Solution: We calculate the probability of drawing exactly 2 red marbles and 2 blue marbles from a total of 16 marbles (7 red, 9 blue), without replacement.

Title: Probability of Drawing Exactly 2 Red and 2 Blue Marbles from a Bag Ã¢ÂÂ Step-by-Step Calculation

When sampling without replacement, understanding the likelihood of mixed outcomes becomes essential in probability and statistics. In this article, we explore how to calculate the probability of drawing exactly 2 red marbles and 2 blue marbles from a total of 16 marbles Ã¢ÂÂ 7 red and 9 blue Ã¢ÂÂ using combinatorial methods.

Understanding the Context

Problem Statement

We have a bag containing:
- 7 red marbles
- 9 blue marbles
Total = 16 marbles

We draw 4 marbles without replacement and want to compute the probability of obtaining exactly 2 red and 2 blue marbles.

[ANSWERED] A bag contains 2 green marbles, 4 red marbles, and 3 blue ...

Image Gallery

Solved: A bag contains 2 red marbles and 4 blue marbles. What is the ...

Solved: A bag contains 3 red marbles, 6 blue marbles and 4 green ...

Solved: A bag contains 5 red marbles 6 blue marbles 3 green marbles 4 ...

Key Insights

Why Use Combinations?

Because the drawing is without replacement and order doesnÃ¢ÂÂt matter, we use combinations to count possible selections. The probability is calculated as:

[
P(2, \ ext{red and } 2, \ ext{blue}) = rac{\ ext{Number of favorable outcomes}}{\ ext{Total number of possible 4-marble combinations}}
]

Step 1: Calculate Total Possible Outcomes

🔗 Related Articles You Might Like:

Final Thoughts

The total number of ways to choose any 4 marbles from 16 is given by the combination formula:

[
inom{n}{k} = rac{n!}{k!(n-k)!}
]

So,

[
inom{16}{4} = rac{16!}{4! \cdot 12!} = 1820
]

Step 2: Calculate Favorable Outcomes

We want exactly 2 red marbles 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
]

Since the selections are independent, multiply the combinations: