Problem 78 - A First Course in Probability

Question

An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly \(n\) selections?

Step-by-Step Solution

Answer

The probability of making exactly n selections until we get 2 white and 2 black balls in a single draw for the first time is given by the expression: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\]

Step 1: Calculate the probability of getting 2 white and 2 black balls in one draw

There are \(\binom{8}{4}\) ways to choose 4 balls out of 8 without regard to the color. We can have 2 white and 2 black balls in one draw, which can be chosen in \(\binom{4}{2}\binom{4}{2}\) ways. Therefore, the probability of getting 2 white and 2 black balls in one draw is: \[P = \frac{\binom{4}{2}\binom{4}{2}}{\binom{8}{4}}\]

Step 2: Calculate the probability of not getting 2 white and 2 black balls in one draw

The complementary event is failing to get 2 white and 2 black balls. We can calculate the probability of this complementary event by subtracting the probability of success from 1: \[P(\text{not getting } 2 \text{ white and } 2 \text{ black balls}) = 1-P\]

Step 3: Evaluate the probabilities

Now, we calculate these probabilities: \[P = \frac{\binom{4}{2}\binom{4}{2}}{\binom{8}{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70} = \frac{18}{35}\] \[P(\text{not getting } 2 \text{ white and } 2 \text{ black balls}) = 1 - \frac{18}{35} = \frac{17}{35}\]

Step 4: Calculate the probability of making exactly n selections

To find the probability of making exactly n selections until we get 2 white and 2 black balls in a single draw for the first time, we have to fail in the first n-1 draws and succeed in the nth draw. Since these events are independent, we can multiply their probabilities: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\] The final answer is given by the expression: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\]

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics that deals with counting, arranging, and grouping items. In the context of this exercise, we are interested in calculating the number of ways in which we can select 4 balls from a total of 8 balls in the urn, which includes both black and white balls.
When we talk about combinations, we use the binomial coefficient notation, represented as \( \binom{n}{k} \), which means choosing \( k \) items from \( n \) total items without considering the order. In this exercise, to choose 4 balls out of 8, we use \( \binom{8}{4} \) since the order of selection doesn't matter.

Specifically for this problem, we not only want to choose 4 balls but also need a certain configuration: 2 white and 2 black. The number of ways to choose 2 white balls out of 4 white ones is \( \binom{4}{2} \), and similarly, the number of ways to choose 2 black balls from 4 black ones is again \( \binom{4}{2} \). Hence, the total number of desirable selections can be calculated as the product of the two combinations: \( \binom{4}{2}\binom{4}{2} \).
The power of combinatorics lies in allowing us to systematically count various configurations even in complex scenarios.

Random Selection

Random selection involves picking items from a set without any specific pattern or predictability. In these problems, we are dealing with probability and chance. The choice of balls from an urn is a classic representation of random selection in probability theory.
Every time we select balls from the urn, each ball has an equal probability of being chosen. This randomness is key to calculating the probability of a certain event occurring, such as drawing exactly 2 white and 2 black balls. The unpredictability of the outcome means that each draw has to consider all possible combinations of selected balls.

To make sure that our selections reflect true randomness, we assume that each ball is equally likely to be picked. This means when a problem states that the draw is random, it confirms no bias or pre-conditions influence the selection process, highlighting the core idea of uniform probability distribution. Ensuring a random selection helps us apply our knowledge of probability accurately to predict outcomes.

Complementary Probability

Complementary probability is a useful concept that involves determining the probability that a certain event does not occur. This principle is often simpler, especially if calculating the direct probability is complex.
For any event \( A \), the probability of the event not happening is given by \( 1 - P(A) \). Here in the urn problem, the event \( A \) is drawing exactly 2 white and 2 black balls. If we know the probability of this happening, \( P(A) \), the complementary probability of not drawing 2 white and 2 black balls is \( 1 - P(A) \).

Understanding this concept allows us to efficiently determine the probability of continuous failure until success. In this exercise, to find the probability of exactly \( n \) selections, recognizing that it entails failing \( n-1 \) times and succeeding on the \( n \)th time, is key. Using complementary probability makes it easy to first calculate all possible successful outcomes and then use subtraction to find the failure rate, streamlining complex probability problems.