A bag contains 3 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.  Choose the correct choice from the following:	A) 19/20	B) 17/20	C) 21/40	D) 7/10	E) 9/40
Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.   Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.   P(A) = (1/2) x (6/10) = 3/10.   P(B) = (1/2) x (4/10) = 1/5.   Hence, the required probability is P(A) + P(B) which is nothing but 7/10   ANSWER:7/10
<result>D</result>
A bag contains 3 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.