# Sweet Probability

“There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.” Hannah takes and eats two random sweets from the bag. It says the probability that she eats two orange sweets is 1/3, then asks pupils to show that n² – n – 90 = 0. The question in Thursday’s Edexcel exam prompted despair in many youngsters, who complained it was far too difficult. A lot of students panicked over this question. However, working step by step it is relatively simple algebra.

You only need to know two simple facts:

1) The definition of probability is successful outcomes / possible outcomes.

2) In a test with independent outcomes. If you know that the probability of event *e* is *P(e)*, and the probability of event *f* is *P(f)* Then the outcome of *e & f* – that is, of having *e* as the outcome of the first trial, and *f* as the outcome of the second, is *P(e)×P(f)*.

The probability that the first sweet is orange is:

**6/n**

Having eaten the first sweet there are now 5 orange sweets and one less sweet in the bag so the probability of the second sweet being orange is:

**5/(n-1)**

Multiplying the two probabilities gives.

It says this is 1/3:

Rearranging and simplifying:

**n (n-1) = 3 x 30**

**n² – n – 90 = 0**

Q.E.D. (sweet!)

The question could have been extended as we can now work out the value of n by solving the quadratic equation.

Using our LEAF system to remember equations we know that if,

**an² + bn + c = 0**

Then

*a=1, b=-1, c=-90*

Substituting in the values gives:

The square root of 361 is 19 so:

**n=10 or n=-9**

As we are dealing with real bags of sweets, not antimatter, then n=10 and since there were 6 oranges sweets to start with there must have been 4 yellow sweets.