Probability of Drawing at Least One White Ball in Three Draws

A classic problem in probability involves a bag containing 6 white, 3 red, and 9 black balls. When drawing three balls one by one with replacement, what is the probability that at least one of the balls drawn is white?

Introduction

Probability problems often require us to consider the possible outcomes and calculate their likelihood. This example showcases a step-by-step approach to solving such problems using the complement rule, which involves finding the probability of the opposite event first and then calculating the required probability as the complement.

Step-by-Step Solution

Step 1: Total Number of Balls

The total number of balls in the bag is:

6 white 3 red 9 black

Total 6 3 9 18 balls.

Step 2: Probability of Drawing a Non-White Ball

The number of non-white balls (red and black combined) is:

3 red 9 black

Non-white balls 3 9 12.

The probability of drawing a non-white ball in one draw is:

Pnon-white Number of non-white balls}{Total number of balls 12}{18} 2}{3}

Step 3: Probability of Drawing Non-White Balls in Three Draws

Since the draws are with replacement, the probability of drawing a non-white ball in all three draws is:

Pall non-white Pnon-white3 (2}{3})3 8}{27}

Step 4: Probability of Drawing at Least One White Ball

Now, we use the complement rule. The probability of drawing at least one white ball is the complement of drawing no white balls:

Pat least one white 1 - Pall non-white 1 - 8}{27} 27 - 8}{27} 19}{27}

Final Result

The probability that at least one of the three balls drawn is white is:

bphoto{frac{19}{27}}

Alternative Approach

Alternatively, we can approach this problem by calculating the probability that none of the three balls drawn is white. This complementary probability can be calculated as follows:

12}{18} 8}{27}

Now, subtract this probability from 1 to get the probability of drawing at least one white ball:

1 - 19}{27}

Conclusion

This detailed step-by-step solution showcases the use of the complement rule and direct calculation methods to solve probability problems involving multiple draws with replacement. Whether you use the straightforward calculation or the alternative method, the final answer is consistent, validating the robustness of the approach.