Picking Colored BallsDate: 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/ |
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