Ask Uncle Colin: a load of balls

Dear Uncle Colin,

I struggled with a problem where you had 5 blue balls and x green balls, and the probability of picking two blue balls out of the bag was $\frac{5}{14}$. I can’t really see where to start!

-Baffled About Likelihood, Lacking Startpoint

Hi, BALLS, thanks for your message!

This is the kind of question that tests your abstract thinking – taking something you know how to do with numbers and asking you to do it with algebraic expressions instead.

With numbers

If you knew what x was – suppose it’s 7 – you’d be able to work out the probability of getting two blue balls out without too much bother: the probability of the first ball being blue would be $\frac{5}{5+7}$ and the second ball also being blue would be $\frac{4}{4+7}$. You’d then multiply those together to get the probability of both, which is $\frac{20}{12 \times 11} = \frac{5}{33}$ after a bit of cancelling.

Sadly, that’s not the answer we’re after, since it’s not $\frac{5}{14}$.

But with algebra…

… you do exactly the same. Instead of 7, you put an x. The probability of the first ball being blue is $\frac{5}{5+x}$ and the second, $\frac{4}{4+x}$. Multiplying those gives a slightly messy $\frac{20}{(5+x)(4+x)}$, and we know that must equal $\frac{5}{14}$.

We have the same thing written two different ways, so we can make an equation: $\frac{20}{(5+x)(4+x)} = \frac{5}{14}$.

Now deal with the fractions

Multiplying each side by the pair of brackets on the bottom on the left, and by 14, gives: 280 = 5(5 + x)(4 + x), which is already a lot nicer – but can be made nicer still if you divide everything by 5.

56 = (5 + x)(4 + x).

The standard approach

The standard way to tackle this is to expand the right hand side and simplify into a quadratic in the usual form, then solve it: 56 = 20 + 9x + x2, so x2 + 9x − 36 = 0.

That factorises as (x + 12)(x − 3)=0, which means x = −12 (not possible, unless someone has invented negative balls) or x = 3.

A bit more piratey

If you know 56 = (5 + x)(4 + x), and spot that the first bracket is one more than the second, you might spot that 5 + x = 8 and 4 + x = 7 would be an excellent solution to the equation – making x = 3 directly.

The downside to this approach is that you have to check there are no others – 5 + x = −7 and 4 + x = −8 would also work, giving x = −12, although that’s impossible (as above).

Hope that helps!

– Uncle Colin

* Edited 2018-11-07 to turn a 5 into a 7.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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