Calculate the probability of drawing one white ball and one black ball in two draws from a bag containing 5 white and 3 black balls, without replacement.

Step 1: Determine the total number of balls. There are 5 white balls and 3 black balls. Total number of balls is 5 + 3 = 8. Step 2: Calculate the probability of drawing a white ball first, then a black ball second. The probability of drawing a white ball first is Number of white ballsTotal number of balls = (5)/(8). After drawing one white ball, there are 4 white balls and 3 black balls left, making a total of 7 balls. The probability of drawing a black ball second is Number of black ballsTotal number of remaining balls = (3)/(7). The probability of this sequence (White then Black) is: P(White then Black) = (5)/(8) × (3)/(7) = (15)/(56) Step 3: Calculate the probability of drawing a black ball first, then a white ball second. The probability of drawing a black ball first is Number of black ballsTotal number of balls = (3)/(8). After drawing one black ball, there are 5 white balls and 2 black balls left, making a total of 7 balls. The probability of drawing a white ball second is Number of white ballsTotal number of remaining balls = (5)/(7). The probability of this sequence (Black then White) is: P(Black then White) = (3)/(8) × (5)/(7) = (15)/(56) Step 4: Calculate the total probability of drawing one ball of each color. To find the probability of drawing one ball of each color, add the probabilities of the two possible sequences (White then Black, or Black then White). P(One of each color) = P(White then Black) + P(Black then White) P(One of each color) = (15)/(56) + (15)/(56) P(One of each color) = (15 + 15)/(56) = (30)/(56) Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2. P(One of each color) = (30 ÷ 2)/(56 ÷ 2) = (15)/(28) The probability of drawing one ball of each color is (15)/(28). 3 done, 2 left today. You're making progress.