So, all over twitter after the maths GCSE last year, kids were kicking off because they thought “Hannah’s Sweets” was an impossible question; however, it was simply a very clever question about probability with some of the usual structure removed, meaning those sitting the paper would be challenged to apply their knowledge.

The question was:

full question

Part (a)

For part (a), students were required to form an equation from the given information, some of which had to be deduced. The key part of the information is that “She eats the sweet”, therefore indicating that the number of sweets is reducing, and the probabilities will change.

The probability Hannah chooses an orange sweet the first time is \frac{6}{n}, as there are 6 orange sweets and the total is n. Since she eats the sweet, the probability she chooses an orange sweet the second time is \frac{5}{n-1}, because there are now only 5 orange sweets, and the total has decreased by 1.

The total probability of Hannah choosing and eating two orange sweets, is found by multiplying the two probabilities together, if you’re unsure of the rules of probability you can find out more here. We’re also told in the question, that the probability she eats two orange sweets is \frac{1}{3}; from this the following equation is formed:

    \[\frac{6}{n}\times\frac{5}{n-1}=\frac{1}{3}\]

This can then be simplified and rearranged to give the required equation:

    \[\frac{6\times5}{n(n-1)}=\frac{1}{3}\]

    \[\frac{30}{n^2-n}=\frac{1}{3}\]

    \[3\times30=1(n^2-n)\]

    \[90=n^2-n\]

    \[n^2-n-90=0\]

 

Part (b)

For part (b) students were required to solve the quadratic equation, n^2-n-90=0; solving the equation will give two answers, but as the c term (the number term) of the equation is negative, one of these numbers will give a negative value for n and can therefore, be rejected. There are many methods for solving quadratics, in this case I have simply chosen two factors of 90, which when added give -1; more information on solving quadratics is available here.

    \[n^2-n-90=0\]

    \[(n-10)(n+9)=0\]

 

    \[(n-10)=0\]

    \[n=10\]

 

    \[(n+9)=0\]

    \[n=-9\]

 

As n must be positive n=-9 is rejected; therefore, the number of sweets in the bag, n, is 10.

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