Three balls marked with l, 2 and 3 are placed in an um. One ball is drawn, its number is noted, then the ball is returned to the um. This process is repeated and then repeated once more. Each ball is equally likely to be drawn on each occasion. If the sum of the numbers noted is 6, then the probability that the ball numbered with 2 is drawn at all the three occasions, is
Since the sum of the noted numbers is 6, the number on the balls are either 1, 2,3 or 2, 2, 2.
So, total number of cases = 3! + 1 = 7
So, the required probability = 1/7