Probability of Drawing Same-Colored Balls from a Box

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SUMMARY

The probability of drawing two balls of the same color from a box containing 2 red, 4 white, and 4 green balls is calculated by considering each color's probabilities. For red, the probability is calculated as $\frac{2}{10} \cdot \frac{1}{9}$. For white and green, the probabilities are both $\frac{4}{10} \cdot \frac{3}{9}$. The total probability of drawing two balls of the same color is the sum of these probabilities: $$\frac{2}{10} \cdot \frac{1}{9} + \frac{4}{10} \cdot \frac{3}{9} + \frac{4}{10} \cdot \frac{3}{9}$$ which simplifies to a definitive result.

PREREQUISITES
  • Basic understanding of probability theory
  • Knowledge of combinatorial counting principles
  • Familiarity with drawing without replacement
  • Ability to perform basic arithmetic operations with fractions
NEXT STEPS
  • Study the concept of conditional probability in depth
  • Learn about combinatorial probability and its applications
  • Explore the concept of drawing with and without replacement
  • Investigate more complex probability scenarios involving multiple events
USEFUL FOR

Students of probability, educators teaching probability concepts, and anyone interested in understanding the mechanics of drawing objects from a finite set.

rymatson406
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A box contains 2 red, 4 white, and 4 green two balls are drawn in succession without replacement. What is the probability that both balls are the same color?
 
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rymatson406 said:
A box contains 2 red, 4 white, and 4 green two balls are drawn in succession without replacement. What is the probability that both balls are the same color?

The probability is the sum of the probabilities:
  • 1. ball red and 2. ball red
  • 1. ball white and 2. ball white
  • 1. ball green and 2. ball green

The probability "1. ball red and 2. ball red":

The probability to pick a red ball from the box which contains $10$ balls, where $2$ of them are red, is $\displaystyle{\frac{2}{10}}$.
Now there are $9$ balls left in the box and $1$ of them is red.
So the probability for the second ball to be red is $\displaystyle{\frac{1}{9}}$.

Therefore the probability "1. ball red and 2. ball red" is equal to $$\frac{2}{10} \cdot \frac{1}{9}$$Do the same for the two other cases, and you get:
$$\frac{2}{10} \cdot \frac{1}{9}+\frac{4}{10} \cdot \frac{3}{9}+\frac{4}{10} \cdot \frac{3}{9}$$
 
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