Question

Four digits $1,2,4$ and $6$ are selected at random to form a four digits numbers. What is the probability that the number so formed, would be divisible by $4$?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

Let the probability of the number so formed after dividing by 4 be denoted by event A.

The total number of numbers that can be formed by 4 digits can be obtained through factorial of 4

$4!=4\times 3\times 2\times 1=24$

So, total likely events=24

Now, favorable outcomes can be obtained as all forming 24 different numbers with 4 digits 1,2,4,6 will be divisible by 4 if they have last two numbers that can be divided by 4 i.e. those numbers which have 12,16,24 and 64 at last will be divisible by 4.

For example 6412, 2416 and so on...

Similarly, 6412 and 4612 are different numbers so, favorable outcomes(m) would be $4\times 2=8$

Lastly, $P\left(A\right)=\frac{m}{n}=\frac{8}{24}=\frac{1}{3}$