Probability-8 (IIT-2004)

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:

Solution:

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 = 4/1155

Answer (d)