Question 1042367
Let {{{R[1]}}} = the event that the first draw is red,
{{{G[1]}}} = the event that the first draw is green,
{{{R[2]}}} = the event that the second draw is red,
{{{G[2]}}} = the event that the second draw is green. 

Then P({{{R[1]}}}&{{{R[2]}}}) = {{{(4/14)*(3/13) = 6/91}}},
P({{{R[1]}}}&{{{G[2]}}}) = {{{(4/14)*(10/13) = 20/91}}},
P({{{G[1]}}}&{{{R[2]}}}) = {{{(10/14)*(4/13) = 20/91}}}, and 
P({{{G[1]}}}&{{{G[2]}}}) = {{{(10/14)*(9/13) = 45/91}}}.

What we're interested at are events {{{R[1]}}}&{{{G[2]}}} and {{{G[1]}}}&{{{R[2]}}}.  Since these are mutually exclusive events, the probability is {{{20/91+20/91 = highlight(40/91)}}}.