Question

A dice is thrown $2n+1$ times, $nϵN$. The probability that faces with even numbers show odd number of times is

• A. $\frac{2n+1}{4n+3}$
• B. less than $\frac{1}{2}$
• C. greater than $\frac{1}{2}$
• D. none of these

Answer 1

The correct option is D none of these

Probability of choosing an even number or an odd number = $\frac{1}{2}$
The even number can come up either once or thrice or 5 times ... or 2n+1 times; the remaining throws should give odd numbers.
Therefore, probability = ${}_1^{(2n+1)}C\ast {\left(\frac{1}{2}\right)}^{(2n+1)}{+}_3^{(2n+1)}C\ast {\left(\frac{1}{2}\right)}^{(2n+1)}+...{+}_{(2n+1)}^{(2n+1)}C\ast {\left(\frac{1}{2}\right)}^{(2n+1)}=\frac{{}_1^{(2n+1)}C{+}_3^{(2n+1)}C+...{+}_{(2n+1)}^{(2n+1)}C}{2^{(2n+1)}}=\frac{2^{2n}}{2^{2n+1}}=\frac{1}{2}$