Probability of Drawing at Least One Green Marble: A Comprehensive Guide

In a box, there are 10 red marbles, 20 blue marbles, and 30 green marbles. If 5 marbles are drawn from the box, what is the probability that at least one will be green? This article aims to provide a clear, step-by-step solution to this problem, understanding the nuances of combinatorial probability.

Understanding the Problem

The total number of marbles in the box is 60. We need to calculate the probability of drawing at least one green marble when 5 marbles are drawn randomly. This can be approached using the principle of complementary probability, which simplifies the calculation as follows:

[ P(text{at least one green}) 1 - P(text{no green}) ]

Calculation

Using Complementary Probability

We start by calculating the probability of drawing no green marbles. If no green marbles are drawn, all 5 marbles must be drawn from the 30 non-green marbles.

The number of ways to choose 5 marbles from 30 non-green marbles is given by:

[ binom{30}{5} ]

The total number of ways to choose 5 marbles from the 60 marbles in the box is:

[ binom{60}{5} ]

Therefore, the probability of drawing no green marbles is:

[ P(text{no green}) frac{binom{30}{5}}{binom{60}{5}} ]

The probability of drawing at least one green marble is then:

[ P(text{at least one green}) 1 - frac{binom{30}{5}}{binom{60}{5}} ]

Let's calculate this step by step. First, we calculate (binom{30}{5}) and (binom{60}{5}).

[ binom{30}{5} frac{30!}{5!(30-5)!} frac{30!}{5! cdot 25!} ] [ binom{60}{5} frac{60!}{5!(60-5)!} frac{60!}{5! cdot 55!} ]

Substituting these values, we get:

[ P(text{no green}) frac{frac{30!}{5! cdot 25!}}{frac{60!}{5! cdot 55!}} frac{30! cdot 55!}{5! cdot 25! cdot 60!} cdot frac{5! cdot 55!}{1} frac{30! cdot 55!}{25! cdot 60!} ]

Finally, the probability of drawing at least one green marble is:

[ P(text{at least one green}) 1 - frac{30! cdot 55!}{25! cdot 60!} ]

Using a calculator or software to compute these values, we can find the exact probability.

Alternative Calculation

Another approach to solving this problem is to consider all the possible ways of drawing at least one green marble. We can calculate the probability of drawing exactly 1, 2, 3, 4, or 5 green marbles and sum these probabilities. However, this method is more complex and time-consuming compared to the first method.

For simplicity, we use the first method, which is more efficient and accurate.

Conclusion

By using the principle of complementary probability, we can simplify the calculation of probability problems involving drawing multiple items from a set. In this case, the probability of drawing at least one green marble from a box containing 10 red, 20 blue, and 30 green marbles is given by the formula:

[ P(text{at least one green}) 1 - frac{binom{30}{5}}{binom{60}{5}} ]

Calculating the exact value requires using a calculator or software, but this method provides a clear and efficient way to solve the problem.

Comments

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