Associated Topics || Dr. Math Home || Search Dr. Math

### Picking Colored Balls

```
Date: 05/05/97 at 21:51:25
From: Connie
Subject: Probability

A bag contains an assortment of red and blue balls. If I draw two
balls at random, the probability that I'll draw two red balls is five
times the probability I'll draw two blue balls. The probability that I
will draw one ball of each color is six times the probability that I
will draw two blue balls. How many red and blue balls are in the bag?

Connie
```

```
Date: 05/06/97 at 08:56:27
From: Doctor Anthony
Subject: Re: Probability

Connie,

Let x = number of red balls and y = number of blue balls.  If we let n
represent the total number of balls in the bag, then x+y = n.

First, what is the probability of drawing one red ball?  It is the
number of red balls in the bag divided by the total number of balls in
the bag, which is x/n.  Once a red ball has been drawn, there is one
less red ball in the bag.  What is the probability of drawing a red
ball in this revised situation?  Well, there are now (x-1) red balls
in the bag and a total of (n-1) balls in the bag. So the probability
of drawing a red ball now is (x-1)/(n-1).

The probability of drawing two red balls in a row is the product of
the two probabilities that we just calculated.  So, the probability of
drawing 2 red balls = x(x-1)/[n(n-1)].

We can do the same type of analysis for the blue balls and find that
the probability of drawing two blue balls = y(y-1)/[n(n-1)].

We are told the first of these probabilities is five times the second
one, which means that x(x-1)/[n(n-1)] = 5y(y-1)/[n(n-1)]. This
simplifies to x(x-1) = 5y(y-1).  We will call this equation (1).

What is the probability of drawing one ball of each color?  It is the
probability of drawing a red ball times the probability of then
drawing a blue ball PLUS the probability of drawing a blue ball times
the probability of then drawing a red ball. We already know that the
probability of drawing a red ball is x/n. After a red ball has been
drawn, there are (n-1) balls left and y blue balls left, so the
probability of drawing a blue ball now is y/(n-1). This means that
the probability of drawing a red ball and then a blue ball is
x/n * y/(n-1) = xy/[n(n-1)].

Similarly, we already know that the probability of drawing a blue ball
is y/n. The probability of drawing a red ball after having drawn a
blue ball is x/(n-1). So the probability of drawing a blue ball and
then a red ball is y/n * x/(n-1) = xy/[n(n-1)].

Now that we have the probability of drawing a red ball and then a blue
ball as well as the probability of drawing a blue ball and then a red
ball, we can add these probabilities to get the probability of drawing
one ball of each color. So, the probability of drawing one ball of
each color is: xy/[n(n-1)] + xy/[n(n-1)] = 2xy/[n(n-1)].

We are told that the probability of drawing one ball of each color is
6 times the probability of picking two blue balls.  We know both of
these values, so we can write:

2xy/[n(n-1)] = 6y(y-1)/[n(n-1)]
2xy = 6y(y-1)
xy = 3y(y-1)
x = 3(y-1)   We will call this equation (2).

Now we can evaluate x and y using equations (1) and (2).

x(x-1) = 5y(y-1)  Equation (1)
3(y-1)(3y-3-1) = 5y(y-1)  Substituting equation (2)
3(3y-4) = 5y       Divide out (y-1)
9y - 12 = 5y
4y = 12
y = 3

We substitute this value for y back into equation (2):

x = 3(y-1)
x = 3(3-1)
x = 3(2)
x = 6

Therefore, we have 6 red balls and 3 blue balls.

-Doctors Anthony and Rachel,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/