91Ó°ÊÓ

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) = \(\frac{number\,of\,white\,balls}{total\,number\,of\,balls}\) = \(\frac{5}{12}\).

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) = \(\frac{7}{12} \times \frac{9}{13} \times \frac{5}{14} \times \frac{7}{16} = \frac{2205}{43008}\).

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) = \(\frac{881}{21504}\).