Statistics & Probability

📊 Mean, Median, and Mode

Part 1 of 7 — Measures of Center, Weighted Averages, and Outliers

Statistics questions on the ACT Math test (45 questions in 50 minutes, 4 answer choices each) are usually short, but they reward students who know exactly which measure is being asked for and which shortcut fits. This part covers the three measures of center, the range, and the two ideas the ACT returns to most often: working with sums and weighted (combined) averages.

The Four Basic Measures

Median with an even count: average the two middle values. For 3, 5, 8, 12, the median is $\frac{5 + 8}{2} = 6.5$. Always sort first — averaging the two middle entries of an unsorted list is one of the most common wrong answers.

A data set can have no mode (every value appears once) or more than one mode (two values tie for most frequent).

Think in Sums: Mean × Count = Total

The single most useful fact about the mean is

$\text{sum} = \text{mean} \times \text{count}$

Almost every "missing value" question becomes easy once you convert means into totals:

Example: The mean of 6 numbers is 15, so their total is 90. If one number is removed and the mean of the remaining 5 is 13, their total is 65, so the removed number was $90 - 65 = 25$.

The shortfall shortcut: To raise a mean of 82 on 4 tests to 85 on 5 tests, the new score must be 85 plus 3 points for each of the 4 earlier tests: $85 + 4(3) = 97$.

Weighted Averages and Combined Means

When groups of different sizes are combined, you cannot simply average the group means. Weight each mean by its group size:

$\text{combined mean} = \frac{n_1 m_1 + n_2 m_2}{n_1 + n_2}$

If 20 students average 75 and 30 students average 85, the combined mean is $\frac{20(75) + 30(85)}{50} = \frac{4050}{50} = 81$, not 80. The combined mean always lands between the two group means and closer to the larger group.

Percent weights work the same way. If homework is 20%, tests 50%, and the final 30% of a grade, then

$\text{grade} = 0.20(\text{homework}) + 0.50(\text{tests}) + 0.30(\text{final})$

Check that the weights add to 100% before you compute.

How Outliers and Changes Affect Each Measure

Because the median ignores how far the extreme values are from the center, it is called resistant. When a data set has an outlier or is strongly skewed (salaries, home prices, wait times), the median usually describes a "typical" value better than the mean. A quick test: if the mean is larger than most of the data values, an outlier is dragging it.

Worked Examples

Question: Leah's first 3 test scores have a mean of 78. What must she score on the 4th test for her 4-test mean to be 82?

Solution:

1. Current total: $3 \times 78 = 234$.
2. Needed total: $4 \times 82 = 328$.
3. Needed score: $328 - 234 = 94$.

Check with the shortfall shortcut: each of the 3 earlier tests is 4 points below 82, so the 4th test must be $82 + 3(4) = 94$. ✓

ACT trap: Answers like 86 or 90 make up the shortfall for only one or two of the earlier tests.

Question: A 10-person team has a mean time of 70 seconds on a drill. Five new members join, and the mean time for all 15 members becomes 74 seconds. What is the mean time of the 5 new members?

Solution:

1. Original total: $10 \times 70 = 700$ seconds.
2. New total: $15 \times 74 = 1110$ seconds.
3. New members' total: $1110 - 700 = 410$, so their mean is $\frac{410}{5} = 82$ seconds. ✓

Why not 78? 78 is what you get if the two groups were the same size (74 is halfway between 70 and 78). The larger original group pulls the combined mean toward 70, so the new members must be farther above 74 to compensate.

Quick Check: Measures of Center 🎯

Work with Totals 🧮

1. The mean of 8 numbers is 12.5. What is the sum of the 8 numbers?

2. The mean of 5 numbers is 20. One number is removed, and the mean of the remaining 4 numbers is 18. What number was removed?

3. Section A has 12 students with a mean score of 80. Section B has 18 students with a mean score of 90. What is the mean score of all 30 students?

ACT-Style Practice

Try each one in under a minute, converting means to totals wherever you can.

ACT Tip: If a question gives you a mean and asks about an individual value, your first move should almost always be to multiply and get a total.

ACT-Style Questions: Weighted Averages and Changes 📋

Key Takeaways

• Mean = sum ÷ count; median = middle of the sorted list (average the two middle values for an even count); mode = most frequent; range = max − min.
• Turn means into totals: $\text{sum} = \text{mean} \times \text{count}$. Missing, removed, and needed values all come from comparing totals.
• Combined means are weighted: $\frac{n_1 m_1 + n_2 m_2}{n_1 + n_2}$. The result sits closer to the larger group. Percent weights work the same way.
• An outlier pulls the mean but barely moves the median, so the median is the better "typical value" for skewed data.
• Adding $c$ to every value shifts the mean and median by $c$ and leaves the range alone; multiplying by $k$ scales all three.

📈 Data Displays and Spread

Part 2 of 7 — Frequency Tables, Histograms, Box Plots, Range, and IQR

The ACT rarely hands you a plain list of numbers. More often the data arrive in a display — a frequency table, a histogram, a dot plot, a stem-and-leaf plot, or a box plot — and you must read the display correctly before any formula helps. This part teaches you to pull the mean, median, and spread out of each display.

Frequency Tables: Every Row Is Repeated Data

A frequency table lists each value once with how many times it occurs.

• Count: add the frequencies: $6 + 9 + 7 + 2 + 1 = 25$ students.
• Mean: multiply each value by its frequency, add, then divide by the count: $\frac{0(6) + 1(9) + 2(7) + 3(2) + 4(1)}{25} = \frac{33}{25} = 1.32$.
• Median: find its position first. With 25 values, the median is the 13th. Count up the frequencies (a running total): students 1–6 have 0 pets, students 7–15 have 1 pet, so the 13th student has 1 pet.
• Mode: the value with the largest frequency (here, 1).

Trap: the median is a data value (1 pet), never its position (13) and never the middle frequency. Likewise, the mean is not the average of the frequencies.

Histograms and Dot Plots

A histogram groups values into intervals (bins), and each bar's height is the number of values in that bin. You can find how many values fall in a range and which bin contains the median, but you usually cannot find exact values, the exact mean, or the exact median. A dot plot shows every value as a dot, so it works like a frequency table.

Stem-and-Leaf Plots

Each value is split into a stem (leading digits) and a leaf (last digit). A row "Stem 5: leaves 0, 3, 3, 7" means the values 50, 53, 53, and 57. Read every leaf as a separate data value; the plot is already sorted, which makes the median easy.

Measures of Spread

Finding quartiles: sort the data and find the median. $Q_1$ is the median of the lower half and $Q_3$ is the median of the upper half. With an odd number of values, leave the overall median out of both halves.

For 3, 5, 7, 8, 10, 12, 13, 15, 18, 20, 24: median = 12, $Q_1$ = median of 3, 5, 7, 8, 10 = 7, $Q_3$ = median of 13, 15, 18, 20, 24 = 18, so IQR $= 18 - 7 = 11$ and range $= 24 - 3 = 21$.

Standard deviation on the ACT is almost always conceptual: you will be asked which data set has the larger standard deviation, not to compute it. Values bunched tightly around the mean → small standard deviation; values spread far from the mean → large standard deviation. Two data sets can have the same mean and very different spreads.

Box Plots (Five-Number Summary)

A box plot shows minimum, $Q_1$, median, $Q_3$, maximum. The box runs from $Q_1$ to $Q_3$, with a line at the median; the whiskers reach the min and max.

A longer box or whisker means the values in that region are more spread out, not that it contains more values. A box plot hides the mean entirely.

How Changes Affect Spread

Worked Examples

Question: Twenty students took a 10-point quiz.

Find the median and the mean.

Solution:

1. Count: $2 + 5 + 8 + 4 + 1 = 20$, so the median is the average of the 10th and 11th scores.
2. Running totals: scores of 6 fill positions 1–2, 7s fill 3–7, 8s fill 8–15. Both the 10th and 11th scores are 8, so the median is 8.
3. Mean: $\frac{6(2) + 7(5) + 8(8) + 9(4) + 10(1)}{20} = \frac{12 + 35 + 64 + 36 + 10}{20} = \frac{157}{20} = 7.85$. ✓

ACT trap: 10.5 is the median's position $\left(\frac{20 + 1}{2}\right)$, not its value.

Question: A box plot of commute times (minutes) has minimum 8, $Q_1$ = 15, median 22, $Q_3$ = 34, and maximum 50. Find the range and IQR, and describe what fraction of commutes take at least 15 minutes.

Solution:

1. Range $= 50 - 8 = 42$ minutes.
2. IQR $= 34 - 15 = 19$ minutes.
3. $Q_1 = 15$ marks the 25th percentile, so about 75% of commutes take at least 15 minutes. ✓

Note: The right part of the box (22 to 34) is longer than the left (15 to 22), so the upper-middle commutes are more spread out — but each part still holds about 25% of the data.

Quick Check: Reading Displays 🎯

Read the Display 🧮

A histogram of 25 test scores has these bars: 0–9: 3 values; 10–19: 7 values; 20–29: 9 values; 30–39: 5 values; 40–49: 1 value.

1. How many scores are 20 or greater?

2. What percent of the scores are less than 20? (Enter a number only.)

3. A stem-and-leaf plot shows: Stem 4: leaves 2, 5, 8; Stem 5: leaves 0, 3, 3, 7; Stem 6: leaves 1, 4. What is the median of these values?

ACT-Style Practice

ACT Tip: Before you compute anything from a display, say out loud what each number in it means: a value, or how many times a value occurs.

ACT-Style Questions: Spread and Comparison 📋

Key Takeaways

• In a frequency table, each value counts as many times as its frequency: mean $= \frac{\sum(\text{value} \times \text{frequency})}{\text{total frequency}}$.
• Find the median's position with running totals, then report the value at that position — never the position itself.
• A histogram tells you which bin holds the median, not its exact value; a stem-and-leaf plot lists every value in order.
• Quartiles: $Q_1$ and $Q_3$ are the medians of the lower and upper halves (leave out the overall median for an odd count). IQR $= Q_3 - Q_1$ covers the middle 50%.
• Box plot: min, $Q_1$, median, $Q_3$, max; each section holds about 25% of the data.
• Standard deviation measures spread around the mean — compare it by eye. Adding a constant leaves all spread measures unchanged; multiplying by $k$ scales them by $k$.

🔢 Counting Principles

Part 3 of 7 — The Multiplication Principle, Cases, Restrictions, and Arrangements

Counting questions ask "how many ways?" The ACT rarely requires a formula you cannot rebuild from one idea: fill the slots and multiply. Probability questions later in this unit also depend on counting the total number of outcomes, so this part is the foundation for Parts 4 and 5.

The Multiplication Principle

If one choice can be made in $a$ ways and a second, separate choice can be made in $b$ ways, the pair of choices can be made in $a \times b$ ways. The rule extends to any number of steps.

Example: 4 shirts, 3 pairs of pants, and 2 pairs of shoes make $4 \times 3 \times 2 = 24$ outfits. Adding ($4 + 3 + 2 = 9$) is the classic wrong answer: it counts single items, not combinations of items.

The Slot Method

Draw one blank for each position, write the number of choices for each blank, and multiply.

Read the repetition rule carefully. "Different," "distinct," or "no repeats" means each slot has one fewer choice than the slot before it. "May repeat" means every slot has the full set of choices.

Restrictions Go First

When one slot has a special condition, fill that slot first, then fill the rest.

• A 3-digit number with distinct digits: the first digit can't be 0 (9 choices), the second can be anything except the first (9 choices, now including 0), the third has 8 choices: $9 \times 9 \times 8 = 648$.
• 5 people in a line with Ana first: Ana's slot has 1 choice, then $4 \times 3 \times 2 \times 1 = 24$ ways for the rest.

Arranging Everything: Factorials

The number of ways to arrange $n$ different objects in a row is

$n! = n \times (n - 1) \times \cdots \times 2 \times 1$

So 5 books can stand on a shelf in $5! = 120$ orders, and the letters of MATH can be arranged in $4! = 24$ ways. By definition, $0! = 1$.

"Or" Means Add (When the Cases Don't Overlap)

If the outcomes split into separate cases that can't happen together, count each case and add.

Example: From town A to town C, a driver can go through town B (4 roads from A to B, then 3 roads from B to C) or take one of 2 direct highways. Through B: $4 \times 3 = 12$ routes. Direct: 2 routes. Total: $12 + 2 = 14$.

Rule of thumb: "and then" (steps in sequence) → multiply; "either…or" (separate cases) → add.

Yes/No Choices: Powers of 2

When each of $n$ items is either included or not, each item is a 2-way choice, so there are $2^n$ possible selections (including selecting nothing).

• 4 coin flips: $2^4 = 16$ possible heads/tails sequences.
• 5 optional pizza toppings: $2^5 = 32$ topping choices, including a plain pizza.

"At Least One": Count the Opposite

Counting "at least one" directly means adding many cases. Instead, use

$\text{(at least one)} = \text{(total)} - \text{(none)}$

Example: 4-digit PINs with at least one repeated digit: all PINs ($10^4 = 10{,}000$) minus PINs with no repeats ($10 \times 9 \times 8 \times 7 = 5{,}040$) = 4,960.

Small Cases: Just List Them

If the total is small (under about 15 outcomes), an organized list or tree diagram is fast and safe — and it is a good way to check a formula you are unsure of.

Worked Examples

Question: A locker code is 3 letters followed by 2 digits. The letters must all be different, and the first digit cannot be 0 (digits may repeat). How many codes are possible?

Solution:

1. Letters: $26 \times 25 \times 24 = 15{,}600$.
2. Digits: first digit 9 choices (1–9), second digit 10 choices: $9 \times 10 = 90$.
3. Multiply the two parts: $15{,}600 \times 90 = 1{,}404{,}000$. ✓

Check: each restriction only lowers one factor. If letters could repeat you would use $26^3$; if 0 were allowed first you would use $10 \times 10$.

Question: How many 4-digit PINs (digits 0–9, leading 0 allowed) contain at least one repeated digit?

Solution:

1. Total PINs: $10^4 = 10{,}000$.
2. PINs with all different digits: $10 \times 9 \times 8 \times 7 = 5{,}040$.
3. At least one repeat: $10{,}000 - 5{,}040 = 4{,}960$. ✓

Why not count directly? "At least one repeat" includes exactly one pair, two pairs, three of a kind, and four of a kind — four separate cases. The complement is one clean calculation.

Quick Check: Fill the Slots 🎯

Count It 🧮

1. In how many ways can the letters of the word MATH be arranged?

2. A password is 1 letter from A through E followed by 2 digits that must be different from each other. How many passwords are possible?

3. Six runners are in a race. In how many ways can first place and second place be awarded (no ties)?

ACT-Style Practice

ACT Tip: Write the slots before you write any numbers. Most counting mistakes come from forgetting a slot or giving a restricted slot too many choices.

ACT-Style Questions: Restrictions and Complements 📋

Key Takeaways

• Multiplication principle: steps in sequence multiply. Draw a slot for each position and write the number of choices in it.
• Repetition: "may repeat" keeps every slot full; "different/distinct" drops one choice per slot.
• Restricted slots first (no leading 0, a person fixed in a spot), then fill the rest.
• Arranging $n$ different objects: $n!$ ways.
• Separate cases ("either…or") add; steps within a case multiply.
• Each item in or out: $2^n$ selections. At least one: total − none.

🎲 Basic Probability

Part 4 of 7 — Simple Probability, Complements, the Addition Rule, and Independent Events

Probability measures how likely an event is, on a scale from 0 (impossible) to 1 (certain). ACT probability questions are built from a small set of rules. The skill being tested is choosing the right rule and the right denominator.

Simple Probability

When all outcomes are equally likely,

$P(\text{event}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$

Example: A bag holds 5 red, 7 blue, and 8 green marbles. $P(\text{blue}) = \frac{7}{20}$. The denominator is all 20 marbles — not the 13 that are not blue.

Bounds check: every probability is between 0 and 1, inclusive. An answer like $\frac{5}{4}$, 1.3, or a negative number is impossible, so eliminate it on sight. Probabilities may be written as fractions, decimals, or percents ($\frac{1}{4} = 0.25 = 25\%$).

The Complement Rule

The complement of $A$ ("not $A$") contains every outcome where $A$ does not happen:

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

If $P(\text{rain}) = 0.3$, then $P(\text{no rain}) = 0.7$. The complement is the fastest route whenever the question says not, neither, or at least one.

"Or": The Addition Rule

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

The subtraction removes the outcomes counted twice — the ones in both events.

Example: In a club of 50, 22 play chess, 17 play piano, and 5 play both. Then $P(\text{chess or piano}) = \frac{22 + 17 - 5}{50} = \frac{34}{50}$, and $P(\text{neither}) = 1 - \frac{34}{50} = \frac{16}{50} = \frac{8}{25}$.

Mutually exclusive events cannot happen together (rolling a 2 and rolling a 5 on one die). For them $P(A \text{ and } B) = 0$, so the rule becomes simply $P(A) + P(B)$.

Number-range version: For integers 1 through 30, there are 10 multiples of 3 and 6 multiples of 5, and the multiples of 15 (15 and 30) are in both groups. So $P(\text{multiple of 3 or 5}) = \frac{10 + 6 - 2}{30} = \frac{14}{30} = \frac{7}{15}$.

Venn Diagrams in Words

Many ACT questions describe overlapping groups without drawing them. Fill in a mental (or scratch) Venn diagram from the inside out:

The same reasoning works with percents: if 55% own a dog, 40% own a cat, and 20% own both, then $55 + 40 - 20 = 75\%$ own at least one and 25% own neither.

"And": Independent Events

Events are independent when one happening does not change the probability of the other (separate coin flips, separate dice, draws with replacement). Then

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

Example: $P(\text{heads on 3 straight flips}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.

Dependent Events: Without Replacement

When items are drawn without replacement, the second draw's probabilities change because the first item is gone. Multiply, but update the counts:

$P(\text{both defective}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$

for 2 bulbs drawn from a box of 8 that contains 3 defective ones. Using $\frac{3}{8} \times \frac{3}{8}$ would treat the draws as if the first bulb were put back.

"At Least One" with Independent Events

$P(\text{at least one}) = 1 - P(\text{none})$

A player who makes 70% of free throws misses 30%. The chance she misses both of 2 shots is $0.3 \times 0.3 = 0.09$, so $P(\text{at least one make}) = 0.91$. Adding $0.7 + 0.7 = 1.4$ gives an impossible probability — a sign you used "or" logic on overlapping events.

Rule Selection at a Glance

Worked Examples

Question: Of 40 students, 18 take art, 15 take music, and 7 take both. A student is chosen at random. Find $P(\text{art or music})$ and $P(\text{neither})$.

Solution:

1. Addition rule: $P(\text{art or music}) = \frac{18 + 15 - 7}{40} = \frac{26}{40} = \frac{13}{20}$.
2. Complement: $P(\text{neither}) = 1 - \frac{13}{20} = \frac{7}{20}$ (14 students). ✓

ACT trap: Forgetting to subtract the 7 gives 33 students in art or music and only 7 in neither — the students in both classes get counted twice.

Question: A drawer holds 4 black socks and 6 white socks. Two socks are taken at random without replacement. What is the probability that at least one is black?

Solution:

1. Use the complement: "at least one black" is the opposite of "both white."
2. $P(\text{both white}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ (one white sock is gone before the second draw).
3. $P(\text{at least one black}) = 1 - \frac{1}{3} = \frac{2}{3}$. ✓

Check: Counting directly needs three cases (black then white, white then black, black then black): $\frac{4}{10} \cdot \frac{6}{9} + \frac{6}{10} \cdot \frac{4}{9} + \frac{4}{10} \cdot \frac{3}{9} = \frac{24 + 24 + 12}{90} = \frac{60}{90} = \frac{2}{3}$.

Quick Check: Probability Rules 🎯

Compute the Probability 🧮 (enter decimals)

1. A spinner has 8 equal sections numbered 1 through 8. What is the probability of landing on a prime number?

2. A fair coin is flipped 3 times. What is the probability of getting heads all 3 times?

3. Events A and B are independent, with $P(A) = 0.6$ and $P(B) = 0.5$. What is $P(A \text{ and } B)$?

ACT-Style Practice

Two dice = 36 ordered outcomes. Treat the dice as a first die and a second die: (2, 5) and (5, 2) are different, equally likely outcomes. So a sum of 7 has 6 ways out of 36 ($\frac{6}{36} = \frac{1}{6}$), not 1 way out of 11 possible sums.

ACT Tip: Before computing, underline the key word — not, or, and, at least one — and match it to its rule from the table above.

ACT-Style Questions: Compound Events 📋

Key Takeaways

• $P = \frac{\text{favorable}}{\text{total}}$ for equally likely outcomes; every probability lies between 0 and 1.
• Complement: $P(\text{not } A) = 1 - P(A)$. Use it for "not," "neither," and "at least one."
• Addition rule: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. Mutually exclusive events have no overlap to subtract.
• Independent events: $P(A \text{ and } B) = P(A) \times P(B)$.
• Without replacement: multiply, but reduce the counts after each draw.
• Sort overlapping groups into both / only A / only B / neither before you divide.

🧩 Combinations and Permutations

Part 5 of 7 — Does Order Matter? Counting Selections and Using Them in Probability

Part 3 counted outcomes by filling slots. This part handles the case where you select some items from a larger group, and the one question that decides everything: does the order of the selection matter?

The Order Test

Ask: if I swap two of the chosen items, do I get a different outcome?

Key signal words: roles, ranks, positions, arrangements, codes → order matters. Groups, teams, committees, sets, selections with no distinct roles → order does not matter.

Permutations: Ordered Selections

The number of ways to choose and arrange $r$ items from $n$ different items is

$P(n, r) = \frac{n!}{(n - r)!} = n \times (n - 1) \times \cdots \text{ (} r \text{ factors)}$

Example: gold, silver, and bronze among 8 runners: $P(8, 3) = 8 \times 7 \times 6 = 336$. This is just the slot method — 8 choices, then 7, then 6.

Combinations: Unordered Selections

The number of ways to choose $r$ items from $n$ when order does not matter is

$C(n, r) = \binom{n}{r} = \frac{n!}{r!\,(n - r)!} = \frac{P(n, r)}{r!}$

Why divide by $r!$? Each group of $r$ items appears $r!$ times in the ordered count, once for each way of arranging it. A 3-person committee from 7 volunteers: $\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35$.

Symmetry shortcut: $C(n, r) = C(n, n - r)$. Choosing 7 questions to answer out of 10 is the same as choosing the 3 to skip: $C(10, 7) = C(10, 3) = 120$.

Handshakes and pairs: if each of 8 people shakes hands with every other person once, the number of handshakes is $C(8, 2) = \frac{8 \times 7}{2} = 28$. Dividing by 2 removes the double count (A with B is the same handshake as B with A).

Most ACT-approved calculators have nPr and nCr keys, but it is often faster to compute small cases by hand.

Selections from Separate Groups

When you choose from two groups at once, count each group's choices and multiply:

$\text{2 of 6 boys and 2 of 5 girls} = C(6, 2) \times C(5, 2) = 15 \times 10 = 150$

Mixed Roles

Some selections have one special role inside an otherwise equal group. A 3-person team from 8 people with one designated captain: choose the captain (8 ways), then the other 2 members ($C(7, 2) = 21$): $8 \times 21 = 168$.

Arrangements with Repeated Letters

To arrange letters when some repeat, divide by the factorial of each repeat count: BOOK has $\frac{4!}{2!} = 12$ arrangements, because swapping the two O's does not create a new word.

Combinations in Probability

For "choose a group at random" probability questions, both the numerator and denominator are combinations:

$P = \frac{\text{number of favorable groups}}{\text{number of possible groups}}$

Example: 2 of 7 students (4 boys, 3 girls) are chosen at random. $P(\text{both girls}) = \frac{C(3, 2)}{C(7, 2)} = \frac{3}{21} = \frac{1}{7}$. The without-replacement method from Part 4 gives the same answer: $\frac{3}{7} \times \frac{2}{6} = \frac{1}{7}$.

Consistency rule: if the denominator counts unordered groups, the numerator must too. Mixing an ordered count with an unordered count gives an answer off by a factor of $r!$.

Worked Examples

Question: A club has 10 members. (a) In how many ways can a president, a vice president, and a treasurer be chosen? (b) In how many ways can a 3-member planning committee be chosen?

Solution:

1. (a) Roles differ, so order matters: $P(10, 3) = 10 \times 9 \times 8 = 720$.
2. (b) No roles, so order does not matter: $C(10, 3) = \frac{720}{3!} = \frac{720}{6} = 120$. ✓

Check: each committee of 3 people can be turned into officers in $3! = 6$ ways, which is exactly why the ordered count is 6 times larger.

Question: A group has 4 boys and 5 girls. Three people are chosen at random. What is the probability that exactly 2 of them are girls?

Solution:

1. Possible groups: $C(9, 3) = \frac{9 \times 8 \times 7}{6} = 84$.
2. Favorable groups: 2 of the 5 girls and 1 of the 4 boys: $C(5, 2) \times C(4, 1) = 10 \times 4 = 40$.
3. Probability: $\frac{40}{84} = \frac{10}{21}$. ✓

ACT trap: using only $C(5, 2) = 10$ in the numerator forgets that the third person must be a boy.

Quick Check: Order or No Order? 🎯

Compute the Count 🧮

1. How many 3-person groups can be chosen from 6 people?

2. In how many ways can a first-place and a second-place winner be chosen from 6 finalists?

3. A sundae comes with any 4 different toppings chosen from 6. How many topping combinations are possible?

ACT-Style Practice

ACT Tip: Find the ordered count with slots first. If order doesn't matter, divide by $r!$ — that one extra step is the whole difference between the two formulas.

ACT-Style Questions: Selections and Probability 📋

Key Takeaways

• The order test: swap two chosen items. A different outcome → permutation; the same outcome → combination.
• $P(n, r) = \frac{n!}{(n - r)!}$ (slots: $n$, then $n - 1$, …, for $r$ slots). $C(n, r) = \frac{P(n, r)}{r!}$.
• $C(n, r) = C(n, n - r)$; pairs from $n$ people: $C(n, 2) = \frac{n(n - 1)}{2}$.
• Separate groups: multiply the combinations for each group. One special role: choose that role first.
• Repeated letters: divide $n!$ by the factorial of each repeat count.
• Group probability: $\frac{\text{favorable groups}}{\text{possible groups}}$, counting both the same way (both unordered or both ordered).

🗂️ Two-Way Tables and Conditional Probability

Part 6 of 7 — Choosing the Right Denominator

Two-way tables are among the most common data displays on the ACT. The arithmetic is easy — one division — but the answer choices are built so that dividing by the wrong total always produces one of them. This part is about picking the right denominator every time.

Anatomy of a Two-Way Table

• Inner cells count people in two categories at once (18 students are in Grade 9 and walk).
• Row and column totals (the margins) count one category (56 students ride the bus).
• The grand total (120) counts everyone.

Three Kinds of Probability

The last row is conditional probability: the condition shrinks the group you are choosing from.

$P(A \mid B) = \frac{\text{number in both } A \text{ and } B}{\text{number in } B}$

Read $P(A \mid B)$ as "the probability of $A$, given $B$."

Spotting the Condition

The condition is whatever the question tells you is already known. Look for phrases like:

• "If a senior is chosen…" → denominator = seniors
• "Given that the student rides the bus…" → denominator = bus riders
• "A student who plays a sport is chosen…" → denominator = athletes
• "Of the students who walk, what fraction…" → denominator = walkers
• "What percent of Grade 10 students…" → denominator = Grade 10

The condition order matters. $P(\text{girl} \mid \text{plays a sport})$ divides by athletes; $P(\text{plays a sport} \mid \text{girl})$ divides by girls. The numerator is the same cell, but the answers differ. Reversing the condition is the most common trap on these questions.

"Or" in a Table

For P(Grade 9 or car), add the Grade 9 total and the car total, then subtract the cell counted in both:

$\frac{60 + 32 - 12}{120} = \frac{80}{120} = \frac{2}{3}$

Completing a Table

Many questions leave cells blank. Every row and column must add to its total, so fill in whatever you can from those sums before answering.

(Bold cells were found by subtraction: $80 - 50$, $90 - 50$, then $70 - 40$.)

Conditional Probability Without a Table

Overlapping-group problems from Part 4 work the same way. If 25 of 60 members swim, 30 run, and 10 do both, then

$P(\text{swims} \mid \text{runs}) = \frac{10}{30} = \frac{1}{3}$

because the condition "runs" leaves only the 30 runners.

When information comes as percents of percents ("60% of customers are adults, and 30% of adults prefer streaming"), build a table for a convenient total such as 1,000 people, fill it with counts, and then divide.

Are Two Variables Related?

Compare the conditional rates, not the raw counts. If 12 of 30 left-handed students (40%) and 48 of 120 right-handed students (40%) wear glasses, the rates are equal, so glasses and handedness appear independent in this group — even though 48 is much larger than 12. In general, $A$ and $B$ are independent when $P(A \mid B) = P(A)$.

Worked Examples

Question: A survey asked 200 students whether they support a later school start.

Find (a) the probability that a randomly chosen senior said Yes, and (b) the probability that a randomly chosen Yes-voter is a senior.

Solution:

1. (a) The condition is "senior," so divide by the 80 seniors: $\frac{63}{80}$.
2. (b) The condition is "said Yes," so divide by the 108 Yes votes: $\frac{63}{108} = \frac{7}{12}$. ✓

ACT trap: $\frac{63}{200}$ answers neither question — it is P(senior and Yes).

Question: Of a streaming service's customers, 60% are adults and 40% are teens. 30% of adults and 70% of teens prefer watching on a phone. If a customer who prefers a phone is chosen at random, what is the probability that the customer is a teen?

Solution:

1. Imagine 1,000 customers: 600 adults and 400 teens.
2. Phone fans: $0.30 \times 600 = 180$ adults and $0.70 \times 400 = 280$ teens, so 460 in all.
3. Given "prefers a phone," divide by 460: $\frac{280}{460} = \frac{14}{23}$. ✓

ACT trap: 0.70 is P(phone | teen), the reverse condition.

Quick Check: Pick the Denominator 🎯

Complete the Table 🧮

A school surveyed 100 students about playing an instrument. Some cells are blank.

1. How many seniors play an instrument?

2. How many students in all do not play an instrument?

3. What percent of seniors play an instrument? (Enter a number only.)

ACT-Style Practice

Use the Walk / Bus / Car table from the lesson (Grade 9: 18, 30, 12; Grade 10: 14, 26, 20; totals 32, 56, 32, 120).

ACT Tip: Before you divide, write the denominator in words ("all students," "car riders," "Grade 10"). Then find that total in the table.

ACT-Style Questions: Conditions and Relationships 📋

Key Takeaways

• No condition: divide by the grand total. "And": one inner cell over the grand total.
• Conditional ("if," "given," "of those who," "a student who…"): divide by the total of the condition's row or column. $P(A \mid B) = \frac{\text{both}}{\text{number in } B}$.
• $P(A \mid B)$ and $P(B \mid A)$ share a numerator but not a denominator — check which group is known.
• "Or" in a table: row total + column total − shared cell.
• Fill blank cells using row and column sums; turn percent information into a table of counts (try 1,000 people).
• Compare rates, not raw counts, to decide whether two variables are related.

🎯 Expected Value and Mixed Review

Part 7 of 7 — Expected Value, Sampling, and Integrated ACT Problems

This final part adds the last major idea in the unit — expected value — and a short look at how data are collected. Then it puts all seven parts together, because ACT statistics questions often chain two skills: find a missing value, then a median; count the groups, then a probability.

Expected Value

The expected value of a random quantity is its long-run average: what you would get per trial, on average, over many repetitions.

$E(X) = x_1 p_1 + x_2 p_2 + \cdots + x_n p_n$

Multiply each possible value by its probability, then add.

From a probability distribution table:

$E(X) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7$

Things to notice:

• The probabilities in a distribution must add to 1. If one is missing, find it by subtraction before computing the expected value.
• The expected value does not have to be a possible outcome — no household has 1.7 cars.
• It is not the plain average of the values (1.5 here) and not the most likely value (2 here). It weights each value by how likely it is.

Equally likely outcomes: the expected value is just the mean of the outcomes. For a fair die, $E = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$.

Games and Net Gain

For a game with a cost to play, find the expected winnings, then subtract the cost:

$\text{expected net gain} = E(\text{winnings}) - \text{cost}$

A game costs 3 points to play and pays 10 points with probability $\frac{1}{4}$ (nothing otherwise). Expected winnings: $10 \cdot \frac{1}{4} = 2.5$ points; expected net gain: $2.5 - 3 = -0.5$ points per play. A negative value means the player loses on average. A game is fair when the expected net gain is 0.

Expected Counts

If each of $n$ independent trials succeeds with probability $p$, the expected number of successes is

$n \times p$

With a 4% defect rate, a batch of 250 items is expected to contain $250 \times 0.04 = 10$ defective items. A player who makes 80% of free throws is expected to make $15 \times 0.8 = 12$ of 15 shots.

Collecting Data: Random Samples

The ACT may ask which survey method gives the most reliable estimate for a population. The best answer is the one that gives every member of the population an equal chance of being chosen.

A larger sample gives a more precise estimate only if it is also chosen randomly; a huge biased sample is still biased.

Choosing the Tool: A Unit Map

Test-Day Habits for This Unit

1. Sort before you find a median. Every time.
2. Convert means to totals when a value is missing, added, removed, or replaced.
3. Name the denominator before dividing in any probability problem.
4. Eliminate impossible probabilities (below 0 or above 1) immediately.
5. Check whether order matters before choosing a counting rule.
6. Use a quick estimate to test your answer: a combined mean must lie between the group means; "at least one" must be at least as large as each single probability.

Worked Examples

Question: A random variable $X$ takes the values 1, 2, 3, and 4 with probabilities 0.2, 0.35, $p$, and 0.15. What is $E(X)$?

Solution:

1. Probabilities add to 1: $p = 1 - (0.2 + 0.35 + 0.15) = 0.3$.
2. $E(X) = 1(0.2) + 2(0.35) + 3(0.3) + 4(0.15) = 0.2 + 0.7 + 0.9 + 0.6 = 2.4$. ✓

ACT trap: Skipping the missing term gives 1.5, and averaging the values 1 through 4 gives 2.5 — neither uses all the probabilities.

Question: The data set 5, 8, 13, $x$ has a mean of 10. What is the median of the data set?

Solution:

1. A mean of 10 for 4 values means a total of 40, so $x = 40 - (5 + 8 + 13) = 14$.
2. In order: 5, 8, 13, 14. The median is $\frac{8 + 13}{2} = 10.5$. ✓

ACT trap: 10 is the mean and 14 is $x$; the question asks for a third quantity. Always reread what is asked after finishing step 1.

Quick Check: Expected Value 🎯

Expected Values 🧮

1. What is the expected value of one roll of a fair six-sided die?

2. A spinner has 4 equal sections labeled 2, 4, 6, and 12. What is the expected value of one spin?

3. A player makes 80% of her free throws. How many makes are expected in 15 attempts?

ACT-Style Practice: Mixed Set

ACT Tip: On multi-step questions, write the intermediate result (a total, a missing value, a count) next to the problem. The wrong answers are often exactly those intermediate numbers.

ACT-Style Questions: Integrated Review 📋

Key Takeaways

• Expected value: $E(X) = \sum x \cdot P(x)$. Probabilities must add to 1; find any missing one first.
• Expected value is a long-run average — not the most likely value and not the plain average of the outcomes.
• Games: expected net gain = expected winnings − cost. Negative means a loss on average; zero means fair.
• Expected count: $n \times p$.
• Representative samples come from random selection out of the whole population; volunteers, convenience samples, and single groups are biased.
• On integrated problems, write down each intermediate result — wrong answer choices are usually those numbers.