them is exactly 3!=6 — every committee can be lined up in 3! orders, which is the whole distinction the order-matters question asks.

How it works

  1. Decide orderYes → permutation. No → combination. This single fork prevents the most common error in the chapter.
  2. Write the smaller binomialPrefer ^nC_r with r \le n-r, using ^nC_r = {}^nC_{n-r} when the selection is large.
  3. Cancel before multiplyingExpand only the factors that do not cancel against the denominator, so ^6C_3 is (6 \times 5 \times 4)/(3 \times 2 \times 1).
Permutation or combination
QuestionToolFormulaTiny check
Order matters?Permutation^nP_r = n!/(n-r)!^6P_3 = 120
Order does not?Combination^nC_r = n!/(r!(n-r)!)^6C_3 = 20
Large selectionUse complement index^nC_r = {}^nC_{n-r}^{20}C_{18} = {}^{20}C_2 = 190

Selecting a Committee (Combination)

In how many ways can a committee of 3 be chosen from 6 people?

  • Order does not matteruse ⁶C₃
  • ⁶C₃ = 6! / (3! 3!)(6 × 5 × 4) / (3 × 2 × 1)
  • (6 × 5 × 4) / (3 × 2 × 1)120 / 6
  • Number of committees20

Pro tip. For selections use combinations; cancel factorials early to keep numbers small. The same count as a permutation would be ^6P_3 = 120, which is 20 \times 3! — every committee counted once for each seating order the problem never asked for.

The number of ways to arrange 3 of 6 distinct books on a shelf is
  1. 120
  2. 20
  3. 216

Shelf order matters, so ^6P_3 = 6 \times 5 \times 4 = 120. Answering 20 is ^6C_3 and drops the 3! arrangements of each selection. Answering 216 is 6^3 and allows repeats the problem never offered.

5Mean, median, mode, and the empirical link

The mean is the arithmetic average \sum x_i / n. The median is the middle value after the list is ordered (or the average of the two middle values when n is even). The mode is the most frequent value. Each answers a different question: the mean balances the total, the median resists outliers, and the mode names the peak.

For a moderately skewed unimodal distribution the three are linked by the empirical relation Mode \approx 3\,Median - 2\,Mean. It is a check and an estimator, not an identity — on a small discrete list it may hold exactly by accident, and when the distribution is symmetric all three coincide.

Figure. Five values as bars with a dashed rule at the mean. The two equal bars at 14 are the mode; the middle bar in the ordered list is also 14, so median, mode and mean sit on the same rule.

How it works

  1. Order the dataSort ascending before reading the median or scanning for a mode.
  2. Compute each measureMean from the total, median from the middle position, mode from the highest frequency.
  3. Use the empirical link when askedGiven any two of mean, median and mode, estimate the third with Mode \approx 3\,Median - 2\,Mean.

Five scores, three measures

For the data 10, 12, 14, 14, 20, find the mean, median and mode, and check the empirical relation.

  • Ordered data10, 12, 14, 14, 20
  • Mean = (10+12+14+14+20)/514
  • Median = middle value; mode = most frequent14; 14
  • 3×Median − 2×Mean = 3×14 − 2×1414 (= mode here)

Pro tip. Here the distribution is peaked at the centre, so mean, median and mode agree and the empirical check is exact. When a question gives mean 20 and median 18 and asks for the mode, the same row yields 3 \times 18 - 2 \times 20 = 14 — and Mode < Median < Mean signals a right skew.

Mean = 20 and median = 18. The empirical estimate of the mode is
  1. 14
  2. 16
  3. 22

Mode \approx 3 \times 18 - 2 \times 20 = 54 - 40 = 14. Answering 16 averages the two given numbers; answering 22 adds them with the wrong signs as 3 \times 18 - 2 \times 16 style arithmetic on the mean.

6Standard deviation measures spread about the mean

Standard deviation is \sigma = \sqrt{\sum (x_i - \bar x)^2 / n}. Square the deviations so that left and right misses do not cancel, average them, then take the square root to return to the original units. A larger \sigma means the data sit farther from the mean; \sigma = 0 means every value equals the mean.

Exam questions use the population form with n in the denominator unless they say "sample" and write n-1. Read the formula sheet before dividing. Comparing two lists with the same mean is the cleanest way to see that \sigma is about spread, not about the centre.

Figure. Both groups share the dashed mean at 5. The left group reaches from 2 to 8 (\sigma = \sqrt{5}); the right group is four bars on the mean (\sigma = 0). Same centre, different spread — that difference is what \sigma reports.

How it works

  1. Find the meanCompute \bar x = \sum x_i / n first; every deviation is measured from it.
  2. Square the deviationsForm (x_i - \bar x)^2 for each value and add them.
  3. Divide by n, then root\sigma = \sqrt{(\sum (x_i - \bar x)^2)/n}. Leave the answer under a root when it is not a perfect square.

Spread of 2, 4, 6, 8

Find the standard deviation of 2, 4, 6, 8 using \sigma = \sqrt{\sum (x_i - \bar x)^2 / n}.

  • Mean = (2+4+6+8)/45
  • Deviations from 5−3, −1, 1, 3
  • Sum of squared deviations9+1+1+9 = 20
  • σ = √(20/4) = √5√5

Pro tip. The set 5, 5, 5, 5 has the same mean and \sigma = 0. Same centre, zero spread — which is why a question that holds the mean fixed and asks which list is "more consistent" is asking only which \sigma is smaller.

For 2, 4, 6, 8 the population standard deviation is
  1. √5
  2. 5
  3. √20

Sum of squared deviations is 20, so \sigma = \sqrt{20/4} = \sqrt{5}. Answering 5 is the mean. Answering \sqrt{20} forgot to divide by n before taking the root.

Notes

  • Classical Probability: For equally likely outcomes, P(E)=\frac{\text{favourable outcomes}}{\text{total outcomes}}, always between 0 and 1, with P(\text{not }E)=1-P(E).
  • Permutation vs Combination: Use permutation ^nP_r=\frac{n!}{(n-r)!} when order matters (arrangements) and combination ^nC_r=\frac{n!}{r!(n-r)!} when it does not (selections).
  • Central Tendency: The mean is the arithmetic average, the median is the middle value of ordered data, and the mode is the most frequent value; the empirical relation is Mode \approx 3\,Median -2\,Mean.
  • Addition & Multiplication Rules: For mutually exclusive events P(A\cup B)=P(A)+P(B); for independent events P(A\cap B)=P(A)\times P(B).
  • Standard Deviation: SD measures dispersion about the mean as \sqrt{\frac{\sum(x_i-\bar x)^2}{n}}; a larger SD means data is more spread out.

Formulas

  • Probability: P(E)=\frac{\text{favourable}}{\text{total}}, P(\bar E)=1-P(E)
  • Permutations: ^nP_r = \frac{n!}{(n-r)!}
  • Combinations: ^nC_r = \frac{n!}{r!(n-r)!}, with ^nC_r = {}^nC_{n-r}
  • Mean =\frac{\sum x_i}{n}; empirical: Mode = 3\,Median - 2\,Mean
  • Addition rule: P(A\cup B)=P(A)+P(B)-P(A\cap B)
  • Standard deviation: \sigma = \sqrt{\frac{\sum(x_i-\bar x)^2}{n}}

Exam traps & shortcuts

  • Decide 'order matters?' first: yes \to permutation, no \to combination; this single question prevents the most common error.
  • Use ^nC_r = {}^nC_{n-r} to shrink big selections, e.g. ^{20}C_{18}={}^{20}C_2 = 190, computed instantly.
  • For 'at least one' probability, compute 1 - P(\text{none}) rather than summing many favourable cases.

Reference tables

Formula sheet
IdeaFormulaWatch
Classical probabilityP(E) = \text{favourable}/\text{total}Reduce the fraction
ComplementP(\bar E) = 1 - P(E)Use for "at least one"
AdditionP(A\cup B) = P(A)+P(B)-P(A\cap B)Drop the last term if exclusive
Independent "both"P(A\cap B) = P(A)\times P(B)Not for exclusive events
Permutation^nP_r = n!/(n-r)!Order matters
Combination^nC_r = n!/(r!(n-r)!)^nC_r = {}^nC_{n-r}
Mean\bar x = \sum x_i / nTotal first
Empirical modeMode \approx 3\,Median - 2\,MeanEstimate, not identity
Standard deviation\sigma = \sqrt{\sum(x_i-\bar x)^2/n}Population n unless told sample
Wording → tool
WordingToolCommon trap
"at least one"1 - P(\text{none})Listing every favourable case
"committee" / "select"CombinationUsing ^nP_r and overcounting by r!
"arrange" / "rank"PermutationUsing ^nC_r and undercounting
"more consistent" / "less scattered"Smaller \sigmaComparing means instead of spreads

Recap

Read only this the night before.

Probability
Favourable over total, then reduce. Complement is 1 minus the event; "at least one" is 1 minus none.
Unions
Add and subtract the intersection. Exclusive means the intersection is empty; independent means multiply for "both".
Counting
Order matters → ^nP_r. Order does not → ^nC_r. Shrink with ^nC_r = {}^nC_{n-r}.
Centre
Mean from the total, median from the ordered middle, mode from frequency. Mode ≈ 3 Median − 2 Mean.
Spread
σ is the root-mean-square deviation from the mean. Same mean can hide very different spreads.

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