Introduction to Probability: Free Response

Start with the outcomes, because everything else is divided by their count. The drum holds $40$ slips, and the setup promises they are alike in every way that matters here: identical slips, folded the same way, stirred, and drawn without looking. That is what drawing at random means, and it is what makes the $40$ slips equally likely. Without that promise there would be nothing to count.

Maya's slips are the favorable ones, and there are $9$ of them, so her probability is the share of the $40$ that her slips take:

$P(\text{Maya}) = \frac{9}{40}.$

That is already in lowest terms, since $9 = 3 \times 3$ and $40 = 2 \times 2 \times 2 \times 5$ have no factor in common. For the decimal, divide:

$9 \div 40 = 0.225,$

and a percent is that decimal multiplied by $100$, so $0.225 = 22.5\%$. The three forms are one number written three ways, sitting a little under a quarter of the way up the scale from $0$ to $1$.

Notice what the answer did not depend on: whose names are on the other slips, or how many people entered. Only two numbers matter, the favorable count and the total number of equally likely outcomes.

Both events use the same formula as part A. Only the favorable count is extreme.

Every slip in the drum carries the name of a member who entered, so this event collects all $40$ outcomes:

$P(\text{an entered member}) = \frac{40}{40} = 1.$

A probability of $1$ is the top of the scale, and it says the event is certain: there is no way for the draw to miss it.

The organiser did not enter, so not one slip carries the organiser's name and the event collects none of the outcomes:

$P(\text{the organiser}) = \frac{0}{40} = 0.$

A probability of $0$ is the bottom of the scale, and it says the event is impossible.

These two are the ends that every other probability sits between, and the counting shows why nothing can escape them. A favorable count is never smaller than $0$ and never larger than the total, so the fraction is never negative and never above $1$. A raffle answer of $1.2$ or $-0.3$ would be a signal that the counting had gone wrong, not a discovery about the drum.

The member has counted the wrong things. Counting gives a probability only when the things counted are equally likely, and the twelve members are not: Maya has $9$ slips in the drum and Priya has $4$, while a stirred drum knows nothing about names. What is equally likely here is a slip, not a member.

So the sample space is the $40$ slips, and each member's probability is the share of the slips that member holds:

$P(\text{Maya}) = \frac{9}{40}, \qquad P(\text{Owen}) = \frac{7}{40}, \qquad P(\text{Priya}) = \frac{4}{40}.$

Those are three different numbers, so no single fraction can serve for every member. As a check that nothing has gone missing, the three named members and the nine others hold $9 + 7 + 4 + 20 = 40$ slips between them, so their probabilities add to $\frac{40}{40} = 1$, the whole draw.

Now the condition. For $\frac{1}{12}$ to be right, every member would have to hold the same number of slips, because only then are the twelve members equally likely and only then can they be counted like the faces of a fair die. This raffle cannot be repaired that way, since $40$ slips do not share equally among twelve members, but a drum of $36$ or $48$ would.

The habit to carry away is that how many possibilities are there is never the whole question. The question is how many equally likely possibilities there are, and the two counts can be very different.

The wedges are all the same size, so the $20$ wedges are the equally likely outcomes and the total is $20$. A prize of some kind means a gift card or a pencil, so the favorable wedges are the two marked 'gift card' together with the five marked 'pencil':

$2 + 5 = 7 \text{ wedges}.$

That addition is the addition principle from the counting lesson: the two groups do not overlap, since no wedge carries two markings, so their sizes add.

Divide the favorable count by the total:

$P(\text{prize}) = \frac{7}{20} = 0.35 = 35\%.$

The fraction is already in lowest terms, since $7$ is prime and does not divide $20$. The decimal comes from $7 \div 20 = 0.35$, or from scaling the fraction to hundredths, since $\frac{7}{20} = \frac{35}{100}$, which reads off the percent at the same time. Read it back: a little over a third of the wheel pays something.

Count directly first. The wedges marked 'try again' are the thirteen that are left, so

$P(\text{no prize}) = \frac{13}{20} = 0.65 = 65\%.$

Now the complement route. 'No prize' is exactly 'not a prize', so subtract from $1$:

$P(\text{no prize}) = 1 - \frac{7}{20} = \frac{20}{20} - \frac{7}{20} = \frac{13}{20}.$

The two agree, and they were never free to do otherwise. Every wedge on the wheel either pays something or does not, never both and never neither, so the two events split the $20$ wedges between them with nothing left over:

$7 + 13 = 20.$

Divide that by $20$ and it says the two probabilities add to $1$, which is the complement rule written the other way round. The direct count and the subtraction are the same counting done from opposite ends.

That is also why the rule subtracts rather than flips. Turning $\frac{7}{20}$ upside down would give $\frac{20}{7}$, which is bigger than $1$ and off the scale, so it cannot be anybody's probability.

The operator is looking at one of the two numbers in a probability and ignoring the other.

Nothing has happened to the prize wedges: there are still seven of them. But a probability is not a count of favorable outcomes, it is their share of all the equally likely outcomes, and the wheel now has more of those. Five extra wedges make $20 + 5 = 25$ in all, so

$P(\text{prize}) = \frac{7}{25} = 0.28 = 28\%.$

Compare that with the wheel as it was by putting both over a common denominator:

$\frac{7}{20} = \frac{35}{100}, \qquad \frac{7}{25} = \frac{28}{100}.$

So the chance of a prize has dropped by seven parts in every hundred spins. The operator's sentence would be true only if a probability were the numerator alone.

A quick way to feel it: the same seven prize wedges now have more company, so every spin has more ways to miss them. Push the idea and it becomes obvious. A wheel of a hundred equal wedges carrying the same seven prizes would pay on only $\frac{7}{100}$ of its spins, and nobody would call that the same chance.

The general statement is worth keeping. Adding outcomes that are not favorable leaves the numerator alone and raises the denominator, which can never make the probability larger, and makes it strictly smaller whenever anything was favorable to begin with.

Every tile carries exactly one of the four shapes, and the four shapes are all there are, so the four events divide the $24$ equally likely tiles between them with nothing left over and nothing counted twice. That is what lets the four probabilities account for the whole draw, and the whole draw is $1$.

Add the three listed probabilities over the common denominator $24$:

$\frac{10}{24} + \frac{6}{24} + \frac{3}{24} = \frac{19}{24}.$

What is left of the one whole belongs to the suns:

$P(\text{sun}) = 1 - \frac{19}{24} = \frac{24}{24} - \frac{19}{24} = \frac{5}{24}.$

Now the check. A probability of $\frac{5}{24}$ over $24$ equally likely tiles means $5$ favorable tiles, and counting tiles directly says the same: the listed shapes take $10 + 6 + 3 = 19$ tiles, leaving $24 - 19 = 5$ for the suns.

The two routes are the same arithmetic seen twice. Subtracting probabilities over a denominator of $24$ is subtracting counts and then dividing by $24$, so the smudge could be repaired either way round.

Take any experiment with $N$ equally likely outcomes. Running it is certain to produce one of them, and certainty is what the number $1$ names on the probability scale. So the whole sample space carries probability $1$, and the outcomes have to share that one whole between them.

Equally likely means no outcome takes a larger share than another. Cut the whole into $N$ equal pieces and hand one to each outcome, and each outcome's probability is

$P(\text{one outcome}) = \frac{1}{N}.$

Adding all $N$ of those shares puts the whole back together, because adding $N$ copies of a number is multiplying it by $N$:

$\frac{1}{N} + \frac{1}{N} + \cdots + \frac{1}{N} = N \times \frac{1}{N} = \frac{N}{N} = 1.$

That is the whole argument, and notice what it never mentions: tiles, shapes, or the number $24$. It uses only that the outcomes are equally likely and that they are all of the outcomes there are, so it holds for a coin, a wheel, a bag or anything else with the same two features.

The same reasoning covers events rather than single outcomes, which is what part A leaned on. If several events between them contain every outcome exactly once, their favorable counts add to $N$, so their probabilities add to $\frac{N}{N} = 1$. On the bag that reads $\frac{10 + 6 + 5 + 3}{24} = \frac{24}{24} = 1$.

Set the counting up first. The experiment has $N$ equally likely outcomes and the event $A$ contains $k$ of them, so by the definition of probability,

$P(A) = \frac{k}{N}.$

The event 'not $A$' is everything else in the sample space. There are $N$ outcomes in all and $k$ of them lie in $A$, so the number left over is $N - k$, and

$P(\text{not } A) = \frac{N - k}{N}.$

That single subtraction is where the whole rule comes from, and it is legitimate because every outcome lies in exactly one of $A$ and 'not $A$', never in both and never in neither. Add the two probabilities:

$\frac{k}{N} + \frac{N - k}{N} = \frac{k + N - k}{N} = \frac{N}{N} = 1.$

Subtract $P(A)$ from both sides of $P(A) + P(\text{not } A) = 1$ and the complement rule is left:

$P(\text{not } A) = 1 - P(A).$

Nothing in that argument used what the outcomes were, only that they were equally likely and that each one was counted once.

Now the check on the bag, with $A$ the event that the tile is a star. Here $N = 24$ and $k = 10$, so $P(A) = \frac{10}{24} = \frac{5}{12}$ and the rule predicts

$P(\text{not a star}) = 1 - \frac{5}{12} = \frac{7}{12}.$

Count instead: the tiles that are not stars number $24 - 10 = 14$, and $\frac{14}{24} = \frac{7}{12}$. The rule and the count agree, which is no surprise, because the rule is that count done once and for all.

One warning the derivation makes plain: the rule subtracts, it does not turn the fraction over. Nothing in $\frac{N - k}{N}$ swaps a numerator for a denominator.

An experimental probability is a record of what happened, not a count of what could happen. It divides the number of times the event occurred by the number of trials:

$P_{\text{exp}}(\text{point up}) = \frac{68}{200}.$

Reduce it. Both numbers are divisible by $4$, since $68 = 4 \times 17$ and $200 = 4 \times 50$:

$\frac{68}{200} = \frac{17}{50}.$

Since $17$ is prime and does not divide $50$, that is lowest terms. For the decimal, scale to hundredths instead of dividing, which is quicker here:

$\frac{17}{50} = \frac{34}{100} = 0.34.$

So about a third of the recorded tosses landed point up. Note carefully what has been measured: this is a statement about the $200$ tosses the class actually made, and so far nothing more.

Counting outcomes gives a probability only when those outcomes are equally likely, and that is a promise about the object, not about how many things the object can do. A fair coin makes the promise because its two faces are alike; a wheel makes it because its wedges are the same size; a fair die makes it because its six faces are the same. In those cases the object's own symmetry does the work, and the words fair and equal are how a problem hands you that promise. Symmetry is not the only route to it: stirring a drum of identical slips, or drawing without looking from a bag of identical tiles, makes the outcomes equally likely by the way the choice is made rather than by the shape of the thing chosen.

A drawing pin has neither route open to it. It is a broad flat head at one end and a thin point at the other, so no symmetry swaps one landing for the other, and no amount of shaking or careful tossing evens out a shape that was never even to begin with. So the observation that it can land in two ways establishes only that there are two outcomes, which is not enough to divide anything by. The class's own record shows the gap between the two claims:

$\frac{1}{2} = 0.5 \qquad \text{against} \qquad \frac{17}{50} = 0.34.$

When the equally likely promise is missing, data takes the place of counting. Toss the pin many times, record what happens, and take the fraction of trials that came out point up. That fraction is the experimental probability, and here it is the only honest route to a number.

This is also the reason nobody needs to toss a fair coin to learn its probability, while everybody has to toss a pin. One object argues from its own symmetry; the other has nothing to argue with and must be measured.

Expecting the same count again misreads what the figure is. The record says how a particular $200$ tosses came out; it puts no obligation on the next $200$ to copy them. A run of trials scatters around the true probability rather than landing on it, and the class has evidence of that in its own data, since the whole class and one group inside it disagree:

$\frac{68}{200} = 0.34 \qquad \text{and} \qquad \frac{6}{25} = 0.24.$

Same pin, two different figures, and neither is a mistake. Each is an honest record of the tosses it counted.

What the class can honestly claim is that $0.34$ is its best estimate of the pin's true probability of landing point up, based on $200$ tosses. What would sharpen the claim is more tosses. As the number of trials grows, the fraction of them that come out point up settles closer and closer to the true value, which is the law of large numbers, and it is exactly why the class's $200$ tosses deserve more trust than any single group's $25$.

Notice what the law does not promise. It says nothing about the next toss, and nothing about any one short run; a further $200$ tosses could easily give a nearby but different count. It promises only that the long-run fraction closes in, which is why a sound report says 'in $200$ tosses we recorded point up $68$ times, an experimental probability of $0.34$' rather than 'point up happens $68$ times in every $200$'.

The two stages do not offer the same choices, because the first marker is kept out of the box, so count them one at a time.

The first choice is from the whole box, which holds $7$ markers. Whatever is taken, the box then holds $6$, so the second choice is from $6$. Since we are keeping track of which came first, the multiplication principle counts the ordered pairs:

$7 \times 6 = 42 \text{ ordered pairs}.$

All $42$ are equally likely, because the markers look identical from the outside and are taken without looking, so no pair is favored over another. That is the promise the whole calculation rests on.

Now the favorable pairs. For both markers to write, the first has to be one of the $4$ that write, and the second has to be one of the $3$ that write and are still in the box:

$4 \times 3 = 12 \text{ ordered pairs}.$

Divide the favorable count by the total:

$P(\text{both write}) = \frac{12}{42} = \frac{2}{7}.$

The fraction reduces because $12$ and $42$ share a factor of $6$. The one thing that made this different from spinning a wheel twice is that a marker had already left the box when the second choice was made, which is why both counts drop by one at the second stage.

'At least one has dried out' is the opposite of 'neither has dried out', and 'neither has dried out' is exactly the event of part A, both markers writing. So the complement rule turns one answer straight into the other:

$P(\text{at least one dried}) = 1 - \frac{2}{7} = \frac{5}{7}.$

The direct count confirms it. Of the $42$ equally likely ordered pairs, $12$ have both markers writing, so every other pair contains at least one dried marker:

$42 - 12 = 30, \qquad \frac{30}{42} = \frac{5}{7}.$

The two agree because every pair falls into exactly one of the two events, which is the counting that the complement rule was built from.

It is worth seeing what the complement saved. Counting 'at least one dried' the direct way, case by case, means adding three separate counts: dried then writing, writing then dried, and dried then dried.

$3 \times 4 + 4 \times 3 + 3 \times 2 = 12 + 12 + 6 = 30.$

Same answer, three times the work, and three chances to forget a case. That is why 'at least one' is the standard signal to look at the opposite event first.

Putting the marker back changes exactly one thing in the count: what the second stage has to offer.

With the marker returned and the box shaken, the second choice faces the same $7$ markers as the first, of which the same $4$ write. So the counts become

$7 \times 7 = 49 \text{ ordered pairs}, \qquad 4 \times 4 = 16 \text{ favorable},$

and the probability that both markers taken write is $\frac{16}{49}$.

To compare the two rules, bring the answers onto a common denominator. Without replacement the answer was $\frac{2}{7}$, and $\frac{2}{7} = \frac{14}{49}$, so replacing the marker lifts the probability from $\frac{14}{49}$ to $\frac{16}{49}$.

The reason is worth saying in words. When the student keeps the first marker and that marker was one that writes, the box left behind is poorer in markers that write: $3$ of the remaining $6$, where the full box held $4$ of $7$. Putting the marker back undoes that, so the second stage is as good as the first was. Taking a favorable item out of the box makes a second favorable item harder to find, and that is the general shape of it.

The formula itself never changed. Both answers are favorable outcomes over equally likely outcomes; only the counting of the second stage moved.