A bag contains 6 red balls, 4 green balls, and 3 blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be red?
Accepted Solution
A:
First ball will be green- 4/13 Second ball will be red-6/12 or 1/2
First ball will be green AND second ball will be red- 4/13x1/2=4/26 or 2/13