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

• A. zero
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is D None of these

Given digits are 1, 2, 3, 4.
Possibilities for units place digit (either 1 or 3) = 2

Let $S$ be the sample space i.e set of all numbers formed by the digits $1,2,3,4$(without repetition).

Let $A$ be an event of getting odd numbers.

If $1$ comes at unit's place, then no. of ways of forming odd number is $3!$.

Similarly, if digit $3$ appears at unit's place then there are $3!$ ways of arranging remaining $3$ digits i.e. $1,2,$ and $4$ to form odd number.

$\therefore$ There will be $3!+3!=6+6=12$ odd numbers which are formed by the digits $1,2,3$ and $4$.

$⟹n\left(A\right)=12$
Number of numbers formed by 1, 2, 3, 4 (without repetitions) $=4!=24=n\left(S\right)$

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$
$\therefore$ Required probability $=\frac{12}{24}=\frac{1}{2}$