A bag contains 4 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.
Answers:
A) 19/20
B) 17/20
C) 3/5
D) 8/10
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) = (4/7) x (3/5) =12/35.   P(B) = (3/7) x (2/5) = 6/35.   Hence, the required probability is P(A) + P(B) which is nothing but 3/5.   ANSWER:3/5
<result>C</result>
A bag contains 4 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.