A bag contains 3 red balls and 2 green balls. If we draw the balls one by one without replacement, what is the probability that the first ball drawn is red and the second ball drawn is green?
Answer
Since these are dependent events, the probability of both events occurring is the product of their individual probabilities. So, the probability that the first ball is red and the second ball is green is \(\frac{3}{5}\times\frac{1}{2}\)
Step 1 ：Calculate the total number of balls, which is \(3 + 2 = 5\)
Step 2 ：Find the probability of drawing a red ball first, which is \(\frac{3}{5}\)
Step 3 ：After drawing a red ball, there are now 4 balls left in the bag, with 2 of them being green. So, the probability of drawing a green ball second is \(\frac{2}{4} = \frac{1}{2}\)
Step 4 ：Since these are dependent events, the probability of both events occurring is the product of their individual probabilities. So, the probability that the first ball is red and the second ball is green is \(\frac{3}{5}\times\frac{1}{2}\)
