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?  Choose the correct answer.
A) 10/21 B) 11/21 C) 5/20 D) 9/20 E) 5/21
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*6)/(2*1)   =42/2= 
<gadget id="calculator">42/2</gadget>
<output>21</output>
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*4)/(2*1)   =20/2= 
<gadget id="calculator">20/2</gadget>
<output>10</output>
10   P(E)=n(E)/n(s)   =10/21   The answer should be 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?