Answer to Question #103887 in Statistics and Probability for nicole

Question

A bag contains an assortment of red and blue balls. If two balls are drawn from it at random, the probability that 2 re balls are drawn is 5 times the probability that 2 blue balls are drawn. Furthermore, the probability that one ball of each color is drawn is 6 times the probability that 2 blue balls are drawn. How many red balls are there in the bag?

Answer 1

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.