Question

A fair die is rolled $(2n+1)$ times. The probability that the faces with even number show odd number of times is

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

Answer 1

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

Probability of showing even numbers in a throw$=\frac{3}{6}=\frac{1}{2}$
$\therefore$ Required probability
${=}^{2n+1}{C}_1.\frac{1}{2}.{\left(\frac{1}{2}\right)}^{2n}{+}^{2n+1}{C}_3.{\left(\frac{1}{2}\right)}^3.{\left(\frac{1}{2}\right)}^{2n+2}+...{+}^{2n+1}{C}_{2n+1}.{\left(\frac{1}{2}\right)}^{2n+1}$
$={\left(\frac{1}{2}\right)}^{2n+1}\left[{}^{2n+1}{C}_1{+}^{2n+1}{C}_3{+}^{2n+1}{C}_{2n+1}\right]$
$=\frac{1}{2^{2n+1}}\times 2^{2n+1-1}=\frac{1}{2}$