# How Many Balls?

A bag contains red and blue balls. You are told the probabilities
of drawing certain combinations of balls. Find how many red and how
many blue balls there are in the bag.

## Problem

A bag contains an assortment of red and blue balls. Two balls are drawn at random.

You are told that the probability of drawing two red balls is 5 times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each colour is 6 times the probability of drawing two blue balls.

How many blue and how many red balls are in the bag?

## Getting Started

Decide on the unknowns and assign letters to them. Take the
information given step by stepand write down
expressions for the probabilities. Can you obtain an
equation and solve it?

## Student Solutions

This is another good solution from Alan of Madras College.

Call the number of red balls $r$ and the number of blue balls $b$. The probability of drawing a red ball is $r/(b+r)$ leaving $b$ blue balls and $r-1$ red balls and so the probability of drawing a second red ball is $(r-1)/(b+r-1)$ and the probability of drawing two red balls is $r(r-1)/(b+r)(b+r-1)$. The question tells us that this is five times the probability of drawing two blue balls.

By a similar argument, the probability of drawing two blue balls is $b(b-1)/(b+r)(b+r-1)$.

Also, after one red ball has been drawn the probability of drawing a blue ball is $b/(b+r-1)$ and, considering that you can draw the balls in either order, the probability of a red and a blue ball is $2rb/(b+r)(b+r-1)$. This is six times the probability of drawing two blue balls.

Hence $$\frac{r(r - 1)}{(b + r)(b + r - 1)} = \frac{5b(b - 1)}{(b + r)(b + r - 1)}$$ and we know $(b + r)(b + r - 1) \neq 0$.

$$\eqalign{ r (r - 1) &= 5b (b - 1) &(1) \\ 2 rb &= 6b (b - 1) &(2)}$$

From equation (2)

$r = 3b - 3$.

Substituting for $r$ in equation (1) and simplifying the equation gives

$b^2 - 4b + 3 = 0$.

The solutions are $b = 1$ or $b = 3$.

In the case $b = 1$ we have $r = 3b -3 = 0$, but we know $r$ is non zero so $b \neq 1$.

Hence $b = 3$ and $r = 6$, that is there were 6 red balls and 3 blue balls in the bag.

## Teachers' Resources

### Why try this problem?

To practice interpreting verbal information, writng down expressions for probabilities, putting the information together to obtain an equation and solving the equation.### Key questions

What are the unknowns?What are the probabilities?

Can we obtain an equation from the information given?

### Possible support

If some learners can't get started suggest they have a bag and they decide on how many red and how many blue balls are in their bag ( say 4 and 7) and then they work out the probability of drawing two red balls and of drawing two blue balls and of drawing one ball of each colour from their bag. Suggest that after that they use letters for the numbers of red and blue balls (which are unknown) and work out the probabilities in a similar way.Also try Fixing The Odds