A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?
Select: A) 2/91 B) 3/91 C) 5/91 D) 6/91 E) 5/69
Total number of balls = 4 + 5 + 6= 
<gadget id="calculator">4 + 5 + 6</gadget>
<output>15</output>
15   Let S be the sample space.   n(S)= Total number of ways of drawing 3 balls out of 15 = 15C3   Let E = Event of drawing 3 balls, all of them are yellow.   n(E) = Number of ways of drawing 3 balls, all of them are yellow   = Number of ways of drawing 3 balls from the total 5 = 5C3   (∵ there are 5 yellow balls in the total balls)   P(E) = n(E)/n(S)   =5C3/15C3   =5C2/15C3 [∵ nCr = nC(n - r). So 5C3 = 5C2. 2/91pplying this for the ease of calculation]   = 2/(7*13)   = 2/91   2/91nswer : Option 2/91
<result>A</result>
A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?