Out of first 100 natural numbers, three numbers are chosen without replacement. The probability that all these numbers are divisible both by 2 and 3 is:
(a)4/11 (b) 4/55 (c) 4/33 (d) 4/1155
Solution follows here:
Out of first 100 natural numbers, three numbers can be chosen in 100C3 ways.
A number to be divisible by 2 and 3, it is to be divisible by LCM of 2 and 3 ie., 6.
Out of first 100 natural numbers, all numbers divisible by 6 are:
6,12,18,……96 => these are 16 in number => number of possibilities of picking three numbers out of these 16 are 16C3
Probability = number of desired possibilities / number of all possibilities
= 16C3 / 100C3 = 16*15*14 / 100*99*98 = 4/1155Answer (d)