The Conditional Probability of Drawing Two Red Balls Given at Least One is Red

In probability theory, the concept of conditional probability is a fundamental tool for understanding and solving various problems involving events that depend on each other. One such interesting problem involves drawing balls from a box where the conditions change mid-operation. In this article, we will explore the scenario of selecting two balls from a box containing a specific mix of red, free, and white balls, and how the probability of drawing two red balls changes under certain conditions. This article will cover:

Understanding the events involvedCalculating the probability without replacementCalculating the probability with replacementConclusion and practical applications

Understanding the Events

A box contains 3 red balls, 2 free balls, and 1 white ball. Two balls are selected one after the other without replacement. We need to find the probability that both balls selected are red given that at least one ball is red. To solve this problem, we will use the concept of conditional probability.

Step 1: Define the Events

Let A be the event that both balls are red.

Let B be the event that at least one ball is red.

We want to find P(A | B), the probability that both balls are red given that at least one ball is red. The formula for conditional probability is:

P(A | B) (frac{P(A cap B)}{P(B)})

Step 2: Calculate (P(A cap B))

The event (A) (both balls are red) implies the event (B) (at least one ball is red). Thus, (P(A cap B) P(A)).

- Total number of balls: 6 (3 red, 2 free, 1 white).

- Number of ways to choose 2 balls from 6: (binom{6}{2} 15).

- Number of ways to choose 2 red balls from 3: (binom{3}{2} 3).

- Therefore, (P(A) (frac{3}{15} frac{1}{5}).

Step 3: Calculate (P(B))

To find (P(B)) (at least one ball is red), we can use the complement:

(P(B) 1 - P(text{no red balls})).

- The only balls that can be selected without getting a red ball are the 2 free balls and the 1 white ball (total 3 balls).

- Number of ways to choose 2 balls from these 3: (binom{3}{2} 3).

- Therefore, (P(text{no red balls}) (frac{3}{15} frac{1}{5}).

- So, (P(B) 1 - (frac{1}{5} (frac{4}{5}).

Step 4: Calculate (P(A | B))

Now we can find (P(A | B)):

(P(A | B) (frac{P(A)}{P(B)} (frac{(frac{1}{5}}{(frac{4}{5}} (frac{1}{4})

Conclusion: The probability that both balls selected are red given that at least one ball is red is (boxed{(frac{1}{4}}).

With Replacement

Alternatively, let's consider the scenario where balls are replaced after each draw. The probability of drawing a red ball on the first draw is 3/7, and the same for the second draw. The probability of drawing two red balls in two draws with replacement is:

((frac{3}{7} times (frac{3}{7} (frac{9}{49} approx 18.367%).

Without replacement, the probability is:

((frac{3}{7} times (frac{2}{6} (frac{6}{42} (frac{1}{7} approx 14.286%).

Conclusion and Practical Applications

The example above demonstrates the importance of understanding the difference between drawing with and without replacement. In real-world scenarios, such calculations can be applied in various fields, including statistics, data analysis, and decision-making processes. Whether you're predicting the outcomes of a game, analyzing market trends, or making strategic business decisions, a solid grasp of probability theory can provide valuable insights.