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Probability, Random Variables, and Probability Distributions

Unit 2 of AP Statistics, worth 15–25% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.

Probability rules, conditional probability and independence, random variables, binomial and geometric distributions, sampling distributions.

How this unit is tested

Start by sorting every problem into one of two worlds: probability rules for events (addition, complement, multiplication, conditional, independence), or random variables and their distributions (discrete RVs, binomial, geometric, sampling distributions). Reading the question type tells you which formulas are even in play. For probability-rule problems, draw a two-way table, Venn diagram, or tree diagram before writing any formula — most errors come from skipping this step and guessing which numbers to combine. Always ask whether events are mutually exclusive (use the simple addition rule) or overlapping (subtract the intersection), and whether they are independent (multiply straight probabilities) or dependent (use the conditional multiplication rule). Never assume independence; verify it by computing P(A|B) and comparing to P(A), or by checking whether the problem states it. For random variables, first decide discrete or continuous, then check whether the situation fits BINS (binomial: binary, independent, number fixed, same probability) or a geometric setting (binary, independent, same probability, counting trials until first success). If neither fits, you're dealing with a general discrete distribution and must compute expected value and variance directly from the probability table. For sampling distributions, identify whether you're tracking a sample proportion (p̂) or a sample mean (x̄), state its center and spread using the correct formula, and check the relevant condition for normality (large counts np≥10 and n(1−p)≥10 for p̂; n≥30 or a stated normal population for x̄ via the Central Limit Theorem) before claiming the distribution is approximately normal.

What you have to know

General Addition Rule
P(A or B) = P(A) + P(B) − P(A and B). If A and B are mutually exclusive, P(A and B) = 0, so P(A or B) = P(A) + P(B).
Conditional Probability and General Multiplication Rule
P(B | A) = P(A and B) / P(A), provided P(A) > 0. Equivalently, P(A and B) = P(A) · P(B | A).
Independence
Events A and B are independent if P(B | A) = P(B) (equivalently P(A | B) = P(A)), which is the same as P(A and B) = P(A) · P(B). Independence must be verified, not assumed.
Binomial Distribution (BINS)
X is binomial with parameters n and p if trials are Binary, Independent, a fixed Number n, with the Same probability p of success. Then P(X = k) = C(n,k) p^k (1−p)^(n−k), mean μ = np, and standard deviation σ = sqrt(np(1−p)).
Geometric Distribution
X counts the number of trials up to and including the first success, with binary, independent trials and constant success probability p. Then P(X = k) = (1−p)^(k−1) p, mean μ = 1/p, and standard deviation σ = sqrt((1−p)/p^2).
Sampling Distributions and the Central Limit Theorem
The sampling distribution of p̂ has mean p and standard deviation sqrt(p(1−p)/n), approximately normal when np ≥ 10 and n(1−p) ≥ 10. The sampling distribution of x̄ has mean μ and standard deviation σ/sqrt(n); by the Central Limit Theorem it is approximately normal for large n (commonly n ≥ 30) regardless of the population's shape.

14 practice questions

  1. At a school, 40% of students play a sport, 30% play an instrument, and 15% do both. What is the probability a randomly chosen student plays a sport or an instrument?
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    Answer. 0.55

    Use the general addition rule since the events overlap: P(sport or instrument) = P(sport) + P(instrument) − P(both) = 0.40 + 0.30 − 0.15 = 0.55. Subtracting the overlap avoids double-counting students who do both.
  2. In a survey of 200 students, 120 are girls and 80 are boys. Of the girls, 90 passed an exam; of the boys, 50 passed. What is P(passed | girl)?
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    Answer. 0.75

    Conditional probability restricts attention to the given group. P(pass | girl) = (number of girls who passed) / (total girls) = 90/120 = 0.75.
  3. Using the same survey (200 students, 120 girls with 90 passing, 80 boys with 50 passing), are 'being a girl' and 'passing the exam' independent? Justify with numbers.
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    Answer. No, they are not independent

    Overall P(pass) = 140/200 = 0.70. But P(pass | girl) = 90/120 = 0.75. Since 0.75 ≠ 0.70, knowing a student is a girl changes the probability of passing, so the events are dependent.
  4. A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability both are red?
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    Answer. 5/14 (approximately 0.357)

    Use the general multiplication rule for dependent events: P(both red) = P(first red) × P(second red | first red) = (5/8)(4/7) = 20/56 = 5/14.
  5. A discrete random variable X has distribution: P(X=0)=0.1, P(X=1)=0.3, P(X=2)=0.4, P(X=3)=0.2. Find E(X).
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    Answer. 1.7

    E(X) = Σ x·P(x) = 0(0.1) + 1(0.3) + 2(0.4) + 3(0.2) = 0 + 0.3 + 0.8 + 0.6 = 1.7.
  6. For the same distribution (P(X=0)=0.1, P(X=1)=0.3, P(X=2)=0.4, P(X=3)=0.2, with μ = 1.7), find the standard deviation of X.
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    Answer. 0.9

    Compute Var(X) = Σ(x−μ)²P(x) = (0−1.7)²(0.1)+(1−1.7)²(0.3)+(2−1.7)²(0.4)+(3−1.7)²(0.2) = 0.289+0.147+0.036+0.338 = 0.81. Then SD(X) = sqrt(0.81) = 0.9.
  7. If X has mean 1.7 and standard deviation 0.9, and Y = 3X + 5, what are the mean and standard deviation of Y?
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    Answer. Mean = 10.1, SD = 2.7

    For a linear transformation, E(aX+b) = a·E(X)+b and SD(aX+b) = |a|·SD(X). So E(Y) = 3(1.7)+5 = 10.1, and SD(Y) = 3(0.9) = 2.7. Adding a constant shifts the mean but never changes the spread.
  8. Test 1 scores have mean 75 and standard deviation 8. Test 2 scores (independent of Test 1) have mean 80 and standard deviation 6. Find the mean and standard deviation of the total score T = Test1 + Test2.
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    Answer. Mean = 155, SD = 10

    Means always add: E(T) = 75+80 = 155. For independent random variables, variances add (not standard deviations): Var(T) = 8² + 6² = 64+36 = 100, so SD(T) = sqrt(100) = 10.
  9. Which scenario describes a binomial random variable?
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    Answer. The number of successes in 10 independent trials, each with probability of success 0.3

    This satisfies BINS: binary outcomes, independent trials, a fixed number of trials (10), and constant probability (0.3) each time. The other options involve counting trials until a success (geometric) or drawing without replacement, which violates independence.
  10. Historically, 25% of emails received by a company are spam. In a random sample of 10 emails, what is P(exactly 2 are spam)?
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    Answer. Approximately 0.282

    X ~ Binomial(n=10, p=0.25). P(X=2) = C(10,2)(0.25)²(0.75)^8 = 45 × 0.0625 × 0.1001 ≈ 0.282.
  11. For the spam-email scenario (n = 10, p = 0.25), find the mean and standard deviation of the number of spam emails in the sample.
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    Answer. Mean = 2.5, SD ≈ 1.369

    For a binomial random variable, μ = np = 10(0.25) = 2.5, and σ = sqrt(np(1−p)) = sqrt(10 × 0.25 × 0.75) = sqrt(1.875) ≈ 1.369.
  12. A quality inspector tests items one at a time; the probability an item is defective is 0.1, and items are independent. What is the probability the first defective item is the 5th one tested?
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    Answer. Approximately 0.0656

    This is a geometric setting with p = 0.1. P(X=5) = (1−p)^(5−1) × p = (0.9)^4 × 0.1 = 0.6561 × 0.1 ≈ 0.0656. The first 4 items must be non-defective, and the 5th must be defective.
  13. A population has proportion p = 0.6 with a characteristic of interest. A random sample of n = 50 is taken. Find the mean and standard deviation of the sampling distribution of p̂, and check whether it is approximately normal.
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    Answer. Mean = 0.6, SD ≈ 0.0693; approximately normal since conditions are met

    The sampling distribution of p̂ has mean p = 0.6 and SD = sqrt(p(1−p)/n) = sqrt(0.6×0.4/50) = sqrt(0.0048) ≈ 0.0693. Checking the normal condition: np = 30 ≥ 10 and n(1−p) = 20 ≥ 10, so the sampling distribution is approximately normal.
  14. A population of light bulb lifetimes is strongly skewed with mean 1000 hours and standard deviation 200 hours. A random sample of n = 64 bulbs is selected and the sample mean x̄ is computed. Describe the shape, mean, and standard deviation of the sampling distribution of x̄.
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    Answer. Approximately normal, mean = 1000 hours, SD = 25 hours

    The sampling distribution of x̄ always has mean μ = 1000 and SD = σ/sqrt(n) = 200/sqrt(64) = 200/8 = 25. Even though the population is skewed, the Central Limit Theorem guarantees the sampling distribution of x̄ is approximately normal because the sample size (n = 64) is large.

What people get wrong

  1. Writing P(A|B) when the problem gives P(B|A), or vice versa — conditional probability is not symmetric. Always identify which event is the 'given' condition before dividing.
  2. Assuming two events are independent because they seem unrelated in context. Instead, compute P(A), P(B), and P(A and B) (or P(B|A)) and check the independence equation numerically.
  3. Using the simple addition rule P(A)+P(B) for events that overlap. Instead, always subtract P(A and B) unless the problem explicitly states or a Venn diagram confirms the events are mutually exclusive.
  4. Applying the binomial formula to a without-replacement scenario drawn from a small population, where trials are not actually independent. Instead, check that the population is much larger than the sample (a common rule of thumb is at least 10 times the sample size) before treating it as binomial.
  5. Adding standard deviations directly when combining independent random variables. Instead, add variances first (Var(X+Y) = Var(X) + Var(Y) for independent X, Y), then take the square root to get the standard deviation of the sum.
  6. Declaring a sampling distribution 'approximately normal' without checking the relevant condition. Instead, verify np ≥ 10 and n(1−p) ≥ 10 for a proportion, or confirm a large enough sample size (or a stated normal population) before invoking the Central Limit Theorem for a mean.

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