There is a national exam taken by all students in British high schools known as the GCSE exam and apparently one recent problem has caused an uproar because it stumped most students. Here it is:

Hannah has 6 orange sweets and some yellow sweets.

Overall, she has n sweets.

The probability of her taking 2 orange sweets is 1/3.

Prove that: n^2-n-90=0

^ is “to the power of”

The problem seems pretty straightforward to me and should be doable by anyone who has been taught the elements of probability (the link above gives the solution) and so I am puzzled by fact that there has been a lot of fuss over it.

Apparently it is the wording of the question that is at issue, especially the use of the word ‘prove’, but I am unsure as to how one would interpret the above wording in such a way as to be confused as to what is expected.

Trebuchet says

One problem I see immediately is that it doesn’t say it’s random. Maybe Hannah LIKES yellow sweets better.

Turi says

“The probability of her taking 2 orange sweets is 1/3.” ? What probability? How many draws? If i take 10 sweets and get two with a probability of 1/3 it is different to when i take only two sweets and get two. Also “prove” has a certain meaning in math and should not be used when “show” is sufficient.

While the problem in it self can be solved (and understood), it is not formulated in a way i would seem fit in an exam, let alone in a national exam.

jimatkins says

One problem with this is how are the sweets drawn? Is this two at once or one and then another? Does replacement occur between the draws? I do probability with my 7th grade students. They get the single events but consecutive events throw them.

Improbable Joe, one of the NEW FOUR HORSEMEN OF GLOBAL ATHEIST THINKY LEADER KINGS EDUCATIONAL FOUNDATION COUNCIL says

The wording seems a bit less than straightforward.

Turi says

I am trying to solve this one. Going from take two sweets draw:

6/n is the probability to take on orange sweet, from the question follows: 6/n + 5/(n-1) = 1/3 or (11n-6)/(n²-n) = 1/3

The second part of the question gives: n²-n = 90

Both result in (11n-6)/90 = 1/3 or 11n-6= 30 or n = 24/11

scratch that. testing shows: this is the wrong answer for n.

Which means: I am wrong (quit possible) or the only understandable version of this question is not the right interpretation.

Chiroptera says

I was able to do this one in my head. Is there something wrong with me?

Chiroptera says

PS, the only ambiguity I can see is whether she takes the candy with replacement or without replacement; however, I assume she intends to take two to eat, so I assume she takes two without replacement.

Chiroptera says

Turi, #4:

6/n + 5/(n-1) = 1/3You need to multiply the two probabilities on the left, not add them.

xykademiqz says

They should have just asked “Find n” or “Calculate n.” Then people would have to set up the equation and also solve it.

Claschx says

This stuff i’m sure would make my old math teacher turn in his grave, since he was very careful with the wording .. You SHOW for a finite (very so) set of given values, you PROVE for any value.

That alone would disqualified the question, apart from the embarassing fact that solving the quadratic equation (n=10 in N) should be taken as a correct answer. This was just a failed attempt to make a trick question.

if n=10 6/10*5/9=30/90 p=1/3

Luc Nerwinski says

Forget the probabilities; you are given a quadratic equation, and the solution can be “proved” by applying the quadratic formula. This gives you n = -9 and n = 10. Since she can’t have a negative number of sweets, she has ten sweets.

As a check, plug the numbers into the hypergeometic distribution--assuming that she picked two sweets at the same time (i..e., without replacement) That formula works out to 15/45 which is the 1/3 probability the problem mentions.

I consider the stuff about the probabilities a dirty trick on the part of the examiner. At the very least, the problem should have stated whether or not she took the candies with replacement.

Rob Grigjanis says

Did it stump most students, or most students who tweet (twitter?)?

I’m amazed that anyone saw any ambiguity in the wording. The information provided is exactly what you need to solve it, so reading ‘taking’ as anything other than ‘choosing at random’ is silly, because it would require more information than was given.

And I don’t get the quibbling about ‘replacements’. She has n sweets. She takes 2. Period.

doublereed says

I just did the problem backwards, as the equation is (n-10)(n+9) = 0.

-9 obviously makes no sense. Check that n = 10 solves the problem:

6/10 * 5/9 = 30/90 = 1/3.

Turi says

@Chiroptera Thank you. Stupid mistake from my side. I still think the question was worded very bad.

Chiroptera says

Turi, #14:

Stupid mistake from my side.Maybe, if you currently enrolled in a statistics course. If it’s been a while since you’ve done any statistics, then maybe not so stupid. You add if you want the probability of either event A

orevent B, assuming mutual exclusive events. You multiply if you want the probability of event Aandevent B (or, to be more precise, event B given A). Not being a statistician myself, I think it isn’t too difficult to mix up the concepts. I probably wouldn’t remember if I hadn’t taught the subject recently.Robert B. says

Once understood, the problem is a bit tricky (in that it involves working backward from the combined probability -- reversed problems always take students a little more practice to master) but completely solvable. However, the wording should have been something like “If she picks two of her sweets at random, the probability that both are orange is 1/3.” I understood what they meant only because it was the only interpretation that led to a unique solution. Granted, I help students with math problems for a living so I spot such things very fast -- but the wording is unclear.

I imagine the test writers were looking for something like this, assuming 1-column proofs are acceptable:

6/n * 5/(n-1) = 1/3

30/(n^2-n)=1/3

90=n^2-n

0=n^2-n-90

That technique of solving for exactly what was asked for rather than what’s most intuitive (in this case, solving for n would be most intuitive) is a common trick on standardized tests, at least in the USA. On the other hand, asking for an algebraic proof would have stumped most USA highschoolers because for some reason proofs are only taught in geometry here, not in any other high school classes. A US standardized test would probably have asked something like “Find the value of n^2-n.”

Robert B. says

Also, Chiroptera, the probability of B given A differs from the probability of both A and B:

P(A&B) = P(A) * P(B|A)

Though I probably wouldn’t have caught that if I hadn’t taught this stuff

veryrecently (as in, within the last two weeks).Chiroptera says

Robert B., #17

What you wrote is exactly what I was thinking when I typed my previous comment. Sorry that I botched what I wanted to say.

Robert B. says

No worries, if this stuff was easy to explain they wouldn’t need us.

gronank says

The word “prove” threw me off at first, don’t quite know why. Possibly because the word proof is something I associate heavily with theorems which is not relevant here. It is an exercise in applying the chain rule, not proving the chain rule.

It would have been clearer if it said: “Show that n satisfies the equation:”

Its a nice problem because it tests knowledge of the chain rule without requiring knowledge of solving quadratic equations (which could be and probably is tested separately).

jd142 says

I was confused before I started. I wasn’t sure if “taking” meant taking 2 at a time or in sequence. Let’s say I have 1 red, 1 yellow, 1 blue, and 1 green treat. I take RY. Since the order doesn’t matter, RY is the same as YR. So I have RY, RB, RG, YB, YG, BG. So the odds are 1/6 if I take 2 at a time. But if I take 1, then draw another, the odds of drawing a red or yellow on the first draw is 2/5. The odds of drawing the remaining color is 1/4. So I think that makes it 2/20 or 1/10. Or are you supposed to add .4 + .25??

But statistics was sooooo long ago and I never use it.

Chiroptera says

jd142, #20:

Now

that’san interesting problem. Let’s apply that problem with the original question where there’s 10 treats, 6 of which are orange. The probability of taking a pair of sweets at one time is found the same way as finding the probability of a poker hand:There are 10 choose 2 possible pairs of treats: 45.

There are 6 choose 2 possible pairs of orange treats: 15.

The probability of choosing 2 orange treats out of the bag is 15/45 = 1/3.

I would have naively assumed that the probability shouldn’t matter whether you take two at the same time or whether you take one at a time (after all, you deal cards one at a time), but my intuition is rather poor when it comes to probability.

Rob Grigjanis says

jd142 @20: Two at a time or in sequence doesn’t matter. The prob of taking two at a time and getting RY is 1/6, as you say.

For one at a time, it’s getting R then Y, or Y then R. Two distinct possibilities, so (1/4)(/1/3) + (1/4)(1/3) = 1/12 + 1/12 = 1/6.

Or: getting R or Y on the first take: 2/4=1/2. Taking the other colour on the second draw; 1/3. (1/2)(1/3)=1/6.jd142 says

@22 and @23 — this is why I get confused by statistics. My only defense is that I know, understand and can explain the Monty Hall problem. And that’s about it. 🙂

Chiroptera says

For completeness, I should also point out that the ‘combination’ method of calculating probabilities also brings us to the solution:

6 choose 2 is 15. n choose 2 is n(n-1)/2. So, setting the probability of choosing two orange treats to 1/3, we get

15/[n(n-1)/2] = 1/3

which brings us to n^2 -- n -- 90 = 0.

Which one is the better way to “think about” the problem? Probably up to the individual.

Poppy says

I’ve never took stats, calculus or algebra back when I was in school so that question is completely unintelligible to me. I can’t make heads or tails out of what is being asked. I’m greatly impressed by the people who could solve it.

Poppy says

Oh poop!!!

“I never took” !!! Darn typos!! Even missed it in preview.

Pen says

I bet it’s a combination of the unexpected word ‘prove’ and the bizarre shift from a concrete situation involving sweets to an answer that’s basically an abstraction loosely based on that situation.

In the given context Hannah seems to have mysteriously acquired the square of the number of sweets she started with, given the original number of sweets to one of her friends, given 90 more to another and found herself left with an empty sweet bag, all before she even took the two orange ones for herself. Most things that start life as concrete problems aimed at 16 year olds make more sense than that.

Not that they should have been thrown by the eccentricity of the examiner, but it seems they were.

scoobygang says

I’m not sure where the wording in the OP came from, but it doesn’t appear to be anything like the wording from the actual exam. The exam used “Show” not “Prove”, included information that she ate the sweets (thus no replacement), and made it clear that only two sweets were being drawn from the bag.

It’s a relatively difficult question for the level of maths being studied, but I don’t think it’s needlessly obtuse or confusing nor intended to be a “trick question” in any way. A student who understands the concepts of probability and algebra taught at this level has all the tools to be able to handle this. Doing so within the stressful context of an exam is, as always, another thing entirely.

BecomingJulie says

The question is poory-phrased. The crucial piece of information, which it doesn’t say anywhere, is that Hannah draws exactly two sweets at random from the bag.

If she had drawn (n -- 4) sweets from the bag, then at least two of them would certainly be orange. There are only (n -- 6) yellow sweets in the bag; so after all of those have been pulled out, the next two sweets drawn can

onlybe orange.If you assume (and we know what assumption is the mother of all of …..) that what they mean is that Hannah draws exactly two sweets at random from the bag, then the probability that the first sweet is orange = 6 / n, and the probability that the second sweet is orange = 5 / (n-1). So the combined probability is equal to the products of the individual probabilities; giving

(6 / n) * (5 / [n-1]) = 1 / 3. Rearranging so as to do all the multiplication before the division, we get

(6 * 5) / (n² -- n) = 1 / 3. Inverting,

(n² -- n) / 30 = 3; and rearranging, we get

n² -- n -- 90 = 0.

Was there a second part to the question? They appear to have left a beautiful juicy quadratic equation hanging there, just ripe for the solving. It’s just too tempting! It factorises easily (if you do enough maths problems, you soon come to recognise obvious factorisations), giving (n + 9) * (n -- 10) = 0; and since Hannah can’t have minus 9 sweets, n must be 10. As a quick check, (6 / 10) * (5 / 9) = 30 / 90 = 1/3.

But that might not have been what they meant …..