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 the correct option:
A) 2/91  B) 1/81  C) 1/8  D) 2/81  E) 3/81
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]   =(5×42×1)/(15×14×13/3×2×1)=5×4/(15×14×13/3)=5×4/5×14×13=4/14×13=2/7×13=2/91   2/91nswer is 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?