Problem 29 - A First Course in Probability

Question

An urn contains \(n\) white and \(m\) black balls, where \(n\) and \(m\) are positive numbers. (a) If two balls are randomly withdrawn, what is the probability that they are the same color? (b) If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color? (c) Show that the probability in part (b) is always larger than the one in part (a).

Step-by-Step Solution

Answer

Short Answer: (a) The probability of drawing two balls of the same color without replacement is given by: \(P(A) = \frac{C(n, 2) + C(m, 2)}{C(n + m, 2)}\) (b) The probability of drawing two balls of the same color with replacement is: \(P(B) = \frac{n^2}{(n + m)^2} + \frac{m^2}{(n + m)^2}\) (c) To show that the probability with replacement (part b) is always larger than without replacement (part a), we showed that the inequality \(P(B) > P(A)\) holds true, and since \(n\) and \(m\) are both positive numbers, this inequality is always true, hence proving that \(P(B) > P(A)\).

Step 1: (Step 1: Define relevant terms and formulas)

We'll first define the necessary terms and formulas for solving the exercise. Probability can be calculated as the number of favorable outcomes divided by the total possible outcomes. The probability of drawing two balls of the same color without replacement is given by: \(P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}\) The probability of drawing two balls of the same color with replacement is given by: \(P(B) = \frac{(\text{Probability of drawing two white balls}) + (\text{Probability of drawing two black balls})}{1}\)

Step 2: (Step 2: Find the probability without replacement - part (a))

To find the probability of drawing two balls of the same color without replacement, we need to find the number of favorable outcomes and total possible outcomes. The total number of possible outcomes is the number of ways to choose two balls from the urn, which is given by the binomial coefficient \(C(n + m, 2)\). The number of favorable outcomes is the sum of the number of ways to choose two white balls (\(C(n, 2)\)) and the number of ways to choose two black balls (\(C(m, 2)\)). The probability of drawing two balls of the same color without replacement is then: \(P(A) = \frac{C(n, 2) + C(m, 2)}{C(n + m, 2)}\)

Step 3: (Step 3: Find the probability with replacement - part (b))

To find the probability of drawing two balls of the same color with replacement, we can use the probabilities of drawing two white balls and two black balls: Probability of drawing two white balls (with replacement): \(P_{white} = \frac{n}{n + m} \cdot \frac{n}{n + m}\) Probability of drawing two black balls (with replacement): \(P_{black} = \frac{m}{n + m} \cdot \frac{m}{n + m}\) The probability of drawing two balls of the same color with replacement is then: \(P(B) = P_{white} + P_{black} = \frac{n^2}{(n + m)^2} + \frac{m^2}{(n + m)^2}\)

Step 4: (Step 4: Compare probabilities - part (c))

To show that the probability in part (b) is always larger than the one in part (a), we need to prove that \(P(B) > P(A)\). Substitute the expressions of \(P(A) = \frac{C(n, 2) + C(m, 2)}{C(n + m, 2)}\) and \(P(B) = \frac{n^2}{(n + m)^2} + \frac{m^2}{(n + m)^2}\), and simplify the inequality as follows: \(\frac{n^2}{(n + m)^2} + \frac{m^2}{(n + m)^2} > \frac{C(n, 2) + C(m, 2)}{C(n + m, 2)}\) After simplifying both sides, we find: \(\frac{nm}{(n + m)(n + m - 1)} > 0\) Since \(n\) and \(m\) are both positive numbers, this inequality is always true. Therefore, the probability of drawing two balls of the same color with replacement is always larger than without replacement: \(P(B) > P(A)\)

Key Concepts

Probability calculation

Understanding probability calculation is essential for solving problems where outcomes are uncertain. Probability measures how likely an event is to occur. In mathematical terms, it is the ratio of favorable outcomes to the total number of possible outcomes. Thus, the formula for probability is given by:

• Probability = Number of favorable outcomes / Total possible outcomes

Let's connect this concept with our urn problem. In part (a) of the exercise, we're looking at the probability of drawing two balls that are the same color without replacement. For that, we compute individual probabilities for each color and sum them. First, find how many ways you can pick two balls from the total, noted as the binomial coefficient \( C(n + m, 2) \). Then, separately find the combinations for each color like \( C(n, 2) \) for white balls and \( C(m, 2) \) for black balls. The formula becomes:

• \( P(A) = \frac{C(n, 2) + C(m, 2)}{C(n + m, 2)} \)

For part (b), where balls are replaced after each draw, the calculation reflects probabilities mixed with multiplication. Each draw remains independent, and you sum the probabilities of events where either both drawn are white or both are black. The formula in this case is:

• \( P(B) = \left( \frac{n}{n + m} \right)^2 + \left( \frac{m}{n + m} \right)^2 \)

Mastering these basic rules is crucial to tackle more intricate probability problems.

Combinatorics

Combinatorics, the mathematics of counting, is a structural component of probability calculations. It helps to figure out how to group and arrange items based on a set of rules. In scenarios like our urn example, combinatorial methods determine how many unique ways you can select balls from the urn, which is essential in calculating probability. For finding how many ways you can pick two balls from \( n + m \) balls, use:

• The binomial coefficient \( C(n+m, 2) = \frac{(n + m)!}{2!(n + m - 2)!} \)

Similarly, if you want to know how many pairs of white balls you can pick from \( n \) white balls:

• \( C(n, 2) = \frac{n!}{2!(n-2)!} \)

This is purely about arrangement choices, and it illustrates the importance of combinatorics in assessing complex real-life problems. By mastering these tools, you can effectively solve a wide variety of problems involving choices or arrangements, like partitioning tasks, organizing events, or forming groups.

Conditional probability

Conditional probability considers the probability of one event occurring given that another event has already happened. However, the urn problems discussed here deal more with how event outcomes change due to changing conditions—whether balls are replaced or not, impacting sequence independence.In probability scenarios with replacement, draws remain independent. This means the result of the second draw does not depend on the first since the ball is returned to the urn. Therefore, conditional rules apply less strictly; the formula used in part (b) reflects this independence:

• \( P(B) = \left( \frac{n}{n+m} \right)^2 + \left( \frac{m}{n+m} \right)^2 \)

On the other hand, when no replacement happens, the outcome of the second event (draw) depends on the first one, altering the sample space. Conditional thinking assists in understanding sequences, helping in cases where events are linked. This makes conditional probability more tangible in complex real-world phenomena where event relationships are key to analysis.