Math Problem Statement
Solution
To solve this problem, let’s define the components and calculate the probability of drawing two counters of the same color.
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Counters in the Bag:
- The total number of black counters is .
- The total number of green counters is .
- The total number of counters is .
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Possible Outcomes: We want to calculate the probability of drawing two counters of the same color. This can happen in two cases:
- Both counters are black.
- Both counters are green.
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Probability of Drawing Two Black Counters:
- The probability of drawing the first black counter is .
- After one black counter is drawn, there are black counters left out of a total of counters.
- Thus, the probability of drawing a second black counter is .
- Therefore, the probability of drawing two black counters is:
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Probability of Drawing Two Green Counters:
- The probability of drawing the first green counter is .
- After one green counter is drawn, there are green counters left out of a total of counters.
- Thus, the probability of drawing a second green counter is .
- Therefore, the probability of drawing two green counters is:
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Total Probability of Drawing Two Counters of the Same Color: Now, we add the probabilities of the two cases (both black or both green): Simplifying the numerator: Simplify further:
Thus, the probability of drawing two counters of the same color is indeed , as required.
Would you like further clarification on any steps?
Related Questions:
- What would the probability be if there were an odd number of black counters?
- How does the probability change if we draw with replacement?
- What if there were green counters instead of ?
- How can we calculate the probability for more than two counters drawn?
- Can we generalize this probability formula for any two colors?
Tip:
When working with probabilities, identify and separate each possible case, then combine the cases carefully.
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Math Problem Analysis
Mathematical Concepts
Probability
Combinatorics
Independent and Dependent Events
Formulas
Probability of drawing two black counters: \(\frac{2n}{4n+1} \times \frac{2n - 1}{4n}\)
Probability of drawing two green counters: \(\frac{2n + 1}{4n + 1} \times \frac{2n}{4n}\)
Theorems
Addition Rule for Probabilities
Multiplication Rule for Independent Events
Suitable Grade Level
Grades 11-12
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