Probability and Statistics: Chapter Test

A gift box is an unordered selection of $3$ pastries from $13$ distinct pastries, so this is a combination.

$\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = \frac{1716}{6} = 286$

Each of the three badge positions offers all $26$ letters again, since a used letter remains available for the next position, so the count is a power.

$26 \times 26 \times 26 = 26^3 = 17{,}576$

Part A and part B differ on two axes at once: order (unordered vs ordered) and repetition (none allowed vs allowed), so it is tempting to credit either one for the division. Isolate them with a counterfactual: suppose the badge instead drew $3$ DIFFERENT letters in order, with no repeats allowed, matching part A's "no repetition" while keeping part B's "ordered". That count is a permutation, built directly by the multiplication principle with no set being collapsed, so it needs no division at all, even though repetition is now forbidden just as in part A.

$P(26,3) = 26 \times 25 \times 24 = 15{,}600 \quad \text{(ordered, no repeats, still no division)}$

That settles it: repetition was never the reason part A divided. Part A divides by $3!$ because it is UNORDERED, a combination, so every set of $3$ pastries was counted once for each of its $3! = 6$ orderings.

$286 \times 3! = 1716 = P(13,3)$

Repetition is a separate question, and it decides something else entirely: whether the stage count in an ORDERED build shrinks with each pick ($P(n,k)$, no repeats) or stays constant ($n^k$, repeats allowed). The real badge, with repetition allowed, uses the constant power $26^3$, but that is a fact about repetition, not about order, and it never involves a division either way.

The doubles are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$, so $|A|=6$. The rolls summing to at least $9$ are $(3,6),(4,5),(5,4),(6,3),(4,6),(5,5),(6,4),(5,6),(6,5),(6,6)$, so $|B|=10$. The only doubles among those are $(5,5)$ and $(6,6)$, so $|A\cap B| = 2$.

$P(A \cup B) = \frac{6}{36} + \frac{10}{36} - \frac{2}{36} = \frac{14}{36} = \frac{7}{18}$

"Neither" is the complement of "a double or a sum of at least $9$," so subtract part A's union probability from $1$.

$P(\text{neither}) = 1 - \frac{7}{18} = \frac{11}{18}$

The classmate's claim would require $A$ and $B$ to share no outcome at all, but $(5,5)$ has a double and a sum of $10$, and $(6,6)$ has a double and a sum of $12$, so both outcomes belong to $A \cap B$. Since $A \cap B \neq \emptyset$, the two events overlap, which is precisely what "mutually exclusive" rules out.

Because the two events overlap, adding $P(A)+P(B)$ counts $(5,5)$ and $(6,6)$ once inside $P(A)$ and once again inside $P(B)$, so it overstates the union by exactly the intersection.

$P(A) + P(B) - P(A \cup B) = \frac{6}{36} + \frac{10}{36} - \frac{14}{36} = \frac{2}{36} = P(A \cap B)$

That is why the addition rule always subtracts the intersection: it corrects for exactly this kind of double counting whenever the events are not disjoint.

Since $195 = 200 - 5$ and $205 = 200 + 5$, both endpoints sit exactly $1$ standard deviation from the mean, which is precisely the rule's central band.

$68\%$

Standardize first: $z = \dfrac{209-200}{5} = 1.8$. The table gives the area to the LEFT, so the area to the right is what remains of the total.

$P(X>209) = 1 - \Phi(1.8) = 1 - 0.9641 = 0.0359$

The two figures answer related but not identical questions. The rule's $2.5\%$ estimates the tail beyond EXACTLY $2$ standard deviations above the mean (half of the $5\%$ left outside the $95\%$ central band). Part B's tail begins at $1.8$ standard deviations, a point CLOSER to the mean than $2\sigma$.

$1.8\sigma < 2\sigma \quad \Longrightarrow \quad \text{more area lies beyond } 1.8\sigma \text{ than beyond } 2\sigma$

A point nearer the center of the curve still has more of the curve's area ahead of it, so it makes sense that part B's exact tail, $3.59\%$, comes out a bit larger than the rule's rounded $2.5\%$ estimate for the more distant $2\sigma$ cutoff. The two figures are close because $1.8$ and $2$ are nearby standard-deviation distances, but they measure different cutoffs, so an exact match was never guaranteed.

LEVEL has $5$ letters with $2$ L's and $2$ E's, so tagging every letter and then dividing out the internal reorderings of the repeated letters gives the count.

$\frac{5!}{2!\,2!} = \frac{120}{4} = 30$

The three prizes are distinct ranks filled without repetition, so this is a permutation of $3$ tickets chosen from $20$.

$P(20,3) = 20 \times 19 \times 18 = 6840$

Part A's five letters include two identical L's and two identical E's. Treating those as temporarily distinct would overcount every visible tag by $2! \times 2! = 4$, once for each way the identical letters could be swapped among their own positions with no visible change, so dividing by $2!\,2!$ removes exactly that overcount.

$30 \times (2! \times 2!) = 120 = 5!$

Part B's three drawn tickets are already distinct numbers, since a raffle drum holds $20$ different tickets and none repeats among the three drawn. Swapping which specific tickets landed in which prize slot always changes the outcome (a different ticket in 1st place is a genuinely different result), so no two ways of listing the same three tickets ever collapse onto the same visible outcome. There is nothing indistinguishable to overcount, so the direct ordered count $20 \times 19 \times 18$ needs no correction.

The mean is the total divided by the count.

$\bar{x} = \frac{8+9+10+11+52}{5} = \frac{90}{5} = 18$

The data is sorted, and with $5$ values the median is the middle one, $10$. Four of the five tutors helped between $8$ and $11$ students, well below the mean of $18$, so the median is the honest description of a typical caseload here.

The mean rate is $\bar{x} = \frac{21+24+28+30+32}{5} = \frac{135}{5} = 27$. The deviations are $-6,-3,1,3,5$, which sum to $0$ as they must, and their squares are $36,9,1,9,25$, summing to $80$.

$\sigma^2 = \frac{80}{5} = 16, \qquad \sigma = \sqrt{16} = 4$

A statistic needs resistance only when the data actually contains something for it to resist. The caseload data, $8, 9, 10, 11, 52$, has one tutor far larger than the rest, exactly the kind of outlier that drags a sensitive statistic like the mean far from where most of the data sits, so the resistant median is the honest choice there.

$\text{mean} = 18, \quad \text{median} = 10, \quad \text{four of five tutors helped between } 8 \text{ and } 11 \text{ students}$

The rate data, $21, 24, 28, 30, 32$, tells a different story, and the right evidence for that is not simply that Tukey's fence test found no outlier: a skewed or bimodal data set can pass that same test and still be poorly summarized by the mean. The real evidence is the SHAPE itself. The mean ($27$) and the median ($28$) sit within a dollar of each other, and every value lies within about one standard deviation ($\sigma = 4$) of the mean, which is what a roughly symmetric, compact data set looks like. In that setting the mean and the standard deviation describe the data faithfully, and there is no reason to reach for a resistant alternative built for a shape this data does not have.

List the $8$ sequences and check membership. $A$ (at least $2$ heads) $=\{HHH,HHT,HTH,THH\}$, so $P(A)=\frac48=\frac12$. $B$ (first flip heads) $=\{HHH,HHT,HTH,HTT\}$, so $P(B)=\frac48=\frac12$. Their intersection is $\{HHH,HHT,HTH\}$, so $P(A\cap B)=\frac38$.

$P(A)P(B) = \frac12 \cdot \frac12 = \frac14 = \frac28 \neq \frac38 = P(A \cap B)$

The two sides disagree, so $A$ and $B$ are dependent.

$D$ (second flip heads) $=\{HHH,HHT,THH,THT\}$, so $P(D)=\frac48=\frac12$. Comparing with $A=\{HHH,HHT,HTH,THH\}$, the shared outcomes are $\{HHH,HHT,THH\}$, so $P(A\cap D)=\frac38$.

$P(A)P(D) = \frac12 \cdot \frac12 = \frac14 = \frac28 \neq \frac38 = P(A\cap D)$

Again the two sides disagree, so $A$ and $D$ are dependent as well.

The classmate's reasoning treats independence as a property of $A$ alone, something $A$ either "has" or "lacks" in general. But independence is always a claim about a PAIR of events, decided by comparing $P(A \cap X)$ with $P(A)P(X)$ for that specific $X$.

$P(A \cap X) = P(A)P(X) \quad \text{(a fact about the PAIR } A, X\text{, checked for that } X \text{ alone)}$

Two events failing that test say nothing about a third event $X$ that was never checked, because the comparison depends on how $X$ overlaps with $A$, which can differ completely from how $B$ or $D$ overlapped with $A$. To settle whether $A$ is independent of some other specific event, the only valid route is to compute $P(A \cap X)$ and $P(A)P(X)$ for that particular $X$ and compare them directly. Nothing about parts A and B, however many times repeated, can substitute for that computation on a new pair.