A bag contains 4 blue and 6 green marbles. Two marbles are selected at random from the bag. If the first marble is NOT replaced before the second marble is drawn, what is the probability of getting blue on the 2nd draw, GIVEN YOU DREW GREEN ON THE FIRST DRAW?

Step-by-step explanation:

probability of event # 1 + probability of event # 2 = probability of picking green then blue

probability of event # 1 - chance of drawing green / total # of the possiblities

chance of drawing green = 6

total # of the possibilities = 4(blue) + 6(blue)

total # of the possibilities = 10

probability of event # 1 - 6/10 or 3/5

probability of event # 2 - chance of drawing blue / total # of the possibilities - 1 green

chance of drawing blue = 4

total # of the possibilities - 1 green = 10 - 1

total # of the possibilities - 1 green = 9

probability of event # 2 - 4/9

probability of picking green then blue = 3/5 + 4/9

3 x 4       12

 =

5 x 9       45

probability of picking green then blue = 12/45 or 4/15

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it