Introduction

Welcome to this article on calculating the probability of drawing a yellow marble on the third draw from a bag of marbles without replacement. This is a fundamental concept in probability theory that is relevant in a wide range of applications, including statistics, data analysis, and even real-life decision-making scenarios.

Understanding the Problem

In this problem, we have a bag containing a total of 12 marbles, with 3 blue, 4 yellow, 2 red, and 3 green marbles. The goal is to calculate the probability of drawing a yellow marble for the first time on the third draw. This is a classic example of a problem in combinatorics and probability theory, specifically focusing on the concept of 'without replacement,' where the probability changes with each draw.

Total Marbles

Total Marbles:
Blue: 3
Yellow: 4
Red: 2
Green: 3
Total marbles 3 4 2 3 12

Step-by-Step Calculation

First Draw

The probability that the first marble drawn is not yellow (i.e., it is blue, red, or green) is calculated as follows:

P(not yellow on 1st) frac{8}{12} frac{2}{3}

Here, there are 8 non-yellow marbles (8 out of 12).

Second Draw

After the first non-yellow marble is drawn, there are now 11 marbles left in the bag, with 7 of those being non-yellow. The probability of drawing again a non-yellow marble is:

P(not yellow on 2nd not yellow on 1st) frac{7}{11}

Here, there are now 7 non-yellow marbles (7 out of 11).

Third Draw

After two non-yellow marbles have been drawn, there are now 10 marbles left in the bag, with 4 of them being yellow. The probability of drawing a yellow marble on the third draw is:

P(yellow on 3rd not yellow on 1st and 2nd) frac{4}{10} frac{2}{5}

Total Probability

To find the total probability of this specific sequence of draws, we multiply the probabilities of each individual step:

P(not yellow on 1st) x P(not yellow on 2nd not yellow on 1st) x P(yellow on 3rd not yellow on 1st and 2nd) (frac{2}{3} times frac{7}{11} times frac{2}{5} frac{2 times 7 times 2}{3 times 11 times 5} frac{28}{165})

Conclusion

The probability of drawing the first yellow marble on the third draw is (frac{28}{165}).

Combinatorial Approach

Alternatively, we can calculate this using combinations. There are 12 marbles in total, and we are choosing 3 marbles in a specific sequence. The probability of choosing two non-yellow marbles first, and then a yellow marble:

P(not yellow on 1st and 2nd) frac{binom{8}{2}}{binom{12}{2}} frac{28}{66} frac{14}{33}

Here, C(n, k) frac{n!}{k!(n-k)!})

After drawing two non-yellow marbles, there are 10 marbles left, with 4 being yellow. The probability of drawing a yellow marble on the third draw is:

P(yellow on 3rd) frac{4}{10} frac{2}{5}

Therefore, the required probability of choosing the first yellow marble on the third draw is:

((frac{14}{33} times frac{2}{5}) (frac{28}{165} approx 0.169697)

Conclusion

This article has walked through the step-by-step calculation of the probability of drawing a yellow marble on the third draw from a bag of 12 marbles, with 4 of them being yellow. The process involves understanding the concept of without replacement and using either direct probability multiplication or combinatorial methods to arrive at the solution. This type of problem is not only an excellent example of probability theory but also a valuable tool in statistical analysis and real-world applications.