Question

A four-digit number is formed by the digits 1,2,3,4 with no repetition. The probability that the number is odd is
$1){\text{ }}zero$
$2){\text{ }}\dfrac{1}{3}$
$3){\text{ }}\dfrac{1}{4}$
$4)$ None of these

Answer 1

Hint : In this question we have to make different numbers by using the given data so this question is completely based on the concepts of permutation. Here we have to make four digit numbers using four numbers and all formed numbers should be odd.

Complete step-by-step answer :
It is very important to first understand the concept of permutation and combination. Basically permutation is the arrangement of the objects or items and combination is the grouping and possible selection of different or similar items or objects.
So, keep in mind that at the unit's place there should be an odd number $1$ or $3$ .Then try to arrange the numbers in the tenth, hundredth and thousandth place. So, to take a rough idea of this question consider four blanks or boxes_,,,_ and fill the number of possibilities in each blank like in the units place we can fill $2$ numbers $1$ or $3$ and in the tens place we have three possible values to fill rest of the $3$ numbers, similarly in hundreds place there are $2$ possibilities and $1$ possibility in thousands place. Then on multiplying all the possibilities we get 12 .After identifying the total number of favorable outcomes ,divide it by the total number of possible outcomes to get the answer.
The given digits are $1$ , $2$ , $3$ and $4$ . From which there are two odd digits $1$ and $3$
Let S be the sample space , set of all numbers formed by the digits $1$ , $2$ , $3$ and $4$ without any repetitions. Therefore, $n\left( S \right) = 4!$
We know that the four digit number will be odd if at the unit place there are odd digits present. And in the question we are required with two odd digits only that is $1$ and $3$ . So there are only two possibilities to fill a unit's place for either $1$ or $3$.
Number of possibilities to fill the ten’s place is 3
Number of possibilities to fill the hundreds place is 2 and
The number of possibilities to fill the thousand’s place is 1
So, the total number of favourable outcomes $= {\text{ 2}} \times {\text{3}} \times 2 \times 1$
$= {\text{ 12}}$
Hence, the probability of getting an odd number is
$P\left( {{\text{getting an odd number}}} \right) = \dfrac{{12}}{{24}} = \dfrac{1}{2}$
Thus the correct option is $4)$ None of these
So, the correct answer is “Option 4”.

Note : To calculate the probability, first determine a single event with a single outcome. Then identify the total number of outcomes that can occur. Then divide the number of events by the number of possible outcomes. P(A) denotes the probability of an event ‘A’ and n(A) is the number of favorable outcomes. Things that we have to keep in mind while solving the given question are the given conditions of the question. Odd number of Four digits is possible if there is an odd number present at the unit’s place.