Probability of Drawing Non-White Balls in a Random Selection
Imagine a bag containing 2 yellow, 2 white, and 2 black balls. We are interested in finding the probability of drawing two balls at random such that none are white. This problem can be approached in several ways, utilizing principles of combinatorics and basic probability theory.
Combinatorial Approach
Starting with the broader combinatorial approach, we realize that the total number of balls in the bag is 6: 2 white, 2 yellow, and 2 black. The total number of ways to draw 2 balls from these 6 is:
[ ^6C_2 frac{6!}{2!(6-2)!} 15 ]
Next, we focus on the favorable outcomes, specifically drawing 2 balls that are not white. This leaves us with 4 non-white balls (2 black 2 yellow). The number of ways to draw 2 balls from these 4 is:
[ ^4C_2 frac{4!}{2!(4-2)!} 6 ]
Therefore, the probability that none of the balls drawn is white is given by the ratio of favorable outcomes to total outcomes:
[ P(text{none is white}) frac{^4C_2}{^6C_2} frac{6}{15} frac{2}{5} 0.4 ]
This calculation demonstrates a fundamental principle in probability theory, which is the direct proportion of favorable outcomes to total possible outcomes.
Alternative Problem Setup
Alternatively, we can ask the question: "What is the probability that both balls are black or white?" This can be loosely interpreted as the complement of the probability that none of the balls are white. The total number of ways to draw 2 balls from the 6 balls:
[ ^6C_2 15 ]
The number of ways to draw 2 balls from the 3 white and 3 non-white balls would be:
[ ^3C_2 ^3C_2 3 3 6 ]
Thus, the probability of drawing 2 balls that are either all white or all non-white (black in this case) is:
[ P(text{both are white or both are non-white}) frac{6}{15} frac{2}{5} ]
Therefore, the probability that at least one of the balls is yellow (none are white) would be the complement of this probability:
[ P(text{none are white}) 1 - P(text{both are white or both are non-white}) 1 - frac{2}{5} frac{3}{5} ]
This alternative approach reinforces the principle that the total probability must sum to 1, and each event's probability can be derived from its complement.
Misleading Setup and Correct Application
An interesting case is presented in a mistaken calculation where balls are assumed to be in multiples of 34, leading to an incorrect total of 342 balls. This scenario, however, does not simplify the problem's core:
Given that the problem contradicts the initial setup, we can reapply the correct combinatorial method to avoid confusion. The correct number of non-yellow balls is 4, and the total number of balls is 6. Hence, the probability that none of the balls drawn is yellow (or, equivalently, white in the context provided) is:
[ P(text{none are yellow}) frac{^4C_2}{^6C_2} frac{6}{15} frac{2}{5} ]
This simplification provides a clear and concise solution, reflecting the principles of combinatorics and probability.
Conclusion
In conclusion, the probability that none of the balls drawn is white in a random selection from a bag of 2 yellow, 2 white, and 2 black balls is ( frac{2}{5} ). Understanding the combinatorial approach and the concept of favorable outcomes over total possible outcomes is key to solving such problems. This solution not only addresses the original problem but also provides a deeper insight into the principles of probability and combinatorics.