Question

A number of six digits is written down at random. Probability that sum of digits of the number be even is

• A. $\frac{1}{2}$
• B. $\frac{3}{8}$
• C. $\frac{3}{7}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Let $x_1{x}_2{x}_3{x}_4{x}_5{x}_6$ be the number then each of $x_1,x_2,x_3,x_4,x_5$ has $9,10,10,10,10$ choices respectively.

Now summing these five digits the sum is either odd or even.

If it is odd, take $x_6\in \{1,3,5,7,9\}$ and if it is even, take $x_6\in \{0,2,4,6,8\}$ so that the sum of six digits become even.

Thus number of desired type of numbers $=9\times {10}^4\times 5.$

Thus $P\left(E\right)=\frac{9\times {10}^4\times 5}{9\times {10}^5}=\frac{1}{2}$