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) 10/21  B) 11/21  C) 2/7  D) 5/7  E) 6/7
Explanation:   Total number of balls = (2 + 3 + 2)= 
<gadget id="calculator">(2 + 3 + 2)</gadget>
<output>7</output>
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.   = 5C2   =	(5 x 4) / (2 x 1)   = 10.   P(E) =	n(E)	/ n(S) = 10/21.   10/21NSWER IS 10/21
<result>A</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?