Question

A number of six digits is written down at random. Probability that sum of digits of the number is 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 $E$ be the event of the digit being even and $O$ be the event of digit being odd.

No. of digits $=10\Rightarrow n\left(S\right)=10$

No. of even digits $=5\Rightarrow n\left(E\right)=5$

No. of odd digits $=5\Rightarrow n\left(O\right)=5$

Then, $P\left(E\right)=p=\frac{5}{10}=\frac{1}{2}$ and $P\left(O\right)=q=\frac{5}{10}=\frac{1}{2}$

Sum of two even number is even, sum of two odd numbers is even but sum of one odd and one even number is odd.

So, for the sum of six numbers to be even:

$E+E+E+E+E+E=E$

$E+E+E+E+O+O=E$

$O+O+O+O+O+O=E$

$O+O+E+E+E+E=E$

$\therefore$ Probability that sum of digits is even $={}^6{C}_6{p}^6{q}^0+{}^6{C}_4{p}^4{q}^2+{}^6{C}_0{p}^0{q}^6+{}^6{C}_2{p}^2{q}^4$

$\Rightarrow P\left(S_E\right)=[1\times (\frac{1}{2}{)}^6\times 1]+[15\times (\frac{1}{2}{)}^4\times (\frac{1}{2}{)}^2]+[1\times 1\times \left(\frac{1}{2}{)}^6\right]+[15\times (\frac{1}{2}{)}^2\times \left(\frac{1}{2}{)}^4\right]$

$\Rightarrow P\left(S_E\right)=\left(\frac{1}{2}{)}^6\right[1+15+1+15]=\frac{1}{64}\times 32=\frac{1}{2}$