Probability of Drawing Three Milk Chocolates from a Box

Introduction

In this article, we will explore the probability of drawing three milk chocolates from a box containing 12 chocolates—4 dark and 8 milk chocolates—without replacement. We will apply combinatorial probability and provide a detailed step-by-step solution.

Problem Statement

Suppose a box contains 12 chocolates, 4 of which are dark, and 8 of which are milk. Three chocolate chips are randomly drawn one after the other without replacement. What is the probability that all three drawn chocolates are milk?

Step-by-Step Solution

Step 1: Total Number of Ways to Draw 3 Chocolates from 12

To find the total number of ways to draw 3 chocolates from a box of 12, we can use the combination formula:

Combination Formula:
(binom{n}{r} frac{n!}{r!(n-r)!})

Here, (n 12) (total chocolates) and (r 3) (chosen chocolates). Therefore:

[binom{12}{3} frac{12!}{3!9!} frac{12 times 11 times 10}{3 times 2 times 1} 220]

Step 2: Number of Ways to Draw 3 Milk Chocolates from 8

The number of ways to choose 3 milk chocolates from the 8 available is also a combination problem:

[binom{8}{3} frac{8!}{3!5!} frac{8 times 7 times 6}{3 times 2 times 1} 56]

Step 3: Calculate the Probability of Drawing 3 Milk Chocolates

The probability (P) of drawing 3 milk chocolates is the number of favorable outcomes divided by the total number of outcomes:

[P(text{3 milk chocolates}) frac{binom{8}{3}}{binom{12}{3}} frac{56}{220}]

Step 4: Simplify the Probability

To simplify (frac{56}{220}), we find the greatest common divisor (GCD) of 56 and 220, which is 4. Dividing both the numerator and the denominator by 4:

[frac{56 div 4}{220 div 4} frac{14}{55}]

Therefore, the probability that all three drawn chocolates are milk is:

[boxed{frac{14}{55}}]

Alternative Calculation Method

A second method involves directly calculating the probability step-by-step:

[frac{8}{12} times frac{7}{11} times frac{6}{10}]

Let's simplify these fractions before multiplying:

[frac{2}{3} times frac{7}{11} times frac{3}{5}]

We can cross-simplify the 3s to get:

[frac{2 times 7}{11 times 5} frac{14}{55}]

This confirms our earlier result.

Practical Interpretation

When we calculate the individual probabilities for each draw, we have:

First draw: (frac{8}{12} frac{2}{3} approx 66.67%) Second draw: (frac{7}{11} approx 63.64%) Third draw: (frac{6}{10} frac{3}{5} 60%)

Multiplying these probabilities together:

[frac{2}{3} times frac{7}{11} times frac{3}{5} approx frac{14}{55} approx 25.45%]

This confirms that the probability of drawing three milk chocolates in a row is approximately 25.45%.

Conclusion

Using both the combinatorial approach and the step-by-step probability calculation, we have determined that the probability of drawing three milk chocolates from the box is (frac{14}{55}) or approximately 25.45%. This shows the power of combinatorial methods in solving real-world probability problems.