Question

A three-digit number is formed by using number 1, 2, 3 and 4. The probability that the number is divisible by 3, is
(A) $\dfrac{2}{3}$
(B) $\dfrac{2}{7}$
(C) $\dfrac{1}{2}$
(D) $\dfrac{3}{4}$

Answer 1

Hint:-According to the divisibility rule of number 3, it says that “the sum of all the digits of the given number should be multiple of 3”. For example,37 is a number. Sum of all the digits of $37 = 3 + 7 = 10.$
Now, as 10 is not a multiple of 3 or is not divisible by3, the number 37 is also not divisible by 3.

Complete step by step solution:
Now, to find the combination of three-digit numbers which sum up to a number (which is a multiple of 3), it has to be checked that which case gives the sum up to the number divisible by 3.
Number of ways of formation of three-digit number using 1, 2, 3 and 4 (without repetition)
\[\]$ = {4_{{P_3}}} = \dfrac{{4!}}{{\left( {4 - 3} \right)!}} = 4! = 24\;{\mathop{\rm ways}\nolimits} $
We know, any multi-digit number is divisible by 3, if the sum of its digits is a multiple of 3.
(i) 1 + 2 + 3 =6
This combination can give the expected number with a number of ways
 = 3!
= 6 ways.
(ii) 2 + 3 + 4 = 9
This combination can give the expected number with a number of ways
= 3!
= 6 ways.
The combination such as $\left( {3 + 4 + 1} \right)$or $\left( {4 + 1 + 2} \right)$ will not give the expected number.
$\therefore $ Number of three-digit numbers that sum up to give 3 multiples
$ = 6 + 6 = 12$ ways
$\therefore $ Probability of that number (without repetition )
Probability=$\dfrac{{No of favourable outcomes}}{{Total outcomes}}$
$ = \dfrac{{12}}{{24}} = \dfrac{1}{2}$
The correct option is (C).

Note: In this type of probability problem at first the divisibility rule of every number solved be kept in mind. If the problem is solved using the assumption that the three-digit numbers are formed with repetition, the result obtained will be different. Number of ways of formation of a three-digit number using 1, 2, 3 and 4 can also be obtained by simple $4! = 24$ ways.