Question:

<div class="solution"> <p><strong>Step 1:</strong> Let's define the problem and the events involved. We have an urn containing 6 black balls and an unknown number of white balls, which is less than or equal to 6. We are given that three balls are drawn successively without replacement, and all of them are found to be white. We need to find the probability that a black ball will be drawn in the next draw.</p> <p><strong>Step 2:</strong> To solve this, we first need to understand the possible scenarios for the number of white balls in the urn. Since we've drawn 3 white balls already, the urn must contain at least 3 white balls. Let's denote the event of the urn containing at least 3 white balls as \(E_3\). The total number of white balls can be 3, 4, 5, or 6. We will calculate the probability of drawing a black ball in the next draw given that 3 whites have already been drawn, using the concept of conditional probability.</p> <p><strong>Step 3:</strong> The probability of drawing a black ball in the next draw, given that 3 whites have already been drawn, can be calculated using Bayes' theorem. We need to find \(P(B|E_3)\), where \(B\) is the event of drawing a black ball. This involves calculating the probabilities of having 3, 4, 5, or 6 white balls given that 3 white balls have been drawn, and then finding the overall probability of drawing a black ball in the next draw.</p> <p><strong>Step 4:</strong> Let's calculate the probability of drawing a black ball in the next draw given that we have \(k\) white balls, where \(k = 3, 4, 5, 6\). If there are \(k\) white balls, the total number of balls is \(6 + k\), and after drawing 3 white balls, we have \(6 + k - 3 = 3 + k\) balls left. The probability of drawing a black ball from these is \(\frac{6}{3+k}\) because there are 6 black balls.</p> <p><strong>Step 5:</strong> To apply Bayes' theorem, we also need the prior probabilities of having 3, 4, 5, or 6 white balls before any draw is made. Assuming each possibility is equally likely, the prior probability for each scenario is \(\frac{1}{4}\) since there are 4 scenarios (3, 4, 5, or 6 white balls). However, after drawing 3 white balls, these probabilities need to be updated based on the likelihood of observing 3 white balls in each scenario.</p> <p><strong>Step 6:</strong> The likelihood of drawing 3 white balls in a row for each scenario can be calculated using combinations since the order matters in sequential draws without replacement. For \(k\) white balls, the probability of drawing 3 white balls in a row is \(\frac{k}{6+k} \cdot \frac{k-1}{6+k-1} \cdot \frac{k-2}{6+k-2}\). We calculate this for \(k = 3, 4, 5, 6\) and use these to update our probabilities.</p> <p><strong>Step 7:</strong> After calculating the updated probabilities for each scenario and the probability of drawing a black ball given each scenario, we can find the overall probability of drawing a black ball in the next draw by summing the products of these probabilities.</p> <p><strong>Step 8:</strong> Let's calculate the specific probabilities for each \(k\). For \(k=3\), the probability of drawing 3 whites in a row is \(\frac{3}{9} \cdot \frac{2}{8} \cdot \frac{1}{7}\), and the probability of then drawing a black ball is \(\frac{6}{3+3} = \frac{6}{6} = 1\). For \(k=4\), the probability is \(\frac{4}{10} \cdot \frac{3}{9} \cdot \frac{2}{8}\), and the probability of then drawing a black ball is \(\frac{6}{4+3} = \frac{6}{7}\). We repeat this process for \(k=5\) and \(k=6\), and then apply Bayes' theorem to find the overall probability.</p> <p><strong>Step 9:</strong> Applying Bayes' theorem involves calculating the posterior probabilities of each scenario given that 3 white balls have been drawn, and then using these to find the weighted average