A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?  Pick one:
A) 11/21	B) 10/21	C) 3/5	D) 2/7	E) 5/7
Total number of balls =7   Let S be the sample space.   Then, n(S)	= Number of ways of drawing 2 balls out of 7= 7C2 =(7 x 6)(2 x 1)= 21.   Let E = Event of drawing 2 balls, none of which is blue.   n(E)	= Number of ways of drawing 2 balls out of (2 + 3) balls.   n(E)= 5C2=(5 x 4)(2 x 1)= 10.   P(E) =n(E)=10.   n(S)	21   Let E = Event of drawing 2 balls, none of which is blue.   n(E)	= Number of ways of drawing 2 balls out of (2 + 3) balls.= 5C2=(5 x 4)(2 x 1)= 10.   P(E)=N(E)/N(S)=10/21   So the answer is option 10/21
<result>B</result>
A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?