Question:

Step 1: Define probabilities and stopping condition. There are 20 cards, with 10 labelled A and 10 labelled B. Cards are drawn with replacement. The probability of drawing an A-card is $P(A) = \frac{10}{20} = \frac{1}{2}$. The probability of drawing a B-card is $P(B) = \frac{10}{20} = \frac{1}{2}$. The drawing process stops when the second A-card is obtained. Step 2: State the favorable condition. The second A-card appears before the third B-card. This condition is met if, at the moment the second A-card is drawn, the total number of B-cards drawn is less than 3. This means there are either 0, 1, or 2 B-cards drawn before or at the same time as the second A-card. Step 3: Calculate the probability for each case. Let $P(k \text{ B's})$ denote the probability that exactly $k$ B-cards have been drawn when the second A-card appears. For the second A-card to appear after exactly $k$ B-cards, the sequence of draws must contain $k$ B's and one A in the first $k+1$ positions, followed by an A in the $(k+2)$-th position. The number of ways to arrange $k$ B's and one A in $k+1$ positions is given by the binomial coefficient $\binom{k+1}{k}$ (or $\binom{k+1}{1}$), which simplifies to $k+1$. Since each draw has a probability of $\frac{1}{2}$ for either A or B, any specific sequence of $k$ B's and two A's has a probability of $\left(\frac{1}{2}\right)^{k+2}$. Thus, the general formula for $P(k \text{ B's})$ is: $$P(k \text{ B's}) = \binom{k+1}{1} \left(\frac{1}{2}\right)^{k+2} = (k+1) \left(\frac{1}{2}\right)^{k+2}$$ We calculate the probabilities for $k=0, 1, 2$: * **Case 1: 0 B-cards before the second A.** This means the sequence of draws is AA. $$P(0 \text{ B's}) = (0+1) \left(\frac{1}{2}\right)^{0+2} = 1 \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ * **Case 2: 1 B-card before the second A.** This means one B and one A are drawn, followed by an A (e.g., ABA, BAA). $$P(1 \text{ B}) = (1+1) \left(\frac{1}{2}\right)^{1+2} = 2 \cdot \left(\frac{1}{2}\right)^3 = 2 \cdot \frac{1}{8} = \frac{1}{4}$$ * **Case 3: 2 B-cards before the second A.** This means two B's and one A are drawn, followed by an A (e.g., ABBA, BABA, BBAA). $$P(2 \text{ B's}) = (2+1) \left(\frac{1}{2}\right)^{2+2} = 3 \cdot \left(\frac{1}{2}\right)^4 = 3 \cdot \frac{1}{16} = \frac{3}{16}$$ Step 4: Sum the probabilities. The probability that the second A-card appears before the third B-card is the sum of the probabilities of these mutually exclusive cases: $$P(\text{2nd A before 3rd B}) = P(0 \text{ B's}) + P(1 \text{ B}) + P(2 \text{ B's})$$ $$P(\text{2nd A before 3rd B}) = \frac{1}{4} + \frac{1}{4} + \frac{3}{16}$$ To sum these fractions, we find a common denominator, which is 16: $$P(\text{2nd A before 3rd B}) = \frac{4}{16} + \frac{4}{16} + \frac{3}{16} = \frac{4+4+3}{16} = \frac{11}{16}$$