A bag contains 6 red balls and 8 green balls. Two balls are drawn at r...
Problem:
A bag contains 6 red balls and 8 green balls. Two balls are drawn at random one after one with replacement. What is the probability that the first one is green and the second one is red?
Approach:
To solve this problem, we need to find the probability of two independent events occurring successively. The probability of the first ball being green is 8/14, and since we replace the ball after drawing, the probability of the second ball being red is 6/14. We can multiply these probabilities to find the overall probability.
Solution:
Let's break down the problem into steps:
Step 1: Determine the probability of selecting a green ball on the first draw.
- There are a total of 8 green balls, and the bag contains a total of 14 balls (6 red + 8 green).
- Therefore, the probability of selecting a green ball on the first draw is 8/14.
Step 2: Determine the probability of selecting a red ball on the second draw.
- After the first draw, we replace the ball back into the bag, so the total number of balls remains the same.
- There are now 6 red balls and 8 green balls in the bag.
- Therefore, the probability of selecting a red ball on the second draw is 6/14.
Step 3: Multiply the probabilities from Step 1 and Step 2 to find the overall probability.
- The probability of the first ball being green and the second ball being red is the product of the probabilities from Step 1 and Step 2.
- (8/14) * (6/14) = 48/196 = 12/49.
Therefore, the probability that the first ball drawn is green and the second ball drawn is red is 12/49, which is option D.