A goblet contains 2 red marbles, 6 green marbles, and 4 blue marbles.If we choose a marble, then another marble without putting the first one back in the goblet, what is the probability that the first marble will be green and the second will be green as well?

Answer:P(first marble will be green and the second will be green as well)= 5/22Step-by-step explanation:A goblet contains 2 red marbles, 6 green marbles, and 4 blue marbles. Total number of marbles= 2+6+4                                         =12P(first marble is green)=no. of green marbles/total number of marbles                                   =  6/12                                = 1/2P(second marble is green and we didn't put the first marble back)=no. of green marbles left/total marbles left= 5/11Hence, P(first marble will be green and the second will be green as well)        =[tex]\dfrac{1}{2}\times \dfrac{5}{11}[/tex]       = 5/22