Statistics & Probability

Averages, data displays, counting, probability, two-way tables and expected value.

Worked Examples

<details> <summary><b>Example 1: The score needed for a target mean</b></summary>

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.

</details> <details> <summary><b>Example 2: Combining two groups</b></summary>

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.

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Worked Examples

<details> <summary><b>Example 1: Mean and median from a frequency table</b></summary>

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.

</details> <details> <summary><b>Example 2: Reading a box plot described in words</b></summary>

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.

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Worked Examples

<details> <summary><b>Example 1: A code with two different rules</b></summary>

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 $10 \times 10$.

</details> <details> <summary><b>Example 2: At least one repeat</b></summary>

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.

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Worked Examples

<details> <summary><b>Example 1: Overlapping groups and "neither"</b></summary>

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.

</details> <details> <summary><b>Example 2: Two draws without replacement</b></summary>

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): $\frac{4}{10} \cdot \frac{6}{9} + \frac{6}{10} \cdot \frac{4}{9} + \frac{4}{10} \cdot \frac{3}{9} = \frac{24 + 24 + 12}{90} = \frac{60}{90} = \frac{2}{3}$.

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Worked Examples

<details> <summary><b>Example 1: Officers versus a committee</b></summary>

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.

</details> <details> <summary><b>Example 2: A probability built from combinations</b></summary>

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.

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Worked Examples

<details> <summary><b>Example 1: Same cell, two different conditions</b></summary>

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).

</details> <details> <summary><b>Example 2: Percents into a table</b></summary>

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.

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Worked Examples

<details> <summary><b>Example 1: Expected value with a missing probability</b></summary>

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.

</details> <details> <summary><b>Example 2: A two-step data problem</b></summary>

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.

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