Question

If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all the three numbers are divisible by both 2 and 3 is
A. $ \dfrac{4}{{57}} $
B. $ \dfrac{7}{{99}} $
C. $ \dfrac{4}{{1155}} $
D. $ \dfrac{1}{{1100}} $

Answer 1

Hint: Probability of an event is the ratio of no. of favorable outcomes for the vent to the total no. of outcomes possible. A number is divisible by 2 and 3 only when the number is divisible by 6. So find the numbers which are divisible by 6 from the first 100 natural numbers. Select 3 numbers from these obtained numbers using combinations. This will be the total number of outcomes and select 3 numbers from 100 numbers this will be the total no. of outcomes. Using this info, find the required probability.

Complete step-by-step answer:
We are given that three distinct numbers are chosen randomly from the first 100 natural numbers.
We have to find the probability that all the chosen three numbers are divisible by both 2 and 3.
For a number to be divisible by 2 and 3, it first must be divisible by 6.
Therefore from the first 100 natural numbers, the numbers divisible by 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90 and 96.
Total no. of numbers divisible by 6 is 16.
No. of ways in which 3 numbers are selected from these 16 numbers is $ {}_{}^{16}C_3^{} $
No. of ways in which 3 numbers are selected randomly from 100 numbers is $ {}_{}^{100}C_3^{} $
 $ {}_{}^nC_r^{} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Probability that all the chosen three numbers are divisible by both 2 and 3 is $ \dfrac{{{}_{}^{16}C_3^{}}}{{{}_{}^{100}C_3^{}}} $
$ \Rightarrow \dfrac{{\left( {\dfrac{{16 \times 15 \times 14}}{{3 \times 2 \times 1}}} \right)}}{{\left( {\dfrac{{100 \times 99 \times 98}}{{3 \times 2 \times 1}}} \right)}} = \dfrac{{16 \times 15 \times 14}}{{100 \times 99 \times 98}} $
 $ \Rightarrow \dfrac{{3360}}{{970200}} = \dfrac{4}{{1155}} $
So, the correct answer is “ $\dfrac{4}{{1155}} $”.

Note: Here to select the numbers we have used combinations not permutations. Combination is the way of selecting an object or a number from a group of objects or numbers whereas permutations are the ways of arranging the objects in an order. If the order of the objects matter, then permutations must be used or else combinations. Here we have selected randomly without any order, so we have used combinations.