Understanding Expected Value in Probability: Drawing Red Balls from a Bag

Probability theory is a powerful tool that helps us understand and predict the outcomes of random events. One common problem in probability is calculating the expected number of occurrences of a particular outcome in a series of random draws. In this article, we will explore this concept through a real-world problem: drawing red balls from a bag without replacement.

Problem Statement

Suppose we have a bag that contains 6 blue balls and 4 red balls. We randomly draw 3 balls from the bag without replacement. What is the expected number of red balls that will be drawn?

Concept of Expectation in Probability

To solve this problem, we use the concept of expectation, which is the long-run average value of repetitions of the experiment it represents. Let's denote the number of red balls drawn as (X), which can take on the values 0, 1, 2, or 3. We aim to calculate the expected value (E[X]), which is the sum of the products of each possible outcome and its probability.

The expectation can be formulated as:

[ E[X] sum_{k0}^{n} k cdot P(X k) ]

Step-by-Step Solution

Step 1: Calculate Total Balls

The total number of balls is the sum of blue and red balls:

[ 6 text{ blue} 4 text{ red} 10 text{ total} ]

Step 2: Calculate Probabilities

We can find the probability of drawing 0, 1, 2, or 3 red balls by using combinations. The formula for combinations is:

[ binom{n}{k} frac{n!}{k!(n-k)!} ]

Let's calculate the probabilities for each case.

Probability of Drawing 0 Red Balls (3 Blue Balls)

[ P(X 0) frac{binom{6}{3} cdot binom{4}{0}}{binom{10}{3}} frac{20 cdot 1}{120} frac{20}{120} frac{1}{6} ]

Probability of Drawing 1 Red Ball (2 Blue Balls)

[ P(X 1) frac{binom{6}{2} cdot binom{4}{1}}{binom{10}{3}} frac{15 cdot 4}{120} frac{60}{120} frac{1}{2} ]

Probability of Drawing 2 Red Balls (1 Blue Ball)

[ P(X 2) frac{binom{6}{1} cdot binom{4}{2}}{binom{10}{3}} frac{6 cdot 6}{120} frac{36}{120} frac{3}{10} ]

Probability of Drawing 3 Red Balls

[ P(X 3) frac{binom{6}{0} cdot binom{4}{3}}{binom{10}{3}} frac{1 cdot 4}{120} frac{4}{120} frac{1}{30} ]

Step 3: Calculate Expected Value

Now, we can calculate the expected value:

[ E[X] 0 cdot P(X 0) 1 cdot P(X 1) 2 cdot P(X 2) 3 cdot P(X 3) ]

[ E[X] 0 cdot frac{1}{6} 1 cdot frac{1}{2} 2 cdot frac{3}{10} 3 cdot frac{1}{30} ]

[ E[X] 0 frac{1}{2} frac{6}{10} frac{3}{30} ]

[ E[X] frac{1}{2} frac{3}{5} frac{1}{10} ]

[ E[X] frac{5}{10} frac{6}{10} frac{1}{10} frac{12}{10} 1.2 ]

Conclusion

The expected number of red balls drawn from the bag is 1.2. This means that if we repeat the experiment many times, the average number of red balls drawn will be 1.2.

Alternative Approach

Another way to approach this problem is by considering the individual probability of drawing a red ball in each draw. The probability of drawing a red ball from a bag of 10 balls is:

[ frac{4}{10} 0.4 text{ (probability of red ball on each draw)} ]

Since we are drawing 3 balls, the expected number of red balls is:

[ 3 cdot 0.4 1.2 ]

This alternate approach supports our earlier calculation and confirms the expected number of red balls.