AP Statistics Unit 4: Probability Random Variables and Distributions — Free Review Games.

The complement rule: P(A) + P(A') = 1.

For independent events: P(A and B) = P(A)*P(B) = 0.15.

Discrete random variables have countable (often integer) possible values.

All probabilities must sum to exactly 1.

For mutually exclusive events, there is no overlap: P(A or B) = P(A) + P(B).

The mean of X, denoted E(X), is the expected value.

E(X) = np = 10(0.4) = 4.

The general addition rule accounts for overlap between events.

If A and B are independent, knowing B does not change the probability of A.

The geometric distribution counts the trials needed to get the first success.

\(SD = \sqrt{np(1-p)} = \sqrt{20 \cdot 0.3 \cdot 0.7} = \sqrt{4.2}\).

P(A|B) = P(A and B) / P(B), the probability of A given B occurred.

Independence means the joint probability equals the product of individual probabilities.

Binomial requires fixed n trials. Continuing until first success is geometric.

E(X+Y) = E(X) + E(Y) = 5 + 3 = 8 regardless of independence.

By definition, probabilities must satisfy \(0 \leq P(E) \leq 1\). A probability of \(0\) means the event is impossible, and \(1\) means it is certain — both are valid values. Choice C incorrectly excludes \(0\) and \(1\). Choice A permits negative values, which have no meaning in probability. Choice D has no upper bound.

Mutually exclusive events cannot both occur at the same time, so \(P(A \cap B) = 0\). Choice A, \(P(A) \cdot P(B)\), is the multiplication rule for independent events — independence and mutual exclusivity are different concepts. Choice B gives \(P(A \cup B)\) for mutually exclusive events, not \(P(A \cap B)\).

A discrete random variable takes countable values. The number of heads in 10 flips can only be \(0, 1, 2, \ldots, 10\) — a finite, countable set. Height, time, and weight are continuous random variables because they can take any value within an interval, making them uncountably infinite.

In \(B(n, p)\), \(p\) is the probability of success on each individual trial, which must remain constant. The parameter \(n\) is the number of trials. The expected number of successes is \(E(X) = np\), which is derived from the parameters — it is not a parameter itself.

The complement rule states \(P(A^c) = 1 - P(A)\). Therefore \(P(A^c) = 1 - 0.72 = 0.28\). The events A and \(A^c\) are mutually exclusive and exhaustive, so their probabilities must sum to \(1\). Choice A simply restates \(P(A)\), which would imply \(P(A) + P(A^c) = 1.44 > 1\), an impossibility.

The general multiplication rule is \(P(A \cap B) = P(A \mid B) \cdot P(B)\), valid for any events A and B. Choice A is a special case that applies only when A and B are independent, since independence means \(P(A \mid B) = P(A)\). Choice D has the wrong sign and produces a non-sensical result for most events.

A random variable is a function that assigns a numerical value to each outcome in a sample space. For example, if \(X\) = number of heads in two coin flips, then \(X\) assigns \(0\) to TT, \(1\) to HT or TH, and \(2\) to HH. Its value is not fixed — it depends on the random outcome. Choice B describes the expected value \(E(X)\), which is a single number derived from the random variable.

For a binomial distribution, \(\text{Var}(X) = np(1-p)\). Here \(\text{Var}(X) = 15 \cdot 0.2 \cdot 0.8 = 2.4\). A common error is confusing variance with mean: \(E(X) = np = 15(0.2) = 3.0\), which is choice A. Always use \(np(1-p)\) for variance, not just \(np\).

The mean of a geometric distribution is \(E(X) = \frac{1}{p}\). With \(p = 0.25\), \(E(X) = \frac{1}{0.25} = 4\). Intuitively, if each trial has a 25% chance of success, you expect to need about 4 trials before the first success. Choice A is simply \(p\) itself, not the expected value.

The expected value is \(E(X) = \sum x \cdot P(X=x)\). Computing: \(E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7\). Choice A (\(2.5\)) is the simple arithmetic midpoint of 1 and 4, which ignores the probability weights. Expected value is a weighted average, not a simple average of the possible values.

For independent events, \(P(A \cap B) = P(A) \cdot P(B) = 0.4 \times 0.3 = 0.12\). Choice A (\(0.70\)) is \(P(A) + P(B)\), which would equal \(P(A \cup B)\) only if the events were mutually exclusive. Choice B (\(0.58\)) is \(P(A \cup B) = P(A)+P(B)-P(A \cap B)=0.4+0.3-0.12=0.58\), a different quantity.

For a linear transformation, \(\text{Var}(aX + b) = a^2 \cdot \text{Var}(X)\). The constant \(b\) shifts the entire distribution but does not change its spread, so it contributes \(0\) to the variance. Therefore \(\text{Var}(3X+5) = 3^2 \cdot 4 = 9 \cdot 4 = 36\). Choice A (\(12\)) comes from multiplying by \(a=3\) instead of \(a^2=9\).

The binomial probability formula is \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\). The binomial coefficient \(\binom{n}{k}\) counts the number of distinct ways to arrange \(k\) successes among \(n\) trials. Choice A omits this coefficient, giving the probability of just one specific ordered sequence of \(k\) successes. Choice C swaps the exponents on \(p\) and \((1-p)\).

Use the complement rule: \(P(X \geq 1) = 1 - P(X = 0)\). Since \(P(X=0) = \binom{4}{0}(0.3)^0(0.7)^4 = (0.7)^4 = 0.2401\), we get \(P(X \geq 1) = 1 - 0.2401 = 0.7599\). This complement approach is far more efficient than summing \(P(X=1)+P(X=2)+P(X=3)+P(X=4)\) separately. Choice A (\(0.30\)) is simply \(p\), which is only \(P(X=1)\) in a single trial.

For independent random variables, \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y) = 9 + 16 = 25\). Variances always add for independent variables, even when subtracting, because subtracting a random quantity still increases unpredictability. Choice A (\(7\)) is the incorrect result of subtracting variances, which does not reflect how variability actually combines.

Conditional probability is defined as \(P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.4} = 0.75\). Since \(P(A \mid B) = 0.75 \neq P(A) = 0.6\), knowing B occurred does change the probability of A, confirming A and B are not independent. Choice C confuses \(P(A \cap B)\) with \(P(A \mid B)\) — these are different quantities.

For a geometric distribution, \(P(X = k) = (1-p)^{k-1} \cdot p\). The first success on trial 3 requires exactly two failures followed by one success: \((0.6)^2(0.4) = 0.144\). Choice C resembles the binomial formula and incorrectly counts arrangements — in the geometric setting there is only one valid sequence (FF...FS), so no combinatorial coefficient is needed.

The mean of a geometric distribution is \(E(X) = \frac{1}{p}\). Setting \(\frac{1}{p} = 3\) and solving gives \(p = \frac{1}{3}\). Choice C (\(3\)) confuses \(E(X)\) with \(p\) by returning the mean as the probability. Choice A (\(0.25\)) would give \(E(X) = 4\), not \(3\).

Use the complement: \(P(X \geq 2) = 1 - P(X=0) - P(X=1)\). \(P(X=0) = (0.7)^5 = 0.16807\). \(P(X=1) = \binom{5}{1}(0.3)^1(0.7)^4 = 5(0.3)(0.2401) = 0.36015\). Therefore \(P(X \geq 2) = 1 - 0.16807 - 0.36015 \approx 0.4718\). Choice B (\(0.3087\)) is \(P(X=1)\) alone, a common error when students forget to include \(P(X=0)\) in the complement.

Use the variance identity: \(\text{Var}(X) = E(X^2) - [E(X)]^2\). Solving for \(E(X^2)\) gives \(E(X^2) = \text{Var}(X) + [E(X)]^2 = 3 + 4^2 = 3 + 16 = 19\). Choice C (\(16\)) is \([E(X)]^2\) alone, ignoring the variance term. This reflects the key fact that \(E(X^2) \neq [E(X)]^2\) whenever \(\text{Var}(X) > 0\).

The normal approximation to the binomial is valid when \(np \geq 10\) and \(n(1-p) \geq 10\). These conditions ensure the binomial distribution is sufficiently symmetric with adequate probability mass away from 0 and \(n\). Choice A (\(n \geq 30\)) is the Central Limit Theorem rule of thumb for sample means, not the binomial approximation. Crucially, the adequacy depends on both \(n\) and \(p\) together, not \(n\) alone.

If A and B are mutually exclusive, \(P(A \cap B) = 0\). For independence, we would need \(P(A \cap B) = P(A) \cdot P(B)\). Since \(P(A) > 0\) and \(P(B) > 0\), their product \(P(A) \cdot P(B) > 0 \neq 0\). This is a contradiction, so mutually exclusive events with positive probabilities cannot be independent. Choice A is therefore always false in this situation.

By the memoryless property of the geometric distribution, \(P(X=5 \mid X>3) = P(X=2)\). After 3 failures, the process resets — the probability of success on the very next trial is still \(p\), so the first success occurring on the 5th overall trial is equivalent to success on the 2nd trial from the new starting point: \(P(X=2) = (0.8)^1(0.2) = (0.8)(0.2)\). You can verify directly: \(\frac{P(X=5)}{P(X>3)} = \frac{(0.8)^4(0.2)}{(0.8)^3} = (0.8)(0.2)\).

For independent \(X\) and \(Y\): \(\text{Var}(2X - 3Y + 1) = (2)^2\text{Var}(X) + (-3)^2\text{Var}(Y)\). The constant \(+1\) contributes zero variance. So \(\text{Var} = 4(4) + 9(1) = 16 + 9 = 25\). Choice A (\(13\)) results from using \(2(4) + 1(3) - 2 = 13\), incorrectly not squaring the coefficients. Note that expected value calculations are not needed here — variance of a sum involves squared coefficients only.

Apply Bayes' Theorem. Let S = studied. First find \(P(\text{Pass})\) using the law of total probability: \(P(\text{Pass}) = P(\text{Pass}\mid S)P(S) + P(\text{Pass}\mid S^c)P(S^c) = (0.9)(0.6) + (0.4)(0.4) = 0.54 + 0.16 = 0.70\). Then \(P(S \mid \text{Pass}) = \frac{P(\text{Pass}\mid S) \cdot P(S)}{P(\text{Pass})} = \frac{0.54}{0.70} \approx 0.771\). Choice A (\(0.540\)) is \(P(\text{Pass} \cap S)\), not the conditional probability we need.

The complement rule states \(P(A^c) = 1 - P(A) = 1 - 0.65 = 0.35\). Since every outcome either belongs to \(A\) or not, \(P(A) + P(A^c) = 1\) always. Choice D, \(0.325\), is simply \(P(A)/2\), which has no probabilistic meaning here.

Numbers strictly less than 3 are 1 and 2 — two favorable outcomes out of 6 equally likely outcomes. So \(P(X < 3) = \frac{2}{6} = \frac{1}{3}\). A common error is to include 3 itself, giving \(\frac{3}{6} = \frac{1}{2}\) (choice C), but the question says strictly less than 3.

Mutually exclusive events cannot occur together, so \(P(A \cap B) = 0\). The general addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) therefore simplifies to \(P(A) + P(B)\). Choice B is the general (unsimplified) addition rule. Choice A is the multiplication rule for independent events — a completely different concept.

\(E(X) = \sum x \cdot P(X = x)\), a weighted average where each outcome is weighted by its probability. Choice A describes the mode, choice C describes the median, and choice D describes the standard deviation — all distinct measures from the expected value.

The four BINS conditions for a binomial distribution are: Binary outcomes, Independent trials, fixed Number of trials, and Same probability \(p\) on every trial. Choice D directly violates the last condition — a changing success probability disqualifies the binomial model. The other three choices correctly describe required binomial conditions.

A geometric distribution models the number of trials until the first success, where each trial has the same probability \(p\) of success and trials are independent. Choice A describes the binomial distribution. Choices C and D describe long-run averages and joint probabilities, neither of which defines a geometric random variable.

For a binomial random variable, \(E(X) = np = 20 \times 0.35 = 7\). Choice B, \(4.55\), is approximately \(\sqrt{np(1-p)}\), the standard deviation. Choice D, \(13\), equals \(n(1-p) = 20 \times 0.65\), which would be the expected number of failures rather than successes.

A fundamental probability axiom requires that all probabilities in a distribution sum to exactly 1: \(\sum P(X = x) = 1\). Expected value can be zero or negative (choice A is false). No single outcome must exceed probability \(0.5\) — for example, a fair die has each outcome at \(1/6\) (choice C is false). Variance can be less than 1 or even 0 for a constant random variable (choice D is false).

The conditional probability formula gives \(P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = 0.50\). A frequent error is to report \(P(A \cap B) = 0.2\) directly (choice C), confusing joint and conditional probability. Notice here that \(P(A \mid B) = P(A) = 0.5\), which confirms that \(A\) and \(B\) are independent events.

For a binomial random variable, \(\text{Var}(X) = np(1-p) = 15 \times 0.4 \times 0.6 = 3.6\). Choice A, \(6\), is \(E(X) = np = 15 \times 0.4\), not the variance. The variance formula multiplies the mean \(np\) by the additional factor \((1-p)\), which is easy to omit.

Using the binomial formula: \(P(X = 2) = \binom{5}{2}(0.2)^2(0.8)^3 = 10 \times 0.04 \times 0.512 = 0.2048\). Choice B, \(0.1024\), results from omitting the binomial coefficient \(\binom{5}{2} = 10\). The coefficient accounts for all possible orderings of 2 successes among 5 trials.

\(E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7\). Choice A, \(2.50\), is the simple unweighted average \(\frac{1+2+3+4}{4}\), which ignores the probabilities entirely. Expected value must weight each outcome by its probability.

The general addition rule gives \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.3 - 0.15 = 0.65\). Choice B, \(0.80\), results from forgetting to subtract \(P(A \cap B)\), which would double-count the overlapping region. The subtraction corrects for outcomes counted in both \(P(A)\) and \(P(B)\).

First find the variance: \(\text{Var}(X) = E(X^2) - [E(X)]^2 = 34 - 25 = 9\). Then \(\text{SD}(X) = \sqrt{9} = 3\). Choice B, \(9\), is the variance itself, not the standard deviation. Choice D, \(29\), incorrectly computes \(E(X^2) - E(X) = 34 - 5\) instead of subtracting \([E(X)]^2\).

The geometric probability formula is \(P(X = k) = (1-p)^{k-1} \cdot p\). For \(k = 4\): \(P(X = 4) = (0.75)^3 \times 0.25 = 0.421875 \times 0.25 \approx 0.1055\). Choice D, \(0.4219 \approx (0.75)^3\), forgets to multiply by \(p\). Choice C, \(0.0625 = (0.25)^2\), uses the wrong formula entirely.

Apply the law of total probability: \(P(D) = P(D \mid A) \cdot P(A) + P(D \mid B) \cdot P(B) = 0.03 \times 0.6 + 0.05 \times 0.4 = 0.018 + 0.020 = 0.038\). Choice B, \(0.040\), is the simple unweighted average of 3% and 5%, ignoring that machine A produces more items. Weighting by production share is essential.

By linearity of expectation: \(E(2X - Y + 4) = 2E(X) - E(Y) + 4 = 2(3) - 7 + 4 = 6 - 7 + 4 = 3\). Choice B, \(10\), comes from computing \(2(3) + 4\) and omitting \(-E(Y)\). Linearity of expectation holds for any random variables, independent or not.

Compute each term: \(P(X=0) = (0.7)^6 \approx 0.1176\), \(P(X=1) = 6(0.3)(0.7)^5 \approx 0.3025\), \(P(X=2) = 15(0.3)^2(0.7)^4 \approx 0.3241\). Summing: \(P(X \leq 2) \approx 0.1176 + 0.3025 + 0.3241 = 0.7443\). Choice D, \(0.3241\), is only \(P(X = 2)\). A cumulative probability requires adding all terms from \(X = 0\) through the target value.

First find \(P(B)\) using the law of total probability: \(P(B) = P(B \mid A)P(A) + P(B \mid A^c)P(A^c) = 0.6(0.3) + 0.2(0.7) = 0.18 + 0.14 = 0.32\). Then Bayes' theorem gives \(P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)} = \frac{0.18}{0.32} = 0.5625\). Choice D, \(0.18\), is \(P(A \cap B)\) — the joint probability, not the conditional.

For independent random variables, \(\text{Var}(aX + bY) = a^2\text{Var}(X) + b^2\text{Var}(Y)\). The sign of the coefficient does not matter because variance squares it: \(\text{Var}(2X - 3Y) = (2)^2(5) + (3)^2(3) = 20 + 27 = 47\). Choice B, \(29\), incorrectly uses coefficients \(2\) and \(3\) without squaring. Choice D is negative, which is impossible for any variance.

For a geometric distribution, \(P(X > k) = (1-p)^k\), because exceeding \(k\) trials means the first \(k\) trials are all failures. \(P(X > 3) = (0.8)^3 = 0.512\). Choice B, \(0.488 = 1 - 0.512\), is \(P(X \leq 3)\). Choice C, \(0.4096 = (0.8)^4\), is \(P(X > 4)\), using the wrong exponent.

For choice A: \(np = 50 \geq 10\) and \(n(1-p) = 50 \geq 10\) — both conditions satisfied. For choice B: \(np = 0.5 < 10\). For choice C: \(np = 1 < 10\). For choice D: \(np = 2 < 10\). When \(p\) is near 0 or 1, a very large \(n\) is needed. Only a moderate \(p\) near \(0.5\) allows smaller \(n\) to satisfy both conditions simultaneously.

\(P(X = 6) = \binom{8}{6}(0.7)^6(0.3)^2 = 28 \times 0.117649 \times 0.09 \approx 0.2965\). Choice B results from using \((0.7)^5\) instead of \((0.7)^6\), mismatching the exponent to the count. Always use \(p^k \cdot (1-p)^{n-k}\) with \(k = 6\) successes and \(n - k = 2\) failures.

Expand the expression: \(E[(X-1)^2] = E[X^2 - 2X + 1] = E(X^2) - 2E(X) + 1\). Since \(\text{Var}(X) = E(X^2) - [E(X)]^2\), we get \(E(X^2) = 3 + 4 = 7\). Therefore \(E[(X-1)^2] = 7 - 2(2) + 1 = 4\). Choice B, \(3\), is \(\text{Var}(X)\) itself — note that \(E[(X - \mu)^2] = \text{Var}(X)\) only when the shift equals \(E(X)\), but here the shift is 1, not \(E(X) = 2\).

\(P(\text{Exercise} \mid \text{Diet}) = \frac{P(\text{Exercise} \cap \text{Diet})}{P(\text{Diet})} = \frac{0.15}{0.30} = 0.50\). Choice C, \(0.45\), is \(P(\text{Exercise})\) — the unconditional probability, which ignores the given condition. Choice D, \(0.15\), is the joint probability \(P(\text{Exercise} \cap \text{Diet})\), not divided by \(P(\text{Diet})\) as the formula requires.