Probability of Drawing One Red and One Blue Marble from a Bag

Imagine a scenario where you have a bag containing 5 red marbles and 8 blue marbles. You are tasked with calculating the probability of drawing one marble of each color without replacing the first marble. This article will guide you through the steps to solve this problem and explore the concept of probability in such a situation.

Understanding the Scenario

The bag contains a total of 13 marbles, with 5 red and 8 blue. We are interested in calculating the probability of drawing one red and one blue marble in two draws without replacement. This means once a marble is drawn, it is not returned to the bag before the next draw.

Step-by-Step Calculation

Method 1: Sequential Probability Calculation

One approach to solving this problem is to calculate the probability of each event in sequence and then multiply these probabilities. The first event is drawing a blue marble, and the second event is drawing a red marble.

1. The probability of drawing a blue marble first:

P(Blue) Number of blue marbles / Total number of marbles 8/13

2. After drawing a blue marble, the total number of marbles left is 12, and the number of red marbles is 5. The probability of drawing a red marble second:

P(Red) Number of red marbles / Remaining total number of marbles 5/12

3. The combined probability of these two events is:

P(Blue then Red) P(Blue) * P(Red) (8/13) * (5/12) 40/156 20/39

Method 2: Using Combinations

Another method is to use combinations to calculate the probability directly. The total number of ways to draw 2 marbles from 13 is given by the combination formula (binom{13}{2}).

The number of favorable outcomes (one red and one blue) is given by the product of the combinations of 5 red marbles and 8 blue marbles:

Number of favorable combinations (binom{5}{1} times binom{8}{1}) 5 * 8 40

The total number of ways to choose 2 marbles out of 13 is:

Total combinations (binom{13}{2}) (13 * 12) / (2 * 1) 78

The required probability is then:

P Number of favorable combinations / Total combinations 40 / 78 20 / 39

Conclusion

Through these calculations, we have determined that the probability of drawing one red and one blue marble in two draws without replacement is 20/39. Both methods yield the same result, confirming the accuracy of the calculations.

Final Answer

The probability of drawing one red and one blue marble from a bag containing 5 red and 8 blue marbles (without replacement) is 20/39 or approximately 0.5128.

Additional Calculations for Verification

As verification, let's perform the same calculations for drawing a red marble first and a blue marble second:

1. Probability of drawing a red marble first:

P(Red) 5/13

2. After drawing a red marble, the total number of marbles left is 12, and the number of blue marbles is 8. The probability of drawing a blue marble second:

P(Blue) 8/12 2/3

3. The combined probability of drawing a red and a blue marble in that order is:

P(Red then Blue) P(Red) * P(Blue) (5/13) * (2/3) 10/39

This result, 10/39, is indeed the same as the previous calculation, confirming the consistency of the problem solution.