Probability of Drawing Red Marbles Without Replacement

This article delves into the intriguing question of how many marbles must be drawn in sequence to ensure a 75% probability of drawing a red marble. We will use a jar containing 15 white and 45 red marbles as an example to explore this problem using probability theory.

Understanding the Problem

Suppose you have a jar with two colors of marbles: 15 white and 45 red. The task is to determine the minimum number of marbles that need to be drawn, without replacement, to guarantee a 75% chance of drawing at least one red marble.

Probability Calculation: Not Drawing a Red Marble

To solve this problem, we will start by finding the probability of not drawing a red marble in each draw. This approach helps us to calculate the complementary probability, which is the probability of drawing at least one red marble in the required number of draws.

Initial Probability Calculation

We begin by calculating the probability of not drawing a red marble in the first draw. The total number of marbles initially is 60 (15 white 45 red).

Probability of not drawing a red marble (Pnot red1): [text{P(not red1)} frac{15}{60} frac{1}{4} 0.25]

Subsequent Probability Calculation

Once a white marble has been drawn, the total number of remaining marbles is 59. The probability of not drawing a red marble in the second draw, given that a white marble has been drawn in the first draw, is calculated as:

[text{P(not red2 | not red1)} frac{14}{59}]

Combining these probabilities, the probability of not drawing a red marble in the first two draws is:

[text{P(not red1 and not red2)} frac{15}{60} times frac{14}{59} frac{15 times 14}{60 times 59} frac{1514}{60 times 59} approx 0.059]

Continuing this process, the probability of not drawing a red marble in the first three draws is:

[text{P(not red1 and not red2 and not red3)} frac{15}{60} times frac{14}{59} times frac{13}{58} frac{151413}{60 times 59 times 58} approx 0.019]

By the time we reach the fourth draw, the probability of not drawing a red marble significantly decreases. Calculating this probability step by step is cumbersome, but it helps us understand the trend.

Complementary Probability

The probability of drawing at least one red marble is the complement of the probability of not drawing a red marble. Thus, we can express this as:

[text{P(at least one red)} 1 - text{P(not red1 and not red2 and not red3 ... and not redn)}]

Using the calculations from above:

[text{P(at least one red)} 1 - 0.25 times frac{14}{59} times frac{13}{58} times frac{12}{57} times frac{11}{56} times frac{10}{55} times frac{9}{54}]

Through this calculation, we can see that by the time we have drawn 4 marbles, the probability of having at least one red marble is quite high. In fact, by the time we have drawn 4 marbles, we can be quite sure of drawing a red marble.

Conclusion

In conclusion, by drawing 4 marbles without replacement from a jar containing 15 white and 45 red marbles, we can be 75% sure of drawing at least one red marble. This probability ensures a high likelihood of success, making it a practical and reliable method for solving similar probability problems.

For more in-depth studies on probability and combinatorics, refer to further reading on the topic.