A bag contains 666 red balls, 444 green balls, and 333 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?
Calculate total number of balls to start:
Total Balls = Red Balls + Green Balls + Blue Balls
Total Balls = 666 + 444 + 333
Total Balls = 1,443
Calculate the probability of drawing a green ball on the first pick:
P(Green) = Green Balls / Total Balls
P(Green) = 444/1443
P(Green) = 0.30769
Calculate the probability of drawing a red ball on the second pick (without replacement):
Total Balls decrease by 1, since we do not replace. So Total Balls = 1,443 - 1 = 1,442
P(Red) = Red Balls / Total Balls
P(Red) = 666/1442
P(Red) = 0.46186
Now, we want the probability of Green, Red in that order.
Since each event is independent, we multiply the event probabilities
P(Green, Red) = P(Green) * P(Red)
P(Green, Red) = 0.30769 * 0.46186
P(Green, Red) = 0.14211
Calculate total number of balls to start:
Total Balls = Red Balls + Green Balls + Blue Balls
Total Balls = 666 + 444 + 333
Total Balls = 1,443
Calculate the probability of drawing a green ball on the first pick:
P(Green) = Green Balls / Total Balls
P(Green) = 444/1443
P(Green) = 0.30769
Calculate the probability of drawing a red ball on the second pick (without replacement):
Total Balls decrease by 1, since we do not replace. So Total Balls = 1,443 - 1 = 1,442
P(Red) = Red Balls / Total Balls
P(Red) = 666/1442
P(Red) = 0.46186
Now, we want the probability of Green, Red in that order.
Since each event is independent, we multiply the event probabilities
P(Green, Red) = P(Green) * P(Red)
P(Green, Red) = 0.30769 * 0.46186
P(Green, Red) = 0.14211