Hannah and her Sweets: that EdExcel Examination Question…

You may or may not know that yesterday there was a bit of a Twitterstorm of students complaining about an “unfairly difficult” examination question on the GCSE Mathematics paper set by EdExcel.

This is the question:

There are n sweets in a bag. Six of the sweets are orange. The rest of the sweets are yellow.

Hannah takes a sweet from the bag. She eats the sweet. Hannah then takes at random another sweet from the bag. She eats the sweet.

The probability that Hannah eats two orange sweets is 1/3. Show that n²-n-90=0.

Not sure what all the fuss is about. Seems very straightforward. The question tells you that 6/n × 5/(n-1)=1/3 whence the equation follows by a trivial rearrangement. In fact I’m a little surprised the question didn’t go on to ask the students to solve the quadratic equation n²-n-90=0 to show that n=10…

I don’t really know what is on the GCSE Mathematics syllabus these days. In fact I never did GCSE Mathematics, I did O-level Mathematics which was quite a different thing. You can see the papers I took – way back in 1979 – here.


17 Responses to “Hannah and her Sweets: that EdExcel Examination Question…”

  1. yeah…if that is the question is does look straightforward!

  2. I agree. What’s the fuss? I’d be interested on thoughts as to what the other solution (n=-9) ‘means’ though.

    • telescoper Says:

      Minus nine sweets might be better for Hannah’s waistline…

    • Chris Brunt Says:

      n = – 10 is a solution as long as you agree to count negative sweets by -1, -2, -3… so that:

      (-6/-10) * (-5/-9) = 1/3

      This seems “sensible” and I like it because all the probabilities are positive so presumably this can happen (bear with me…)

      n = -9 doesn’t work, because you are counting negative sweets with positive numbers and the probability gets all out of whack: +6/-9 is a negative probability, though for pairs of sweets taken simultaneously you could have a positive probability I suppose…

      e.g. starting with -9, after you take out an orange sweet you’d have -10 left, of which 5 are orange according to the original formulation of the solution. What does that even mean?

      If you could take positive sweets out of a negative bag, you could continue doing this indefinitely. e.g. starting with -10 you’d have -11 after taking a positive orange sweet out. Of these -11, you’d have -7 oranges because your extracted positive orange adds one more negative orange . You would be more likely to get another orange after the first!

      This version of negative sweets is probably not actually good for the waistline after all…

      • Thank. An interesting reply, that triggers several thought.

        I fear you have just broken mathematics, by finding that -10 is a solution to n^2-n-90=0! Cue the tabloid scream…

        Taking positive sweets from a negative bag, and the probability increasing at each pick… that has to have some link to vacuum energy and inflation, surely?

      • telescoper Says:

        I think these sweets are imaginary. That makes this problem all the more complex

      • Chris Brunt Says:

        -9 works if you have -5 oranges to start and you draw postive orange….

        (-5/-9) * (-6/-10) = 1/3

      • …though the question does state that (presumably +) “six of the sweets are orange”, so I feel we should have a n=-9 solution that is compatible with that requirement. I’m still struggling.

      • Anton Garrett Says:

        Is a negative sweet a sour?

      • Chris Brunt Says:

        reply to Alex re: …the question does state that (presumably +) “six of the sweets are orange”

        I think the n=-9 solution with +6 oranges means that there are -15 yellows, whatever that means! If you pick an orange the first time, you’re left with -10 sweets, 5 of which are orange. This at least fits in with putting together the numbers in the quadratic as:

        (6 / -9) * (5 / -10) = 1/3

        Does that help?

        It’s a stretch to interpret 6/-9 as a probability (but notice that picking two oranges has a positive “probability”, and satisfies the quadratic – that’s why it works I suppose.)

        A correction to drawing positive sweets from a bag of negative sweets. Above I had accidentally assumed that the probability of drawing a positive orange is proportional to the number of negative oranges in the bag, which is of course unsupportable and frankly silly. Not that that is any hindrance with mathematics…

  3. This question is indeed trivial. But one of the other sample questions one can do after following the link has demonstrably wrong answer … (not a math question).

  4. The question did go on to ask for n apparently. See this tweet from a teacher (though what he is going on about regarding “modelling”, I can’t help with) https://mobile.twitter.com/danielstucke/status/606765342514401280?p=p

  5. Here is a video solution for the probability problem about Hannah’s Sweets http://handakafunda.com/hannah-sweets-gcse-math-problem-on-probability/

  6. Phillip Helbig Says:

    Prejudice! Everywhere prejudice! It seems everyone is assuming base 10.

  7. Charlie Deacon Says:

    The answer is 10 surely? n = 10

  8. N is clearly 16.

    The problem only states that the second sweet was taken at random. Hannah clearly preferred orange sweets and carefully chose to begin with an orange one. Therefore 5/(n-1) = 1/3.


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