A bag contains 2 yellow, 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
A) 1/2
B) 10/21
C) 9/11
D) 7/11
E) 11/15
Total number of balls = 2 + 3 + 2= 
<gadget id="calculator">2 + 3 + 2</gadget>
<output>7</output>
7   Let S be the sample space.   n(S)= Total number of ways of drawing 2 balls out of 7 = 7C2   Let E = Event of drawing 2 balls , none of them is blue.   n(E) = Number of ways of drawing 2 balls , none of them is blue   = Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2   (∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2= 
<gadget id="calculator">7 - 2</gadget>
<output>5</output>
5)   P(E)= n(E)/n(S)   =5C2 / 7C2   = (5×42×1)/(7×62×1)   =(5×4)/(7×6)   =10/21   Ans- 10/21
<result>B</result>
A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?