Problem 14 - A First Course in Probability

Question

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that (a) the first 2 balls selected are black and the next 2 are white; (b) of the first 4 balls selected, exactly 2 are black.

Step-by-Step Solution

Answer

The probabilities for the two scenarios are: (a) P(First 2 Black and Next 2 White) = \(\frac{2205}{43008}\) (b) P(Exactly 2 Black in the First 4 Balls) = \(\frac{881}{21504}\).

Step 1: Define the Initial Probabilities

Before we deal with the specific scenarios, we need to determine the initial probabilities for drawing a black ball or white ball from the urn. The probability of drawing a black ball is given by P(B) = \(\frac{number\,of\,black\,balls}{total\,number\,of\,balls}\) = \(\frac{7}{12}\). The probability of drawing a white ball is given by P(W) = \(\frac{number\,of\,white\,balls}{total\,number\,of\,balls}\) = \(\frac{5}{12}\).

Step 2: Calculate the Probability for Scenario (a)

In this scenario, we want the probability of selecting the first two balls black and the next two white. We can represent this as P(BBWW). Since these are dependent events with replacement, we can apply the multiplication rule and calculate the probability as follows: 1. Draw the first black ball: P(B) = \(\frac{7}{12}\). 2. Replace the black ball and add two more black balls: Now we have 9 black balls and 12 total balls. 3. Draw the second black ball: P(B) = \(\frac{9}{13}\). 4. Replace the black ball and add two more black balls: Now we have 5 white balls and 14 total balls. 5. Draw the first white ball: P(W) =\(\frac{5}{14}\). 6. Replace the white ball and add two more white balls: Now we have 7 white balls and 15 total balls. 7. Draw the second white ball: P(W) = \(\frac{7}{16}\). Now we can multiply these probabilities to get the probability of scenario (a): P(BBWW) = P(B) × P(B) × P(W) × P(W) = \(\frac{7}{12} \times \frac{9}{13} \times \frac{5}{14} \times \frac{7}{16} = \frac{2205}{43008}\).

Step 3: Calculate the Probability for Scenario (b)

In this scenario, we want the probability of selecting exactly two black balls from the first four selected balls. This can happen in multiple ways: BBWW, BWBW, WBBW, WBWB, or WWBB. However, since the probabilities of each combination are the same, we can calculate the probability for one combination and multiply it by the number of possible combinations. To make it easier, we calculate the probability of the BWBW combination and add the previously computed probability for the BBWW combination. Then, we multiply the result by 2. Probability of BWBW combination: 1. Draw the first black ball: P(B) = \(\frac{7}{12}\). 2. Replace the black ball and add two more black balls: Now we have 9 black balls and 12 total balls. 3. Draw the first white ball: P(W) = \(\frac{5}{13}\). 4. Replace the white ball and add two more white balls: Now we have 7 white balls and 14 total balls. 5. Draw the second black ball: P(B) = \(\frac{9}{14}\). 6. Replace the black ball and add two more black balls: Now we have 11 black balls and 15 total balls. 7. Draw the second white ball: P(W) = \(\frac{7}{16}\). Now multiply these probabilities to get the probability of BWBW: P(BWBW) = P(B) × P(W) × P(B) × P(W) = \(\frac{7}{12} \times \frac{5}{13} \times \frac{9}{14} \times \frac{7}{16} = \frac{2205}{43008}\). The probability of scenario (b) is given by P(2 Black in 4) = 2 × ( P(BBWW) + P(BWBW) ) = 2 × (\(\frac{2205}{43008} + \frac{2205}{43008}\)) = \(\frac{881}{21504}\). The probabilities for the two scenarios are: (a) P(First 2 Black and Next 2 White) = \(\frac{2205}{43008}\) (b) P(Exactly 2 Black in the First 4 Balls) = \(\frac{881}{21504}\).

Key Concepts

Dependent Events Probability

Understanding the probability of dependent events is crucial in scenarios where the outcome of one event affects the probability of subsequent events. In our exercise, selecting a ball changes the makeup of the urn, therefore subsequent selections are not independent. The key to solving such problems is recognizing the change in probabilities after each event occurs.

For instance, when a black ball is drawn and replaced with two additional black balls, the probability of drawing another black ball increases, because there are more black balls than there were originally. The probability of each subsequent draw must be recalculated based on the new composition of the urn. The Multiplication Rule for Probabilities applies to dependent events as well but iteratively considers the changing odds after each draw. The advice here is to remember to recalculate probabilities at each step, reflecting the altered conditions after each selection.

Probability with Replacement

When tackling the probability with replacement, we imagine each draw from the urn as a separate 'experiment' where the initial conditions are restored, plus the additional changes stipulated by the problem — in this case, adding two more balls of the same color after each draw. This type of problem often confuses students because despite the 'replacement', the probabilities do not stay constant due to the extra balls being added.

Applying this concept to the exercise highlights the process of systematic adjustment to the probability after each draw. After each black ball is drawn and replaced with two additional black balls, the urn's composition is altered, hence the name 'with replacement'. This changing probability scenario is a hallmark feature of many real-world situations. The takeaway is to factor in the added balls after each event, updating your probability calculations accordingly.

Combinatorics in Probability

Combinatorics plays a significant role in probability when we want to find the number of ways something can occur. In the given problem, scenario (b) asks for the probability of drawing exactly two black balls in four draws. Here, combinatorics helps us to understand the different sequences of black and white draws that lead to the desired outcome, such as BBWW, BWBW, WBBW, WBWB, or WWBB.

Once a single sequence probability is calculated — for example, the BWBW combination — we can exploit the symmetry in the problem by recognizing that each sequence has an equal chance of occurring. Therefore, calculating the total probability involves multiplying the probability of one sequence by the number of possible sequences. It's a crucial reminder that in probability, often times it's not just about single event outcomes, but about the arrangement and ordering of multiple outcomes that combinatorics helps to enumerate.