Introduction to Probability: Free Response

Each purchase offers the same ten flavors, and the machine's restocking promise, uniform and with no memory of what happened before, means the ten choices at one purchase do not depend on the flavors chosen earlier. That independence is exactly what the multiplication principle needs, so it counts the ordered lists of three flavors:

$|S| = 10 \times 10 \times 10 = 10^3 = 1000.$

Because nothing about the machine favors one list of three flavors over another, all $1000$ of these ordered lists are equally likely, which is what lets a later probability be found simply by counting.

Now count the lists with no chocolate purchase. Each of the three purchases must land on one of the nine non-chocolate flavors, and that restriction applies at every stage independently of the others:

$9 \times 9 \times 9 = 9^3 = 729.$

"At least one chocolate purchase" is the opposite of "no chocolate purchase at all," which part A already counted, so the complement rule turns that count straight into the probability wanted:

$P(\text{at least one chocolate}) = 1 - \frac{729}{1000} = \frac{1000 - 729}{1000} = \frac{271}{1000}.$

Since $271$ is prime and does not divide $1000$, the fraction is already in lowest terms. Dividing gives the decimal:

$\frac{271}{1000} = 0.271.$

So across three purchases from this machine, a little over a quarter of the time at least one of them is chocolate.

Split "at least one chocolate purchase" into the three cases that make it up. For exactly one chocolate purchase, the chocolate purchase can be any one of the three positions in the list, and the other two purchases must both avoid chocolate:

$3 \times 9 \times 9 = 243.$

For exactly two, the two chocolate positions can be chosen in $3$ ways out of the three purchases, and the remaining purchase must avoid chocolate:

$3 \times 9 = 27.$

For exactly three, every purchase is chocolate, which is a single list. Adding the three cases:

$243 + 27 + 1 = 271,$

which matches part B's answer, as it has to, since the two routes count the same event two different ways.

But the direct route needed three separate counts and an addition, while the complement route needed one count, "no chocolate at all," and one subtraction. The general pattern is about the shape of the event rather than about this machine: "at least one" (or "at least $k$") is usually a sprawling union of several disjoint cases, while its complement, "none" (or "fewer than $k$"), is often a single tidy count. Whenever that asymmetry holds, the complement rule is the shorter path. It stops being a guaranteed advantage once the complement itself splits into several cases, since then neither route is obviously shorter.

"Or" is a union, so the addition rule applies with $A$ the treadmill users and $B$ the free-weight users. Simply adding $84$ and $65$ would count every member who used both pieces of equipment twice, once inside each group, so the rule subtracts the overlap once to undo that:

$P(A \cup B) = \frac{84}{200} + \frac{65}{200} - \frac{30}{200} = \frac{84 + 65 - 30}{200} = \frac{119}{200}.$

Since $119 = 7 \times 17$ shares no factor with $200 = 2^3 \times 5^2$, the fraction is already in lowest terms, and dividing gives

$\frac{119}{200} = 0.595.$

"Neither" is the opposite of "treadmill or weights," so the complement rule turns part A's answer straight into this one:

$P(\text{neither}) = 1 - \frac{119}{200} = \frac{200 - 119}{200} = \frac{81}{200} = 0.405.$

Check it by counting instead. Of the $200$ members, $119$ fall inside the union (treadmill, weights, or both), so the remaining $200 - 119 = 81$ fall outside both groups, and $\frac{81}{200}$ matches. The two routes have to agree, because every one of the $200$ members is either inside the union or outside it, never both and never neither.

The staff member's arithmetic is correct but the rule applied is the wrong one. Adding $\frac{84}{200}$ and $\frac{65}{200}$ counts every member who used both pieces of equipment inside BOTH of those fractions, so the $30$ overlapping members are counted twice, and the naive total $\frac{149}{200}$ overstates the union by exactly $\frac{30}{200}$, the size of the overlap.

Now push the same naive method with a weight-room count of $150$ instead of $65$, treadmills still at $84$:

$\frac{84}{200} + \frac{150}{200} = \frac{234}{200} = 1.17.$

A probability above $1$ describes an event more certain than certainty itself, which is nonsense: $234$ counted "slots" cannot come from a survey of $200$ members. So this second example does not merely repeat the first mistake, it exposes it. Whenever the naive addition is applied to two events large enough to overlap heavily, it can produce a value outside the $[0, 1]$ scale, and a value outside that scale is a certificate that the method used to reach it cannot be correct in general, whether or not a particular pair of numbers happens to stay under $1$ by luck.

The whole roster of $240$ employees is the sample space, and $150$ of them work in Engineering, so

$P(\text{Engineering}) = \frac{150}{240} = \frac{5}{8} = 0.625.$

Learning that the selected employee works in Engineering rules out everyone else on the roster, so the sample space shrinks from all $240$ employees down to the $150$ in Engineering. That shrinking is exactly why the conditional probability divides by $150$: the outcomes outside Engineering are no longer part of the world being counted. Of those $150$, $60$ work remotely, so

$P(\text{remote} \mid \text{Engineering}) = \frac{60}{150} = \frac{2}{5} = 0.4.$

For the selected employee to work in Engineering and remotely, the employee has to work in Engineering (probability $\frac{5}{8}$), and then, among Engineering employees, has to be one of the remote ones (probability $\frac{2}{5}$). The multiplication rule multiplies exactly those two probabilities:

$P(\text{Engineering} \cap \text{remote}) = \frac{5}{8} \cdot \frac{2}{5} = \frac{10}{40} = \frac{1}{4} = 0.25.$

Check it against the roster directly. There are $60$ employees who are both in Engineering and remote, out of $240$ in all:

$\frac{60}{240} = \frac{1}{4},$

which matches, exactly as it must, since the multiplication rule is only the conditional definition rearranged.

Across the whole roster, $60$ Engineering employees and $54$ Sales employees work remotely, so $114$ of the $240$ employees are remote in all:

$P(\text{remote}) = \frac{114}{240} = \frac{19}{40} = 0.475.$

Compare the joint probability from part B against the product independence would predict:

$P(\text{Engineering})P(\text{remote}) = \frac{5}{8} \cdot \frac{19}{40} = \frac{95}{320} = \frac{19}{64} \approx 0.297,$

while part B found $P(\text{Engineering} \cap \text{remote}) = \frac{1}{4} = 0.25$. Since $0.25 \neq 0.297$, department and remote status are dependent, not independent: knowing which department an employee works in genuinely changes the chance that employee is remote.

That dependence is visible directly in the two conditional rates. Engineering's remote rate is $P(\text{remote} \mid \text{Engineering}) = 0.4$, below the company-wide $0.475$, while Sales's is $P(\text{remote} \mid \text{Sales}) = \frac{54}{90} = 0.6$, above it. If department told you nothing about remote status, both conditional rates would equal the company-wide rate; instead they straddle it in opposite directions, which is exactly what the failed independence test predicts.

List the outcomes directly from the die's six faces. "Even" collects $\{2, 4, 6\}$, so $P(A) = \frac{3}{6} = \frac{1}{2}$. "Greater than $4$" collects $\{5, 6\}$, so $P(B) = \frac{2}{6} = \frac{1}{3}$. The only face in both lists is $6$, so $A \cap B = \{6\}$ and $P(A \cap B) = \frac{1}{6}$.

Now compare that against the product, which is what independence actually asks:

$P(A)P(B) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.$

The two sides agree, $P(A \cap B) = P(A)P(B) = \frac{1}{6}$, so $A$ and $B$ are independent. Nothing here appealed to whether "even" and "greater than $4$" feel related; the equation is what decided it.

"Even" and "odd" cannot both describe the same roll, so $A \cap C = \emptyset$ and $P(A \cap C) = 0$. The product of the two individual probabilities, though, is positive:

$P(A)P(C) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$

Since $0 \neq \frac{1}{4}$, $A$ and $C$ fail the independence test: they are dependent.

Conditioning shows exactly how dependent. Because $P(C) > 0$, dividing is allowed, and

$P(A \mid C) = \frac{P(A \cap C)}{P(C)} = \frac{0}{1/2} = 0,$

while $P(A) = \frac{1}{2}$ on its own. So learning that the roll was odd does not merely nudge the chance of "even," it collapses it to zero, which is the opposite of the "no information" behavior independence describes. This is exactly the lesson's theorem: $A$ and $C$ are disjoint, both have positive probability, so they cannot be independent, and the reason is that a disjoint event is maximally informative rather than uninformative.

Test the classmate's claim first. $D = \{3, 6\}$, so $P(D) = \frac{2}{6} = \frac{1}{3}$, and $A \cap D = \{6\}$ (the only roll that is both even and a multiple of $3$), so $P(A \cap D) = \frac{1}{6}$. Compare:

$P(A)P(D) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} = P(A \cap D).$

The two sides agree, so $A$ and $D$ genuinely are independent. But the classmate did not compute anything; the argument was that the two properties "feel" unconnected. That is not a proof, only a guess, and it happened to land on the right answer.

To see the guess fail, look for two events with the same flavor of "unrelated-sounding" properties that are NOT independent. Take $F$, the roll is $1$ or $2$, and $G$, the roll is a multiple of $3$. Being small and being a multiple of $3$ sound at least as unconnected as being even and being a multiple of $3$. But

$F = \{1, 2\}, \qquad G = \{3, 6\}, \qquad F \cap G = \emptyset,$

so $P(F \cap G) = 0$, while

$P(F)P(G) = \frac{2}{6} \cdot \frac{2}{6} = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \neq 0.$

$F$ and $G$ are dependent, and in fact they are mutually exclusive, which by the lesson's theorem is as far from independent as two events with positive probability can get. So a pair of properties that "feel" just as unrelated as $A$ and $D$'s produced the opposite verdict. That is the whole point: whether two events are independent is a numerical fact about $P(A \cap B)$ versus $P(A)P(B)$, and no amount of staring at what the events describe can substitute for computing it.

Each of the nine visitors is assigned independently and uniformly from the same $30$ codes, so the multiplication principle counts the ordered lists of nine codes:

$|S| = \underbrace{30 \times 30 \times \cdots \times 30}_{9 \text{ factors}} = 30^9.$

All of these lists are equally likely, because every assignment is uniform and independent of every other visitor's.

For all nine codes to be different, assign the codes one visitor at a time and let each new visitor avoid every code already used. The first visitor has $30$ choices, the second has $29$ left to avoid a repeat, and so on down to the ninth visitor, who must avoid the $8$ codes already taken:

$30 \times 29 \times 28 \times 27 \times 26 \times 25 \times 24 \times 23 \times 22,$

a product of $9$ falling factors, exactly the shape the birthday problem in the lesson used for its own group and its own pool of categories.

"At least two visitors share a code" is the opposite of "all nine codes are different," so the complement rule applies exactly as it did for the birthday problem in the lesson. Divide the all-different count by the total to get the probability that all nine differ, computing it as a chain of nine fractions:

$P(\text{all different}) = \frac{30 \times 29 \times \cdots \times 22}{30^9} = 1 \cdot \frac{29}{30} \cdot \frac{28}{30} \cdot \frac{27}{30} \cdot \frac{26}{30} \cdot \frac{25}{30} \cdot \frac{24}{30} \cdot \frac{23}{30} \cdot \frac{22}{30} \approx 0.264.$

The complement rule finishes it:

$P(\text{at least two share}) = 1 - 0.264 = 0.736.$

So with only nine visitors and thirty codes, a shared code is already more likely than not.

Both problems are decided by the same complementary-counting argument, all outcomes different versus at least one repeat, so the mechanism is identical. What differs is the ratio between how many things are being placed and how many places there are to put them. Nine visitors drawn from only thirty codes is a substantial fraction of the category pool, while a group of people is a tiny fraction of the days in a year. A small pool relative to the group size leaves each new arrival with noticeably fewer safe, still-unused codes to land on, so the falling factors in the all-different count shrink much faster here than they would against a far larger denominator.

The other way to see it is to count comparisons rather than individuals. It is not any one visitor's code that has to "land right," it is every pair of visitors that gets a chance to collide, and the number of pairs in a group of $n$ grows like

$\binom{n}{2} = \frac{n(n-1)}{2},$

roughly the square of $n$. So doubling the group size roughly quadruples the number of chances for a match, and shrinking the category pool raises the chance that any one of those pairs collides. Both effects point the same way here: a group that is a meaningful fraction of a small category pool produces far more collision opportunities, per person, than the same size group does against a much larger pool of categories, which is exactly why nine visitors against thirty codes clears one half so much faster than the classic birthday problem does, even though that problem needs a noticeably larger group of people to get there, because its pool of days is so much bigger.