A bag contains 5 red and 3 black balls. One is drawn at

Solution

Let R1 be the event that the first ball drawn is red, and R2 be the event that the second ball drawn is red. We want to find the probability P(R2 ∣ R1), which is the probability that the second ball drawn is red given that the first ball drawn was red. Initially, the bag contains: Number of red balls = 5 Number of black balls = 3 Total number of balls = 5 + 3 = 8 In the first draw, one ball is drawn at random and it is red. After the first draw, the state of the bag is: Number of red balls remaining = 5 - 1 = 4 Number of black balls remaining = 3 Total number of balls remaining = 8 - 1 = 7 Now, for the second draw, we want to find the probability that the ball drawn is red, given the new contents of the bag. The number of favorable outcomes (drawing a red ball) is 4. The total number of possible outcomes (drawing any ball) is 7. The probability of drawing a red ball in the second draw, given that a red ball was drawn in the first draw, is: