Question: A box contains 7 red marbles and 9 blue marbles. If four marbles are drawn without replacement, what is the probability that exactly two are red? - Parker Core Knowledge
Understanding the Probability of Drawing Exactly Two Red Marbles from a Box (7 Red & 9 Blue)
Understanding the Probability of Drawing Exactly Two Red Marbles from a Box (7 Red & 9 Blue)
When faced with a probabilistic scenario like drawing marbles from a bag, understanding the underlying math not only satisfies curiosity but also strengthens critical thinking. A particularly common and clear example involves calculating the likelihood of selecting exactly two red marbles out of four drawn without replacement from a box containing 7 red and 9 blue marbles. In this article, weâÃÂÃÂll break down the problem step by step to explain how to calculate this probability and why it matters.
Understanding the Context
The Problem: Drawing 4 Marbles Without Replacement
We are given:
- 7 red marbles
- 9 blue marbles
- Total marbles = 7 + 9 = 16
- Number of marbles drawn = 4 (without replacement)
- We want exactly 2 red and 2 blue marbles
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Key Insights
Why Use Combinations?
Since the marbles are drawn without replacement, each draw affects the composition of the remaining marbles. This is a hypergeometric distribution scenarioâÃÂÃÂideal for calculating probabilities involving finite populations with two outcomes (here, red or blue).
Instead of calculating probabilities for each individual order (like R-R-B-B), we compute all favorable outcomes over total possible outcomes.
Step-by-Step Calculation
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Step 1: Determine total ways to draw 4 marbles from 16
This is the total number of possible combinations:
[
inom{16}{4} = rac{16!}{4!(16-4)!} = rac{16 \ imes 15 \ imes 14 \ imes 13}{4 \ imes 3 \ imes 2 \ imes 1} = 1820
]
Step 2: Count the favorable outcomes (exactly 2 red and 2 blue marbles)
- Ways to choose 2 red marbles from 7:
[
inom{7}{2} = rac{7 \ imes 6}{2 \ imes 1} = 21
]
- Ways to choose 2 blue marbles from 9:
[
inom{9}{2} = rac{9 \ imes 8}{2 \ imes 1} = 36
]
Multiply these to get all favorable combinations:
[
21 \ imes 36 = 756
]
Step 3: Compute the probability
[
P(\ ext{exactly 2 red}) = rac{\ ext{favorable outcomes}}{\ ext{total outcomes}} = rac{756}{1820}
]
Simplify the fraction:
Divide numerator and denominator by 4:
[
rac{756 \div 4}{1820 \div 4} = rac{189}{455}
]
This fraction is in simplest form, though decimal approximation is useful:
