Question

A die is thrown $(2n+1)$ times, $n\in N$. The probability that faces with even numbers show odd number of times is

• A. $\frac{2n+1}{2n+3}$
• B. less that $\frac{1}{2}$
• C. greater than $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (D)

None of these

Probability of showing even number in a throw $=\frac{3}{6}=\frac{1}{2}$

$\therefore$ Required probability

${=}^{2n+1}{C}_1.\frac{1}{2}.{\left(\frac{1}{2}\right)}^{2n}{+}^{2n+1}{C}_3.{\left(\frac{1}{2}\right)}^3.{\left(\frac{1}{2}\right)}^{2n-2}+...{+}^{2n+1}{C}_{2n+1}.{\left(\frac{1}{2}\right)}^{2n+1}$

$={\left(\frac{1}{2}\right)}^{2n+1}\{{}^{2n+1}{C}_1{+}^{2n+1}{C}_3+...{+}^{2n+1}{C}_{2n+1}\}$

$=\frac{1}{2^{2n+1}}\times 2^{2n+1-1}=\frac{1}{2}$