Probability of Drawing Balls of Different Colors from a Bag
Imagine a bag containing a mix of colored balls. This article explores the probability of drawing two balls of different colors from such a bag. We will use combinatorial mathematics to solve a real-world problem and provide a detailed explanation of the solution process.
Problem and Initial Setup
The problem at hand is to find the probability that two balls drawn from a bag are of different colors. The initial setup includes a bag containing 3 red, 4 green, and 7 yellow balls, making a total of 14 balls. We are asked to calculate the probability that the two balls drawn are of different colors.
Favorable Cases: Different Colors
To find the number of favorable cases where the two balls are of different colors, we need to consider all possible combinations of drawing balls of different colors. The number of ways to choose 1 red and 1 green ball, 1 red and 1 yellow ball, or 1 green and 1 yellow ball can be calculated using the combination formula (_n C_r frac{n!}{r!(n-r)!}).
Red and Green: The number of ways to choose 1 red from 3 and 1 green from 4 is:
(3C1 * 4C1 3 * 4 12)
Red and Yellow: The number of ways to choose 1 red from 3 and 1 yellow from 7 is:
(3C1 * 7C1 3 * 7 21)
Green and Yellow: The number of ways to choose 1 green from 4 and 1 yellow from 7 is:
(4C1 * 7C1 4 * 7 28)
Adding these up, the total number of favorable cases is:
(12 21 28 61)
The total number of ways to draw 2 balls from 14 is:
(14C2 frac{14!}{2!(14-2)!} frac{14 * 13}{2 * 1} 91)
The probability that the two balls are of different colors is:
(frac{61}{91})
Complementary Probability: Both Balls of the Same Color
We can also approach the problem using the complementary probability method. Instead of directly calculating the probability of drawing balls of different colors, we calculate the probability of drawing balls of the same color and subtract it from 1.
Both Balls Red:
(5C1 * 4C1 / 14C2 20 / 91)
Both Balls Green:
(4C1 * 3C1 / 14C2 12 / 91)
Both Balls Yellow:
(7C1 * 6C1 / 14C2 42 / 91)
The probability of drawing both balls of the same color is:
(frac{20 12 42}{91} frac{74}{91})
The probability of drawing balls of different colors is:
(1 - frac{74}{91} frac{17}{91})
Conclusion
Using combinatorial mathematics, we can solve real-world probability problems involving the drawing of balls from a bag. The two different methods, direct calculation and complementary probability, provide a clear and comprehensive solution to the problem.